MAT 142 - COLLEGE MATHEMATICS - Partial
Differential Equations Practice Material - Set 2
1. Question: Solve the partial differential equation: ∂u
∂t = 2∂u
∂x subject to the initial condition
u(x, 0) = 3x2.
Ans. Step-by-step solution:
1. We start by separating variables. Let u(x, t) = X(x)T(t).
2. Substitute u(x, t) = X(x)T(t)into the PDE: X(x)dT
dt = 2T(t)dX
dx .
3. Rearranging the terms gives us: 1
T(t)
dT
dt = 2 1
X(x)
dX
dx .
4. Both sides of the equation are equal to a constant, say k. So we have two ODEs:
dT
dt =kT (t)and dX
dx =kX(x)
2.
5. Solve the first ODE: dT
dt =kT (t)⇒T(t) = Aekt.
6. Solve the second ODE: dX
dx =kX(x)
2⇒X(x) = Bekx
2.
7. Now, we have u(x, t) = X(x)T(t) = ABe(kx/2)+kt. We can express this as u(x, t) =
Ce(kx/2)+kt, where C=AB.
8. Apply the initial condition u(x, 0) = 3x2to find the constants Cand k. We get u(x, 0) =
Cekx
2= 3x2.
9. The only way for this to hold is if k= 0 and C= 3. So the solution to the PDE with the
given initial condition is u(x, t) = 3e0+2t= 3e2t.
2. Consider the partial differential equation:
y∂z
∂x +z∂y
∂x =x2
Find the general solution to this PDE.
Ans. Solution: 1. To solve the given partial differential equation, we first recognize that it is
not linear with constant coefficients, so we can use an integrating factor method.
2. Let’s denote the given PDE as:
y∂z
∂x +z∂y
∂x =x2
3. To utilize the integrating factor method, we identify the coefficients of the zterms and y
terms. Let’s rewrite the equation as:
y∂z
∂x +z∂y
∂x −x2= 0
4. Now, we calculate our integrating factor, denoted as µ(x, y, z), from the coefficients of y
and z:
∂µ
∂x =µzzand ∂µ
∂y =µzy
5. Setting these partial derivatives equal, we have: 1
z
dµ
dx =1
y
dµ
dy .
6. Solving this separable differential equation, we obtain:
µ=yz
7. We multiply the given PDE by the integrating factor µ=yz:
y2z∂z
∂x +yz2∂y
∂x =x2yz
8. Next, we recognize that this can be written as a total derivative:
d
dx(y2z) = x2yz
9. Integrating both sides with respect to x, we find:
y2z=x4
4+C(y, z)
10. Therefore, the general solution to the given partial differential equation is:
y2z=x4
4+C(y, z)
where C(y, z)is an arbitrary function of yand z.
3. Question: Solve the following partial differential equation using the method of characteristics:
ux+uy=u, u(x, 0) = x2
Ans. Step-by-step solution:
1. We first rewrite the given partial differential equation as a system of ordinary differential
equations for the characteristics:
dx
dt = 1,
dy
dt = 1,
du
dt =u.
2. Integrate the first two equations to find the characteristic curves:
x(t) = t+C1,
y(t) = t+C2,
where C1and C2are constants of integration.
3. Along the characteristic curves, we have u(t) = u(x(t), y(t)). Using the chain rule, we
can express du
dt in terms of dx
dt and dy
dt :
du
dt =∂u
∂x
dx
dt +∂u
∂y
dy
dt .
4. Substituting the expressions for x(t)and y(t)into the above equation, we get
du
dt =∂u
∂x +∂u
∂y .
5. Comparing this with the given PDE ∂u
∂x +∂u
∂y =u, we see that du
dt =u, which simplifies to
du
u=dt.
6. Integrating both sides of the above equation gives ln |u|=t+C3, where C3is another
constant of integration.
7. Exponentiating both sides yields u=Cet, where C=eC3.
8. Finally, using the initial condition u(x, 0) = x2, we can determine the value of C:
x2=Ce0⇒C=x2.
9. Therefore, the solution to the partial differential equation ux+uy=u, u(x, 0) = x2is
u(x, y) = x2e(x+y).
4. Question: Solve the partial differential equation uxx −4uxy + 4uyy = 0 subject to the initial
conditions u(x, 0) = x2and uy(x, 0) = 2x.
Ans. Step-by-step solution: 1. We begin by finding the characteristic equation associated with
the given PDE. The characteristic equation is given by:
dx
1=dy
−4=du
4
which simplifies to:
dx =−4dy = 4du
Therefore, the characteristic equations are 4y+c1=xand 4u+c2=y, where c1and c2are
arbitrary constants.
2. Next, we express uin terms of xand yfrom the characteristic equations. From 4y+c1=x
and 4u+c2=y, we have:
u=1
4(y−c2) = 1
4(x−c1
4−c2)=1
16x−1
4c1−c2
3. Now, we apply the initial conditions to determine the values of c1and c2. From u(x, 0) = x2
and uy(x, 0) = 2x, we get:
x2=1
16x−1
4c1−c2
2x=1
4
Solving these equations simultaneously, we find c1=−15 and c2=−4.
4. Substituting the values of c1and c2back into the expression for u, we obtain the solution
to the PDE:
u=1
16x+ 15 −4
u=1
16x+ 11
5. Question: Find the general solution to the partial differential equation ∂2u
∂x2−∂u
∂t = 0.
Ans. Step-by-step solution: 1. Let’s assume the general solution takes the form u(x, t) =
X(x)T(t). Substituting this into the given PDE, we get:
∂2(XT )
∂x2−∂(XT )
∂t = 0
2. Simplifying this equation gives us:
X′′(x)T(t)−XT ′(t) = 0
3. Dividing by XT (assuming XT = 0) gives:
X′′(x)
X(x)=T′(t)
T(t)=λ
4. This separates the PDE into two ordinary differential equations:
X′′(x)−λX(x) = 0 and T′(t)−λT (t) = 0
5. The solutions to these ODEs are:
X(x) = Acos(√λx)+Bsin(√λx)
T(t) = Ceλt
where A,B, and Care constants to be determined. 6. Since the constants A,B, and Care
arbitrary, combining the solutions gives the general solution to the PDE:
u(x, t) = (Acos(√λx)+Bsin(√λx))eλt
7. This is the general solution to the given partial differential equation ∂2u
∂x2−∂u
∂t = 0 where λ
can take on any real value.
6. Question: Consider the partial differential equation (PDE) given by:
ut=uxx + 2ux
subject to the initial condition u(x, 0) = x2and boundary conditions u(0, t) = 1 and u(1, t) = 4.
Find the solution u(x, t)for 0< x < 1and t > 0.
Ans. Step-by-step solution: 1. First, let’s assume that the solution can be written as u(x, t) =
X(x)T(t), and substitute this into the PDE to get:
X(x)T′(t) = X′′(x)T(t)+2X′(x)T(t)
2. Dividing both sides by u(x, t), we get:
T′(t)
T(t)=X′′(x)+2X′(x)
X(x)
3. Since the left side of the equation is only a function of tand the right side is only a
function of x, both sides must be equal to a constant. Let’s call this constant λ.
4. We then have two ordinary differential equations to solve:
T′−λT = 0
and
X′′ + 2X′−λX = 0
5. Solving the first ODE for T(t)gives:
T(t) = Ceλt
where Cis a constant to be determined.
6. Solving the second ODE for X(x)gives: Let’s first solve X′′ + 2X′−λX = 0. The
characteristic equation is m2+ 2m−λ= 0. Solving this gives:
m=−1±√1 + λ
Therefore, the solutions for X(x)are of the form:
X(x) = Ae(−1+√1+λ)x+Be(−1−√1+λ)x
where Aand Bare constants to be determined.
7. Applying the boundary conditions u(0, t) = 1 and u(1, t) = 4 gives:
For x= 0:
X(0) = A+B= 1
For x= 1:
X(1) = Ae(−1+√1+λ)+Be(−1−√1+λ)= 4
8. Now, using the initial condition u(x, 0) = x2gives:
X(x)T(0) = x2
Substitute in the solutions for X(x)and T(t), then substitute t= 0, and equate to x2to
solve for the constants Aand B.
9. Finally, substitute the determined A,B, and λvalues back into the expression for u(x, t)
to obtain the solution.
7. Question: Solve the following partial differential equation:
ux+uy= 0
Ans. Step-by-step solution:
1. We start by writing the given partial differential equation in the form of a characteristic
equation:
dx
1=dy
1=du
0
Solving dx
1=dy
1, we get x−y=c1. Let this be equation (1).
2. Next, we solve dx
1=du
0, which gives u=c2. Let this be equation (2).
3. Finally, solving dy
1=du
0, we get v=c3. Let this be equation (3).
4. Now, we combine equations (1), (2), and (3) to obtain the general solution:
u(x, y) = f(x−y)
where fis an arbitrary function.
Therefore, the general solution to the partial differential equation ux+uy= 0 is u(x, y) =
f(x−y), where fis an arbitrary function.
8. Question: Solve the following partial differential equation:
∂u
∂x + 2∂u
∂y = 0
Ans. Step-by-step solution: 1. We can tackle this first-order linear partial differential equation
by recognizing that it is of the form:
∂u
∂x + 2∂u
∂y = 0
2. To solve this equation, we can use the method of characteristics. Let dx and dy be increments
in the xand ydirections, respectively, along a characteristic curve. 3. Along the characteristic
curve, du = 0 since the equation does not involve u. Therefore, the equation for the characteristic
curves is given by:
dx
1=dy
2
4. Solving this characteristic equation, we find:
2dx =dy
5. Integrating both sides, we get:
2x=y+C
where Cis an arbitrary constant. 6. Now, we have u=f(C)along the characteristic curves.
Rearranging the equation 2x=y+Cfor C, we get C= 2x−y. 7. Therefore, the general
solution to the partial differential equation is:
u(x, y) = f(2x−y)
where fis an arbitrary function.
9. Question: Consider the partial differential equation
ut= 4uxx
subject to the boundary conditions u(0, t) = u(π, t)=0and the initial condition u(x, 0) =
sin(2x). Find the solution u(x, t)using the method of separation of variables.
Ans. Solution: We will seek a solution of the form u(x, t) = X(x)T(t). Substituting this into
the PDE gives
X(x)T′(t) = 4X′′(x)T(t)
Dividing by u(x, t)we have
T′(t)
4T(t)=X′′(x)
X(x)=−λ
where λis a constant. Let’s solve the two ODEs separately. 1. Solving the ODE in t:
This gives us T′(t) = −4λT (t)which has the solution
T(t) = c1e−4λt
2. Solving the ODE in x:The ODE X′′(x) + λX(x) = 0 has the general solution
X(x) = c2cos(√λx)+c3sin(√λx)
Applying the boundary conditions u(0, t) = 0 and u(π, t) = 0 gives
X(0) = c2= 0
X(π) = c3sin(√λπ)= 0
For nontrivial solutions, we must have sin(√λπ)= 0, which implies that √λ=nfor
n= 1,2,3, . . .. Therefore, the eigenvalues are λ=n2.
So, the solution in space is
Xn(x) = cnsin(nx)
where cnis a normalization constant.
Finally, the general solution to the problem is given by
u(x, t) = ∞
∑
n=1
cne−4n2tsin(nx)
Applying the initial condition u(x, 0) = sin(2x), we find that c2= 1,cn= 0 for n= 2.
Therefore, the solution to the initial value problem is
u(x, t) = e−16tsin(2x)
10. Question: Solve the following partial differential equation using the method of separation
of variables:
∂u
∂t = 9∂2u
∂x2
subject to the boundary conditions: u(0, t) = u(π, t) = 0 and the initial condition: u(x, 0) =
x(π−x).
Ans. Solution: Step 1. We begin by assuming a solution of the form u(x, t) = X(x)T(t)and
substituting it into the given PDE:
X(x)T′(t) = 9X′′(x)T(t)
Step 2. Dividing both sides by 9XT , we have:
T′(t)
9T(t)=X′′(x)
X(x)
Step 3. Since the left-hand side of the equation depends only on tand the right-hand side
depends only on x, they must be equal to a constant −λ. We thus have two ordinary differential
equations:
T′(t)+9λT (t) = 0 and X′′(x) + λX(x) = 0
Step 4. Solving the first ODE gives: T(t) = c1e−9λt, where c1is a constant.
Step 5. The second ODE has solutions depending on the sign of λ:
- For λ < 0:X(x) = c2cos(√−λx)+c3sin(√−λx)
- For λ= 0:X(x) = c2x+c3
- For λ > 0:X(x) = c2e√λx +c3e−√λx
Step 6. Apply the boundary conditions u(0, t) = u(π, t) = 0:
- We obtain the eigenvalues λn=n2for n∈N- The corresponding eigenfunctions are
Xn(x) = sin(nx)
Step 7. Using the initial condition u(x, 0) = x(π−x) = ∑∞
n=1 cnsin(nx) = x(π−x), we
have:
cn=2
π∫π
0
x(π−x)sin(nx)dx
Step 8. The final solution is given by the series expansion:
u(x, t) = ∞
∑
n=1
e−9n2t(2
π∫π
0
x(π−x)sin(nx)dx)sin(nx)
11. Question 11: Solve the following partial differential equation using the method of separation
of variables:
∂u
∂t =k∂2u
∂x2
subject to the boundary conditions u(0, t) = u(π, t)=0, and the initial condition u(x, 0) =
sin(x).
Ans. Solution: We will begin by assuming a solution of the form u(x, t) = X(x)T(t).
Substituting this into the given partial differential equation, we get:
T′(t)
kT (t)=X′′(x)
X(x)=−λ
where λis a separation constant.
1. Solving the time component:
Solving the time component gives us the ordinary differential equation:
T′(t)
kT (t)=−λ=⇒T′(t) + kλT (t) = 0
This has the solution T(t) = Ae−kλt, where Ais a constant.
2. Solving the spatial component:
Solving the spatial component gives us the ordinary differential equation:
X′′(x) + λX(x) = 0
Subject to the boundary conditions X(0) = X(π) = 0, the solutions are Xn(x) = sin(nx)
where n= 1,2,3, ... and λ=n2.
3. Forming the general solution:
The general solution is then given by the linear combination of the separated solutions:
u(x, t) = ∞
∑
n=1
Bne−kn2tsin(nx)
4. Applying the initial condition:
Applying the initial condition u(x, 0) = sin(x)gives us:
sin(x) = ∞
∑
n=1
Bnsin(nx)
Multiplying both sides by sin(mx)and integrating from 0 to π, we get:
Bm={2
πif mis odd
0if mis even
5. Final solution:
Therefore, the final solution to the partial differential equation is:
u(x, t) = 2
π
∞
∑
n=1,3,5,...
e−kn2tsin(nx)
12. Question:
Consider the partial differential equation
ut= 4uxx −6ux,
where u(x, t)is a function of two variables xand t.
Determine whether the given PDE is parabolic, hyperbolic, or elliptic. Justify your answer.
Ans. Step-by-step solution:
1. To determine the type of the PDE, we can utilize the discriminant ∆ = B2−4AC where
the PDE is of the form
Auxx +Buxt +Cutt +Dux+Eut+F u = 0.
2. From the given PDE, we have A= 4,B= 0,C= 0,D=−6,E= 1, and F= 0.
3. Substituting the values in the discriminant formula, we have
∆ = (0)2−4(4)(0) = 0.
4. Depending on the sign of the discriminant, we can classify the PDE as: - ∆>0⇒
Hyperbolic - ∆ = 0 ⇒Parabolic - ∆<0⇒Elliptic
5. Since the discriminant ∆=0in this case, the given PDE is categorized as a parabolic
partial differential equation.
13. Question: Solve the following partial differential equation using the method of separation
of variables:
∂u
∂t =k∂2u
∂x2
where kis a constant.
Ans. Step-by-step solution:
1. Assume a solution of the form u(x, t) = X(x)T(t).
2. Substitute the separation of variables into the PDE to get:
T′(t)
kT (t)=X′′(x)
X(x)=−λ
where λis a separation constant.
3. Solve the ODE for T(t):
T′(t) + kλT (t) = 0 ⇒T(t) = A·e−kλt
4. Solve the ODE for X(x):
X′′(x) = −λX(x)
The general solution for X(x)is:
X(x) = B·cos (√λx)+C·sin (√λx)
5. Combine X(x)and T(t):
u(x, t) = e−kλt (B·cos (√λx)+C·sin (√λx))
6. Apply boundary conditions or initial conditions to determine the constants A,B,C, and
λ.
14. Question: Solve the following partial differential equation using the method of separation
of variables:
ut= 4uxx
subject to the boundary conditions u(0, t) = u(π, t) = 0 and the initial condition u(x, 0) =
x(π−x).
Ans. Step-by-step solution:
1. Separation of Variables: Assume u(x, t) = X(x)T(t)and substitute this into the given
PDE:
X(x)T′(t) = 4X′′(x)T(t)
Divide both sides by uto get:
T′(t)
4T(t)=X′′(x)
X(x)=−λ2
This gives us two ordinary differential equations:
X′′(x) + λ2X(x) = 0
T′(t)+4λ2T(t) = 0
2. Solving the ODE for X(x):The solutions to the ODE for X(x)are of the form
X(x) = Acos(λx) + Bsin(λx).
Applying the boundary conditions u(0, t) = u(π, t) = 0, we get:
X(0) = A= 0
X(π) = Bsin(λπ) = 0
This implies that λ=nfor n= 1,2,3, . . . and the corresponding eigenfunctions are Xn(x) =
sin(nx).
3. Solving the ODE for T(t):The solution to the ODE for T(t)is T(t) = Ce−4n2twhere
Cis a constant to be determined later.
4. General Solution: By linearity, the general solution to the PDE is given by:
u(x, t) = ∞
∑
n=1
Cne−4n2tsin(nx)
5. Applying the Initial Condition: To determine the constants Cn, we use the initial
condition u(x, 0) = x(π−x). This leads to:
x(π−x) = ∞
∑
n=1
Cnsin(nx)
Multiplying both sides by sin(mx)and integrating over [0, π]gives:
∫π
0
x(π−x)sin(mx)dx =Cm∫π
0
sin2(mx)dx
Solving the integral on the left-hand side, we can find the coefficients Cm.
Therefore, the solution to the given partial differential equation is:
u(x, t) = ∞
∑
n=1
(someexpression)e−4n2tsin(nx)
15. Question:
Find the general solution to the partial differential equation
uxx −4uxy + 4uyy = 0
where uis a function of xand y.
Ans. Solution:
1. We start by assuming a solution of the form u(x, y) = X(x)Y(y). Substituting this form
into the given partial differential equation, we get
X′′(x)Y(y)−4X′(x)Y′(y)+4X(x)Y′′(y) = 0
2. Next, we divide the whole equation by X(x)Y(y)to separate xand yterms, yielding
X′′(x)
X(x)−4X′(x)
X(x)
Y′(y)
Y(y)+ 4Y′′(y)
Y(y)= 0
3. Rearranging, we have
X′′(x)
X(x)−4X′(x)
X(x)= 4Y′′(y)
Y(y)
4. The left-hand side of the equation depends only on xwhile the right-hand side depends
only on y. Since they are equal to a constant, say −λ, we get two ordinary differential equations:
{X′′(x)−4X′(x) + λX(x) = 0
Y′′(y)−4λY (y) = 0
5. Solving the second equation, we find
Y(y) = c1e2y+c2e2y
where c1and c2are constants.
6. For the first equation, we have three cases depending on the value of λ:
Case 1: λ= 0
The solution is X(x) = c3+c4x.
Case 2: λ > 0
The resulting solution is X(x) = c5e2x+c6e−2x.
Case 3: λ < 0
The solution is X(x) = c7cos(√|λ|x)+c8sin(√|λ|x).
7. Combining X(x)and Y(y), we obtain the general solution
u(x, y) = (c3+c4x)(c1e2y+c2e2y)+(c5e2x+c6e−2x)(c1e2y+c2e2y)+(c7cos(√|λ|x)+c8sin(√|λ|x))(c1e2y+c2e2y)
16. Question 16: Solve the following partial differential equation using the method of separation
of variables:
∂u
∂t =k∂2u
∂x2
subject to the boundary conditions u(0, t) = 0 and u(L, t) = 0, and the initial condition
u(x, 0) = f(x).
Ans. Solution:
We assume the solution to be of the form u(x, t) = X(x)T(t).
Substitute this into the given PDE:
T′(t)X(x) = kX′′(x)T(t)
Divide by kX(x)T(t)to separate variables:
T′(t)
kT (t)=X′′(x)
X(x)=−λ
This gives us two ordinary differential equations:
1. T′(t)
kT (t)=−λ(1) 2. X′′(x) + λX(x) = 0 (2)
From the boundary conditions u(0, t) = 0 and u(L, t) = 0, we have X(0) = X(L) = 0.
Solving Equation (2) with these boundary conditions gives the eigenfunctions Xn(x) =
sin (nπx
L)and the corresponding eigenvalues λn=(nπ
L)2for n= 1,2,3, ....
Now we solve Equation (1) with λ=(nπ
L)2:
T′(t)
kT (t)=−(nπ
L)2
This is a simple first-order ODE, whose solution is Tn(t) = Ae−k(nπ
L)2t.
Thus, the general solution to the PDE is:
u(x, t) = ∞
∑
n=1
Bnsin (nπx
L)e−k(nπ
L)2t
where Bncan be determined by the initial condition u(x, 0) = f(x)using Fourier series.
17. Question 17:
Consider the partial differential equation given by
uxx + 4uxy + 4uyy = 0
with initial conditions
u(x, 0) = x2
and
uy(x, 0) = ex
.
Find the solution
u(x, y)
to the partial differential equation.
Ans. Solution:
We first recognize that the given partial differential equation can be expressed in terms of the
differential operator
∆ = ∂2
∂x2+ 2 ∂2
∂x∂y +∂2
∂y2
. Therefore, we have
∆u= 0
.
To solve this equation, we will use the method of characteristics. Let
p(x, y) = ux(x, y)
and
q(x, y) = uy(x, y)
. We will introduce new variables
ξ=x+y
and
η=x−y
in order to simplify the partial differential equation.
1. Compute the characteristic equations:
The characteristic equations are given by
dx
ds = 1
,dy
ds = 1
,du
ds =p
,dp
ds = 4q
,dq
ds = 4p
.
2. Solve the characteristic equations:
Integrating the first two equations with initial conditions
x(0) = x0
and
y(0) = y0
, we get
x=s+x0
and
y=s+y0
.
3. Express
u, p, q
in terms of characteristic variables:
Using the method of characteristics, we have
u(x, y) = F(ξ) = F(x+y)
,
p(x, y) = ux(x, y) = F′(ξ) = F′(x+y)
, and
q(x, y) = uy(x, y) = F′(η) = F′(ξ−2y) = F′(x−y)
.
4. Use the initial conditions to find F(s):
Applying the initial conditions
u(x, 0) = x2=F(x)
, we get
F(x) = x2
. Similarly, using
uy(x, 0) = ex=F′(x−0)
, we find
F′(x) = ex
.
5. Final solution:
Therefore, the solution to the partial differential equation is
u(x, y) = F(x+y) = (x+y)2
.
Hence, the solution to the partial differential equation
uxx + 4uxy + 4uyy = 0
with initial conditions
u(x, 0) = x2
and
uy(x, 0) = ex
is
u(x, y) = (x+y)2
.
18. Question 18: Solve the partial differential equation uxx −2uxy +uyy = 0 subject to the
initial conditions u(x, 0) = sin(x)and uy(x, 0) = x.
Ans. Solution: 1. Let’s first find the characteristics of the given PDE by solving the charac-
teristic equation dy/dx = (2 ±√(−2)2−4)/2 = 1. So the characteristics curves are given by
y−x=Cwhere Cis a constant.
2. We can rewrite the given PDE using the characteristics as follows: let u(x, y) = F(y−x),
where Fis an arbitrary function. Then we have the following:
ux=−F′(y−x), uy=F′(y−x)
uxx =F′′(y−x), uyy =F′′(y−x), uxy =−F′′(y−x)
3. Substituting these expressions into the original PDE, we get:
F′′(y−x)−2(−F′′(y−x)) + F′′(y−x) = 0
⇒F′′(y−x) = 0
4. Integrating twice with respect to y−x, we have F′(y−x) = Aand F(y−x) = Ay +B,
where Aand Bare constants.
5. Now, using the initial conditions u(x, 0) = sin(x)and uy(x, 0) = x, we find:
F(−x) = A·0 + B=sin(x)
F′(−x) = A=x
6. Thus, we have B=sin(x)and A=x. Therefore, the solution to the PDE with the initial
conditions is u(x, y) = x·y+sin(x−y).
19. Question: Solve the partial differential equation uxx −2uxy +uyy = 0 subject to the
boundary conditions u(0, y) = sin(y)and ux(x, 0) = ex.
Ans. Solution: To solve the partial differential equation uxx −2uxy +uyy = 0, we will first
find the characteristic equation corresponding to the given equation. Then we will use the given
boundary conditions to find the solution.
1. Find the characteristic equation: The characteristic equation corresponding to the
given PDE is given by dr2−2sdr +ds2= 0, where r=dx
dt and s=dy
dt . Solving the characteristic
equation, we get (r−s)2= 0. Thus, the characteristics are defined by the equation r−s= 0,
or equivalently, by x−y=Cwhere Cis a constant.
2. Express the solution in terms of characteristic variables: Let v(x, y) = u(x, x −y)
be a new function defined in terms of the characteristic variables. Thus, we have vx=ux−uy
and vy=−uy. Differentiating with respect to xand y, we get
vxx =uxx −2uxy +uyy and vyy =uyy.
Substitute the above expressions into the given PDE to get
vxx −2vxy +vyy = 0.
3. Solve the transformed PDE: The transformed PDE simplifies to vyy = 0, which has
the general solution v(A, B) = f(A) + g(B), where A=x−yand B=xare the characteristic
variables. Thus, the solution to the transformed PDE is of the form v(x, y) = f(x−y) + g(x).
4. Apply the boundary conditions: Using the first boundary condition u(0, y) = sin(y),
we have
u(0, x −x) = f(0) + g(0) = sin(x).
This implies that g(0) = sin(x)since f(0) is a constant. Using the second boundary condition
ux(x, 0) = ex, we have
vy(x, x) = ux(x, 0) −uy(x, 0) = ex.
Differentiating v(x, y) = f(x−y) + g(x)with respect to yand setting y= 0, we find
f′(x) = ex.
Integrating both sides, we get f(x) = ex+C1.
5. Determine the final solution: Substitute f(x) = ex+C1and g(0) = sin(x)back into
the general solution form v(x, y) = f(x−y) + g(x)to get
v(x, y) = ex−y+C1+sin(x).
Since we defined v(x, y) = u(x, x−y), the final solution to the given PDE subject to the boundary
conditions is
u(x, y) = ex−(x−y)+C1+sin(x) = ey+C1+sin(x).
where C1is a constant of integration.
Therefore, the solution to the partial differential equation uxx −2uxy +uyy = 0 subject to
the boundary conditions u(0, y) = sin(y)and ux(x, 0) = exis u(x, y) = ey+C1+sin(x).
20. Question: Solve the following partial differential equation using the method of separation
of variables:
ut= 4uxx, u(x, 0) = x(1 −x), u(0, t) = 0, u(π, t) = 0
Ans. Step-by-step solution:
1. Assume a separation of variables: Let u(x, t) = X(x)T(t). Substituting into the
PDE, we have:
X(x)T′(t) = 4X′′(x)T(t)
Dividing by u, we get:
T′(t)
4T(t)=X′′(x)
X(x)
2. Solve the time ODE: Solving the time ODE T′(t)
4T(t)=λ, we get T(t) = c1e4λt, where λ
is a constant.
3. Solve the spatial ODE: Solving the spatial ODE X′′(x) = λX(x), we have:
X(x) = c2sin(√λx)+c3cos(√λx)
Applying the boundary conditions u(0, t) = 0 and u(π, t) = 0, we have:
X(0)T(t) = 0 =⇒c3= 0
and
X(π)T(t) = 0 =⇒√λπ =nπ =⇒λ=n2, n = 1,2,3, . . .
4. Form the general solution: The general solution is given by:
u(x, t) = ∞
∑
n=1 (cnsin(nx)e−4n2t)
5. Apply the initial condition to find the coefficients: Applying the initial condition
u(x, 0) = x(1 −x), we have:
x(1 −x) = ∞
∑
n=1
cnsin(nx)
Multiplying both sides by sin(mx)and integrating over [0, π], we get:
cm=2
π∫π
0
x(1 −x)sin(mx)dx
6. Calculate the coefficients: Calculating the coefficients cn, we get:
cn={8
πn3if nis odd
0if nis even
7. Final solution: The solution to the given partial differential equation is:
u(x, t) = ∞
∑
n=1 (8
πn3sin(nx)e−4n2t)
21. Question 21: Solve the partial differential equation ∂u
∂t = 4∂2u
∂x2subject to the initial condition
u(x, 0) = 2x2and the boundary conditions u(0, t) = 0 and u(1, t) = 0.
Ans. Solution: 1. We first assume a solution of the form u(x, t) = X(x)T(t)and substitute
it into the given PDE.
2. This gives us T′(t)
T(t)= 4X′′ (x)
X(x).
3. Rearranging terms, we obtain T′(t)
T(t)= 4X′′ (x)
X(x)=−λ, where λis a constant (separation
constant).
4. So we have two ordinary differential equations to solve: T′(t) + λT (t)=0and X′′(x) +
λ
4X(x) = 0.
5. The general solution to T′(t) + λT (t) = 0 is T(t) = Acos(√λt)+Bsin(√λt).
6. For the boundary conditions, u(0, t) = 0 gives X(0)T(t) = 0, which implies X(0) = 0.
7. Similarly, u(1, t) = 0 gives X(1)T(t) = 0, leading to X(1) = 0.
8. This gives us a countable set of eigenvalues λn= (2n−1)2π2with corresponding eigen-
vectors Xn(x) = sin((2n−1)πx).
9. The general solution for X′′(x) + λ
4X(x) = 0 is then X(x) = ∑∞
n=1 Cnsin((2n−1)πx),
where Cnare constants to be determined.
10. Now, using the initial condition u(x, 0) = 2x2, we have X(x)T(0) = 2x2. This gives us
X(x) = 2x2.
11. Substituting X(x) = 2x2into the general solution above, we find the particular solution
u(x, t) = ∑∞
n=1 Cnsin((2n−1)πx)(2 cos((2n−1)πt)).
12. Applying the boundary conditions, we solve for the coefficients Cnand obtain the final
solution u(x, t) = 8 ∑∞
n=1
1
(2n−1)3π3sin((2n−1)πx)cos((2n−1)πt).
22. Question: Solve the following partial differential equation using the method of characteris-
tics:
ux+xyuy=y
Ans. Step 1. We can rewrite the given partial differential equation as a system of ordinary
differential equations (ODEs) using the method of characteristics:
Let x(t),y(t), and u(t)be functions of tsuch that
dx
dt = 1, x(0) = x0,
dy
dt =xy, y(0) = y0,
du
dt =y, u(0) = u0.
Step 2. Solve the ODE dx
dt = 1:
x(t) = t+x0.
Step 3. Solve the ODE dy
dt =xy:
y(t) = y0et2/2.
Step 4. Solve the ODE du
dt =y:
u(t) = u0+∫t
0
y(τ)dτ =u0+∫t
0
y0eτ2/2dτ.
Step 5. Substitute the initial conditions to find u(t):
u(t) = u0+y0∫t
0
eτ2/2dτ =u0+y0et2/2 ∫t
0
e−τ2/2dτ.
Step 6. Using the error function, we have:
u(t) = u0+y0et2/2√π
2erf (t
√2).
Therefore, the general solution to the given partial differential equation is:
u(x, y) = u0+y0e(x−x0)2/2√π
2erf (x−x0
√2).
23. Question: Solve the partial differential equation
∂u
∂t −4∂u
∂x = 0
subject to the initial condition u(x, 0) = sin(x).
Ans. Solution:
We begin by writing the given partial differential equation in the standard form of a first-order
linear partial differential equation:
∂u
∂t −4∂u
∂x = 0
This is a first-order linear partial differential equation, so we can solve it by the method of
characteristics. Let x=x(t)be the path of a characteristic curve.
1. Finding the characteristic equations:
The characteristic equations are given by the system of ordinary differential equations
dx
dt =−4,du
dt = 0
Solving the first equation gives x(t) = −4t+C1for some constant C1. We will determine
C1using the initial condition u(x, 0) = sin(x).
2. Applying the initial condition:
From the initial condition u(x, 0) = sin(x), we have u(x(0),0) = u(C1,0) = sin(C1). Since
x(0) = C1, we can substitute into the characteristic equation to find that C1=x(0) = x.
3. Solving for u(x, t):
Now we know that x=−4t+x, so we have x=x−4t, which implies t= 0.
Therefore, the solution to the partial differential equation is
u(x, t) = sin(x−4t)
24. Find the general solution to the partial differential equation
∂2u
∂x2−∂u
∂t = 0
subject to the initial conditions u(0, t) = f(t)and ux(0, t) = g(t).
Ans. To solve the given partial differential equation, we can first assume a separable solution
of the form u(x, t) = X(x)T(t).
1. Plug this into the PDE to get:
X′′(x)T(t)−X(x)T′(t) = 0
Divide by X(x)T(t)to separate variables:
X′′(x)
X(x)=T′(t)
T(t)=λ
where λis a constant. This gives two ordinary differential equations:
X′′(x)−λX(x) = 0 and T′(t)−λT (t) = 0
2. Solve the ODE for X(x):
The characteristic equation is r2−λ= 0, which has solutions r=±√λ. So the general
solution for X(x)is:
X(x) = c1e√λx +c2e−√λx
3. Solve the ODE for T(t):
The ODE for T(t)has the general solution:
T(t) = c3eλt
4. Combine the solutions for X(x)and T(t):
The general solution to the PDE with separable variables is given by the product of the
solutions for X(x)and T(t), so:
u(x, t) = X(x)T(t) = (c1e√λx +c2e−√λx)(c3eλt)
5. Apply the initial conditions:
From u(0, t) = f(t), we have:
f(t) = (c1+c2)(c3eλt)
And from ux(0, t) = g(t), we have:
g(t) = √λ(c1−c2)(c3eλt)
6. Solve the system of equations to find the constants:
Solving the system of equations formed by the initial conditions will give the specific values
of the constants c1,c2, and c3in the general solution.
25. Question:
Consider the partial differential equation
ut= 4uxx
subject to the boundary conditions
u(0, t) = 0
and
u(π, t) = 0,
and the initial condition
u(x, 0) = x(π−x).
Find the solution to this partial differential equation using the method of separation of vari-
ables.
Ans. Step-by-step solution:
1. We begin by assuming that the solution to the partial differential equation can be expressed
as a product of functions of xand t, i.e., u(x, t) = X(x)T(t).
2. Substituting u(x, t) = X(x)T(t)into the partial differential equation ut= 4uxx, we get
X(x)T′(t) = 4X′′(x)T(t).
3. Dividing both sides by X(x)T(t), we obtain
T′(t)
T(t)= 4X′′(x)
X(x).
4. This equation can be separated into two ordinary differential equations:
T′(t)
T(t)=−λ
and
X′′(x) + λX(x) = 0,
where λis a constant (separation constant).
5. Solving the first ODE, we have
T′(t) + λT (t) = 0,
which gives the solution
T(t) = c1e−λt,
where c1is a constant.
6. Solving the second ODE, we have
X′′(x) + λX(x) = 0,
which has solutions depending on the sign of λ:
For λ= 0, the solution is X(x) = c2x+c3.
For λ < 0, the solution is X(x) = c4sin(√−λx)+c5cos(√−λx).
For λ > 0, the solution is X(x) = c6e√λx +c7e−√λx.
7. Applying the boundary condition u(0, t) = 0, we find X(0)T(t) = 0, which implies
X(0) = 0.
8. Applying the boundary condition u(π, t) = 0, we find X(π)T(t) = 0, which implies
X(π) = 0.
9. Using the initial condition u(x, 0) = x(π−x), we substitute u(x, 0) = X(x)T(0) =
x(π−x)into the general solution u(x, t) = X(x)T(t)to determine the constants.
10. Finally, we express the solution u(x, t)as a series by combining the solutions for different
λvalues and the appropriate constants.
26. Question 26:
Solve the partial differential equation:
uxx −4uxy + 4uyy = 0
subject to the boundary condition:
u(0, y) = ey
∂u
∂x (1, y) = ey
Ans. Step-by-step solution:
1. Let’s first find the characteristic equation for the given partial differential equation. The
characteristic equation is given by:
dr2−4rs + 4s2= (r−2s)2= 0
So, we have a repeated characteristic root: r= 2s.
2. Next, we express the general solution in terms of s. The general solution is of the form:
u(x, y) = A(y)e2x+B(y)xe2x
where A(y)and B(y)are arbitrary functions to be determined.
3. To determine A(y)and B(y), we substitute the general solution into the boundary condi-
tions.
4. Applying the first boundary condition u(0, y) = eygives:
A(y) + 0 = ey⇒A(y) = ey
5. Applying the second boundary condition ∂u
∂x (1, y) = eygives:
2A(y)+2B(y) = ey⇒B(y) = 0
6. Therefore, the solution to the partial differential equation subject to the given boundary
conditions is:
u(x, y) = eye2x=ey+2x
27. Question 27: Solve the following partial differential equation using the method of separation
of variables:
∂u
∂t =k∂2u
∂x2
subject to the initial condition u(x, 0) = x(1−x)and boundary conditions u(0, t) = u(1, t) =
0, where kis a constant.
Ans. Solution:
Step 1. Assume a separated solution of the form u(x, t) = X(x)T(t), substitute into the
PDE, and rearrange:
T′(t)X(x) = kX′′(x)T(t)
Divide by kX(x)T(t)to separate the variables:
T′(t)
kT (t)=X′′(x)
X(x)=−λ
This leads to two ordinary differential equations:
T′(t) + kλT (t) = 0, X′′(x) + λX(x) = 0
Step 2. Solve the ODE for T(t):
The solution to T′(t) + kλT (t) = 0 is T(t) = Ae−kλt, where Ais a constant.
Step 3. Solve the ODE for X(x):
The solution to X′′(x) + λX(x) = 0 subject to the boundary conditions X(0) = X(1) = 0 is
Xn(x) = sin(nπx),where λ=n2π2
Step 4. Write the general solution as a sum of the separated solutions:
u(x, t) = ∞
∑
n=1
Bne−kn2π2tsin(nπx)
Step 5. Apply the initial condition u(x, 0) = x(1 −x)to find the coefficients Bn:
x(1 −x) = ∞
∑
n=1
Bnsin(nπx)
Multiplying both sides by sin(mπx)and integrating from 0 to 1 gives:
Bm= 4 ∫1
0
x(1 −x)sin(mπx)dx
Step 6. Calculate the coefficients Bn:
Bn= 4 ∫1
0
x(1 −x)sin(nπx)dx =4
n3π3((−1)n−1)
Step 7. Substitute the coefficients back into the general solution to obtain the final solution
for u(x, t):
u(x, t) = ∞
∑
n=1
4
n3π3((−1)n−1)e−kn2π2tsin(nπx)
28. Question: Consider the partial differential equation (PDE) defined by ∂u
∂t + 4∂u
∂x = 0. Find
the general solution to this PDE.
Ans. Step-by-step solution: 1. We are given the PDE ∂u
∂t + 4∂u
∂x = 0. 2. To solve this first-order
linear PDE, we can use the method of characteristics. Let x=x(s)and t=t(s)be the equations
of the characteristic curves. 3. The characteristic equations are given by:
dx
ds = 4 and dt
ds = 1
Integrating these equations with respect to s, we get:
x= 4s+C1and t=s+C2
where C1and C2are constants of integration. 4. Rewriting the equations in terms of s, we have
u(x(s), t(s)) = f(s), where fis an arbitrary function. 5. Using the chain rule, we can express
∂u
∂t and ∂u
∂x as:
∂u
∂t =df
ds and ∂u
∂x =df
ds ·4
6. Substituting these expressions into the PDE, we get:
df
ds + 4 ·df
ds = 0 =⇒5·df
ds = 0
7. Integrating this equation with respect to s, we find f(s) = C3, where C3is another constant
of integration. 8. Therefore, the general solution to the given PDE is u(x, t) = C3, where C3is
an arbitrary constant.
29. Question:
Consider the partial differential equation ut=k(uxx +uyy)with initial condition u(x, 0) =
xe−x2−y2for x, y ∈Rand t > 0. Solve the partial differential equation subject to the initial
condition.
Ans. Solution:
We are given the partial differential equation ut=k(uxx +uyy). To solve this equation, we
will use separation of variables and Fourier series.
1. Find the solution using separation of variables:
Let u(x, y, t) = X(x)Y(y)T(t). Then, the partial differential equation becomes
XT ′=k(X′′Y+XY ′′).
Dividing by kXY T gives
T′
kT =X′′Y+XY ′′
XY =λ.
This gives us three ordinary differential equations:
T′=kλT, X′′ −λX = 0, Y ′′ −λY = 0.
2. Solve the first ODE:
Solving T′=kλT gives T(t) = Cekλt.
3. Solve the second ODE:
The solution to X′′ −λX = 0 is X(x) = Acos(√λx)+Bsin(√λx).
4. Solve the third ODE:
The solution to Y′′ −λY = 0 is Y(y) = Ccos(√λy)+Dsin(√λy).
5. Combine the solutions:
The general solution to the PDE is
u(x, y, t) = ∞
∑
n=1
∞
∑
m=1
(anm cos(√λnmx)+bnm sin(√λnmx))(cnm cos(√λnmy)+dnm sin(√λnmy))ekλnmt.
6. Apply the initial condition:
Given initial condition u(x, 0) = xe−x2−y2, we can solve for the coefficients and obtain the
final solution.
30. Consider the partial differential equation (PDE) given by:
ut= 4uxx
subject to the boundary conditions:
u(0, t) = 0
u(π, t) = 0
and the initial condition:
u(x, 0) = sin(x)
Determine the solution to the PDE using the method of separation of variables.
Ans. Solution:
1. Write the general solution using separation of variables
Let’s assume a solution of the form u(x, t) = X(x)·T(t).
Substitute this into the PDE ut= 4uxx:
X(x)·T′(t) = 4X′′(x)·T(t)
Divide both sides by X(x)·T(t)to separate variables:
T′(t)
T(t)= 4X′′(x)
X(x)=−λ
where λis a constant.
2. Solve the ODE for X(x)
Solve the ODE 4X′′(x) + λX(x) = 0 with the boundary conditions X(0) = 0 and X(π) = 0.
The general solution to the ODE is given by X(x) = Asin(√λ
4x)+Bcos(√λ
4x).
Applying the boundary conditions gives λ=n2for n∈Z+, and the corresponding eigenfunc-
tions are Xn(x) = sin(nx).
3. Solve the ODE for T(t)
Now solve the ODE T′(t)
T(t)=−λ:
T′(t) + n24T(t) = 0
T′(t)+4n2T(t) = 0
T(t) = Cne−4n2t
where Cnis a constant.
4. Form the general solution
The general solution to the PDE is the sum of the separable solutions:
u(x, t) = ∞
∑
n=1
Cnsin(nx)e−4n2t
5. Find the coefficients using the initial condition
Given the initial condition u(x, 0) = sin(x), we need to find Cnsuch that:
∞
∑
n=1
Cnsin(nx) = sin(x)
Using the orthogonality property of sine functions, we find that Cn=2
π∫π
0sin(x)sin(nx)dx.
Calculating the integral gives Cn=2
π·2(−1)n
n2−1=4(−1)n
π(n2−1) .
Therefore, the solution to the PDE is:
u(x, t) = ∞
∑
n=1
4(−1)n
π(n2−1) sin(nx)e−4n2t
MAT 142 - COLLEGE MATHEMATICS - Partial
Differential Equations Practice Material - Set 2
1. Question: Solve the partial differential equation: ∂u
∂t = 2∂u
∂x subject to the initial condition
u(x, 0) = 3x2.
Ans. Step-by-step solution:
1. We start by separating variables. Let u(x, t) = X(x)T(t).
2. Substitute u(x, t) = X(x)T(t)into the PDE: X(x)dT
dt = 2T(t)dX
dx .
3. Rearranging the terms gives us: 1
T(t)
dT
dt = 2 1
X(x)
dX
dx .
4. Both sides of the equation are equal to a constant, say k. So we have two ODEs:
dT
dt =kT (t)and dX
dx =kX(x)
2.
5. Solve the first ODE: dT
dt =kT (t)⇒T(t) = Aekt.
6. Solve the second ODE: dX
dx =kX(x)
2⇒X(x) = Bekx
2.
7. Now, we have u(x, t) = X(x)T(t) = ABe(kx/2)+kt. We can express this as u(x, t) =
Ce(kx/2)+kt, where C=AB.
8. Apply the initial condition u(x, 0) = 3x2to find the constants Cand k. We get u(x, 0) =
Cekx
2= 3x2.
9. The only way for this to hold is if k= 0 and C= 3. So the solution to the PDE with the
given initial condition is u(x, t) = 3e0+2t= 3e2t.
2. Consider the partial differential equation:
y∂z
∂x +z∂y
∂x =x2
Find the general solution to this PDE.
Ans. Solution: 1. To solve the given partial differential equation, we first recognize that it is
not linear with constant coefficients, so we can use an integrating factor method.
2. Let’s denote the given PDE as:
y∂z
∂x +z∂y
∂x =x2
3. To utilize the integrating factor method, we identify the coefficients of the zterms and y
terms. Let’s rewrite the equation as:
y∂z
∂x +z∂y
∂x −x2= 0
4. Now, we calculate our integrating factor, denoted as µ(x, y, z), from the coefficients of y
and z:
∂µ
∂x =µzzand ∂µ
∂y =µzy
5. Setting these partial derivatives equal, we have: 1
z
dµ
dx =1
y
dµ
dy .
6. Solving this separable differential equation, we obtain:
µ=yz
7. We multiply the given PDE by the integrating factor µ=yz:
y2z∂z
∂x +yz2∂y
∂x =x2yz
8. Next, we recognize that this can be written as a total derivative:
d
dx(y2z) = x2yz
9. Integrating both sides with respect to x, we find:
y2z=x4
4+C(y, z)
10. Therefore, the general solution to the given partial differential equation is:
y2z=x4
4+C(y, z)
where C(y, z)is an arbitrary function of yand z.
3. Question: Solve the following partial differential equation using the method of characteristics:
ux+uy=u, u(x, 0) = x2
Ans. Step-by-step solution:
1. We first rewrite the given partial differential equation as a system of ordinary differential
equations for the characteristics:
dx
dt = 1,
dy
dt = 1,
du
dt =u.
2. Integrate the first two equations to find the characteristic curves:
x(t) = t+C1,
y(t) = t+C2,
where C1and C2are constants of integration.
3. Along the characteristic curves, we have u(t) = u(x(t), y(t)). Using the chain rule, we
can express du
dt in terms of dx
dt and dy
dt :
du
dt =∂u
∂x
dx
dt +∂u
∂y
dy
dt .
4. Substituting the expressions for x(t)and y(t)into the above equation, we get
du
dt =∂u
∂x +∂u
∂y .
5. Comparing this with the given PDE ∂u
∂x +∂u
∂y =u, we see that du
dt =u, which simplifies to
du
u=dt.
6. Integrating both sides of the above equation gives ln |u|=t+C3, where C3is another
constant of integration.
7. Exponentiating both sides yields u=Cet, where C=eC3.
8. Finally, using the initial condition u(x, 0) = x2, we can determine the value of C:
x2=Ce0⇒C=x2.
9. Therefore, the solution to the partial differential equation ux+uy=u, u(x, 0) = x2is
u(x, y) = x2e(x+y).
4. Question: Solve the partial differential equation uxx −4uxy + 4uyy = 0 subject to the initial
conditions u(x, 0) = x2and uy(x, 0) = 2x.
Ans. Step-by-step solution: 1. We begin by finding the characteristic equation associated with
the given PDE. The characteristic equation is given by:
dx
1=dy
−4=du
4
which simplifies to:
dx =−4dy = 4du
Therefore, the characteristic equations are 4y+c1=xand 4u+c2=y, where c1and c2are
arbitrary constants.
2. Next, we express uin terms of xand yfrom the characteristic equations. From 4y+c1=x
and 4u+c2=y, we have:
u=1
4(y−c2) = 1
4(x−c1
4−c2)=1
16x−1
4c1−c2
3. Now, we apply the initial conditions to determine the values of c1and c2. From u(x, 0) = x2
and uy(x, 0) = 2x, we get:
x2=1
16x−1
4c1−c2
2x=1
4
Solving these equations simultaneously, we find c1=−15 and c2=−4.
4. Substituting the values of c1and c2back into the expression for u, we obtain the solution
to the PDE:
u=1
16x+ 15 −4
u=1
16x+ 11
5. Question: Find the general solution to the partial differential equation ∂2u
∂x2−∂u
∂t = 0.
Ans. Step-by-step solution: 1. Let’s assume the general solution takes the form u(x, t) =
X(x)T(t). Substituting this into the given PDE, we get:
∂2(XT )
∂x2−∂(XT )
∂t = 0
2. Simplifying this equation gives us:
X′′(x)T(t)−XT ′(t) = 0
3. Dividing by XT (assuming XT = 0) gives:
X′′(x)
X(x)=T′(t)
T(t)=λ
4. This separates the PDE into two ordinary differential equations:
X′′(x)−λX(x) = 0 and T′(t)−λT (t) = 0
5. The solutions to these ODEs are:
X(x) = Acos(√λx)+Bsin(√λx)
T(t) = Ceλt
where A,B, and Care constants to be determined. 6. Since the constants A,B, and Care
arbitrary, combining the solutions gives the general solution to the PDE:
u(x, t) = (Acos(√λx)+Bsin(√λx))eλt
7. This is the general solution to the given partial differential equation ∂2u
∂x2−∂u
∂t = 0 where λ
can take on any real value.
6. Question: Consider the partial differential equation (PDE) given by:
ut=uxx + 2ux
subject to the initial condition u(x, 0) = x2and boundary conditions u(0, t) = 1 and u(1, t) = 4.
Find the solution u(x, t)for 0< x < 1and t > 0.
Ans. Step-by-step solution: 1. First, let’s assume that the solution can be written as u(x, t) =
X(x)T(t), and substitute this into the PDE to get:
X(x)T′(t) = X′′(x)T(t)+2X′(x)T(t)
2. Dividing both sides by u(x, t), we get:
T′(t)
T(t)=X′′(x)+2X′(x)
X(x)
3. Since the left side of the equation is only a function of tand the right side is only a
function of x, both sides must be equal to a constant. Let’s call this constant λ.
4. We then have two ordinary differential equations to solve:
T′−λT = 0
and
X′′ + 2X′−λX = 0
5. Solving the first ODE for T(t)gives:
T(t) = Ceλt
where Cis a constant to be determined.
6. Solving the second ODE for X(x)gives: Let’s first solve X′′ + 2X′−λX = 0. The
characteristic equation is m2+ 2m−λ= 0. Solving this gives:
m=−1±√1 + λ
Therefore, the solutions for X(x)are of the form:
X(x) = Ae(−1+√1+λ)x+Be(−1−√1+λ)x
where Aand Bare constants to be determined.
7. Applying the boundary conditions u(0, t) = 1 and u(1, t) = 4 gives:
For x= 0:
X(0) = A+B= 1
For x= 1:
X(1) = Ae(−1+√1+λ)+Be(−1−√1+λ)= 4
8. Now, using the initial condition u(x, 0) = x2gives:
X(x)T(0) = x2
Substitute in the solutions for X(x)and T(t), then substitute t= 0, and equate to x2to
solve for the constants Aand B.
9. Finally, substitute the determined A,B, and λvalues back into the expression for u(x, t)
to obtain the solution.
7. Question: Solve the following partial differential equation:
ux+uy= 0
Ans. Step-by-step solution:
1. We start by writing the given partial differential equation in the form of a characteristic
equation:
dx
1=dy
1=du
0
Solving dx
1=dy
1, we get x−y=c1. Let this be equation (1).
2. Next, we solve dx
1=du
0, which gives u=c2. Let this be equation (2).
3. Finally, solving dy
1=du
0, we get v=c3. Let this be equation (3).
4. Now, we combine equations (1), (2), and (3) to obtain the general solution:
u(x, y) = f(x−y)
where fis an arbitrary function.
Therefore, the general solution to the partial differential equation ux+uy= 0 is u(x, y) =
f(x−y), where fis an arbitrary function.
8. Question: Solve the following partial differential equation:
∂u
∂x + 2∂u
∂y = 0
Ans. Step-by-step solution: 1. We can tackle this first-order linear partial differential equation
by recognizing that it is of the form:
∂u
∂x + 2∂u
∂y = 0
2. To solve this equation, we can use the method of characteristics. Let dx and dy be increments
in the xand ydirections, respectively, along a characteristic curve. 3. Along the characteristic
curve, du = 0 since the equation does not involve u. Therefore, the equation for the characteristic
curves is given by:
dx
1=dy
2
4. Solving this characteristic equation, we find:
2dx =dy
5. Integrating both sides, we get:
2x=y+C
where Cis an arbitrary constant. 6. Now, we have u=f(C)along the characteristic curves.
Rearranging the equation 2x=y+Cfor C, we get C= 2x−y. 7. Therefore, the general
solution to the partial differential equation is:
u(x, y) = f(2x−y)
where fis an arbitrary function.
9. Question: Consider the partial differential equation
ut= 4uxx
subject to the boundary conditions u(0, t) = u(π, t)=0and the initial condition u(x, 0) =
sin(2x). Find the solution u(x, t)using the method of separation of variables.
Ans. Solution: We will seek a solution of the form u(x, t) = X(x)T(t). Substituting this into
the PDE gives
X(x)T′(t) = 4X′′(x)T(t)
Dividing by u(x, t)we have
T′(t)
4T(t)=X′′(x)
X(x)=−λ
where λis a constant. Let’s solve the two ODEs separately. 1. Solving the ODE in t:
This gives us T′(t) = −4λT (t)which has the solution
T(t) = c1e−4λt
2. Solving the ODE in x:The ODE X′′(x) + λX(x) = 0 has the general solution
X(x) = c2cos(√λx)+c3sin(√λx)
Applying the boundary conditions u(0, t) = 0 and u(π, t) = 0 gives
X(0) = c2= 0
X(π) = c3sin(√λπ)= 0
For nontrivial solutions, we must have sin(√λπ)= 0, which implies that √λ=nfor
n= 1,2,3, . . .. Therefore, the eigenvalues are λ=n2.
So, the solution in space is
Xn(x) = cnsin(nx)
where cnis a normalization constant.
Finally, the general solution to the problem is given by
u(x, t) = ∞
∑
n=1
cne−4n2tsin(nx)
Applying the initial condition u(x, 0) = sin(2x), we find that c2= 1,cn= 0 for n= 2.
Therefore, the solution to the initial value problem is
u(x, t) = e−16tsin(2x)
10. Question: Solve the following partial differential equation using the method of separation
of variables:
∂u
∂t = 9∂2u
∂x2
subject to the boundary conditions: u(0, t) = u(π, t) = 0 and the initial condition: u(x, 0) =
x(π−x).
Ans. Solution: Step 1. We begin by assuming a solution of the form u(x, t) = X(x)T(t)and
substituting it into the given PDE:
X(x)T′(t) = 9X′′(x)T(t)
Step 2. Dividing both sides by 9XT , we have:
T′(t)
9T(t)=X′′(x)
X(x)
Step 3. Since the left-hand side of the equation depends only on tand the right-hand side
depends only on x, they must be equal to a constant −λ. We thus have two ordinary differential
equations:
T′(t)+9λT (t) = 0 and X′′(x) + λX(x) = 0
Step 4. Solving the first ODE gives: T(t) = c1e−9λt, where c1is a constant.
Step 5. The second ODE has solutions depending on the sign of λ:
- For λ < 0:X(x) = c2cos(√−λx)+c3sin(√−λx)
- For λ= 0:X(x) = c2x+c3
- For λ > 0:X(x) = c2e√λx +c3e−√λx
Step 6. Apply the boundary conditions u(0, t) = u(π, t) = 0:
- We obtain the eigenvalues λn=n2for n∈N- The corresponding eigenfunctions are
Xn(x) = sin(nx)
Step 7. Using the initial condition u(x, 0) = x(π−x) = ∑∞
n=1 cnsin(nx) = x(π−x), we
have:
cn=2
π∫π
0
x(π−x)sin(nx)dx
Step 8. The final solution is given by the series expansion:
u(x, t) = ∞
∑
n=1
e−9n2t(2
π∫π
0
x(π−x)sin(nx)dx)sin(nx)
11. Question 11: Solve the following partial differential equation using the method of separation
of variables:
∂u
∂t =k∂2u
∂x2
subject to the boundary conditions u(0, t) = u(π, t)=0, and the initial condition u(x, 0) =
sin(x).
Ans. Solution: We will begin by assuming a solution of the form u(x, t) = X(x)T(t).
Substituting this into the given partial differential equation, we get:
T′(t)
kT (t)=X′′(x)
X(x)=−λ
where λis a separation constant.
1. Solving the time component:
Solving the time component gives us the ordinary differential equation:
T′(t)
kT (t)=−λ=⇒T′(t) + kλT (t) = 0
This has the solution T(t) = Ae−kλt, where Ais a constant.
2. Solving the spatial component:
Solving the spatial component gives us the ordinary differential equation:
X′′(x) + λX(x) = 0
Subject to the boundary conditions X(0) = X(π) = 0, the solutions are Xn(x) = sin(nx)
where n= 1,2,3, ... and λ=n2.
3. Forming the general solution:
The general solution is then given by the linear combination of the separated solutions:
u(x, t) = ∞
∑
n=1
Bne−kn2tsin(nx)
4. Applying the initial condition:
Applying the initial condition u(x, 0) = sin(x)gives us:
sin(x) = ∞
∑
n=1
Bnsin(nx)
Multiplying both sides by sin(mx)and integrating from 0 to π, we get:
Bm={2
πif mis odd
0if mis even
5. Final solution:
Therefore, the final solution to the partial differential equation is:
u(x, t) = 2
π
∞
∑
n=1,3,5,...
e−kn2tsin(nx)
12. Question:
Consider the partial differential equation
ut= 4uxx −6ux,
where u(x, t)is a function of two variables xand t.
Determine whether the given PDE is parabolic, hyperbolic, or elliptic. Justify your answer.
Ans. Step-by-step solution:
1. To determine the type of the PDE, we can utilize the discriminant ∆ = B2−4AC where
the PDE is of the form
Auxx +Buxt +Cutt +Dux+Eut+F u = 0.
2. From the given PDE, we have A= 4,B= 0,C= 0,D=−6,E= 1, and F= 0.
3. Substituting the values in the discriminant formula, we have
∆ = (0)2−4(4)(0) = 0.
4. Depending on the sign of the discriminant, we can classify the PDE as: - ∆>0⇒
Hyperbolic - ∆ = 0 ⇒Parabolic - ∆<0⇒Elliptic
5. Since the discriminant ∆=0in this case, the given PDE is categorized as a parabolic
partial differential equation.
13. Question: Solve the following partial differential equation using the method of separation
of variables:
∂u
∂t =k∂2u
∂x2
where kis a constant.
Ans. Step-by-step solution:
1. Assume a solution of the form u(x, t) = X(x)T(t).
2. Substitute the separation of variables into the PDE to get:
T′(t)
kT (t)=X′′(x)
X(x)=−λ
where λis a separation constant.
3. Solve the ODE for T(t):
T′(t) + kλT (t) = 0 ⇒T(t) = A·e−kλt
4. Solve the ODE for X(x):
X′′(x) = −λX(x)
The general solution for X(x)is:
X(x) = B·cos (√λx)+C·sin (√λx)
5. Combine X(x)and T(t):
u(x, t) = e−kλt (B·cos (√λx)+C·sin (√λx))
6. Apply boundary conditions or initial conditions to determine the constants A,B,C, and
λ.
14. Question: Solve the following partial differential equation using the method of separation
of variables:
ut= 4uxx
subject to the boundary conditions u(0, t) = u(π, t) = 0 and the initial condition u(x, 0) =
x(π−x).
Ans. Step-by-step solution:
1. Separation of Variables: Assume u(x, t) = X(x)T(t)and substitute this into the given
PDE:
X(x)T′(t) = 4X′′(x)T(t)
Divide both sides by uto get:
T′(t)
4T(t)=X′′(x)
X(x)=−λ2
This gives us two ordinary differential equations:
X′′(x) + λ2X(x) = 0
T′(t)+4λ2T(t) = 0
2. Solving the ODE for X(x):The solutions to the ODE for X(x)are of the form
X(x) = Acos(λx) + Bsin(λx).
Applying the boundary conditions u(0, t) = u(π, t) = 0, we get:
X(0) = A= 0
X(π) = Bsin(λπ) = 0
This implies that λ=nfor n= 1,2,3, . . . and the corresponding eigenfunctions are Xn(x) =
sin(nx).
3. Solving the ODE for T(t):The solution to the ODE for T(t)is T(t) = Ce−4n2twhere
Cis a constant to be determined later.
4. General Solution: By linearity, the general solution to the PDE is given by:
u(x, t) = ∞
∑
n=1
Cne−4n2tsin(nx)
5. Applying the Initial Condition: To determine the constants Cn, we use the initial
condition u(x, 0) = x(π−x). This leads to:
x(π−x) = ∞
∑
n=1
Cnsin(nx)
Multiplying both sides by sin(mx)and integrating over [0, π]gives:
∫π
0
x(π−x)sin(mx)dx =Cm∫π
0
sin2(mx)dx
Solving the integral on the left-hand side, we can find the coefficients Cm.
Therefore, the solution to the given partial differential equation is:
u(x, t) = ∞
∑
n=1
(someexpression)e−4n2tsin(nx)
15. Question:
Find the general solution to the partial differential equation
uxx −4uxy + 4uyy = 0
where uis a function of xand y.
Ans. Solution:
1. We start by assuming a solution of the form u(x, y) = X(x)Y(y). Substituting this form
into the given partial differential equation, we get
X′′(x)Y(y)−4X′(x)Y′(y)+4X(x)Y′′(y) = 0
2. Next, we divide the whole equation by X(x)Y(y)to separate xand yterms, yielding
X′′(x)
X(x)−4X′(x)
X(x)
Y′(y)
Y(y)+ 4Y′′(y)
Y(y)= 0
3. Rearranging, we have
X′′(x)
X(x)−4X′(x)
X(x)= 4Y′′(y)
Y(y)
4. The left-hand side of the equation depends only on xwhile the right-hand side depends
only on y. Since they are equal to a constant, say −λ, we get two ordinary differential equations:
{X′′(x)−4X′(x) + λX(x) = 0
Y′′(y)−4λY (y) = 0
5. Solving the second equation, we find
Y(y) = c1e2y+c2e2y
where c1and c2are constants.
6. For the first equation, we have three cases depending on the value of λ:
Case 1: λ= 0
The solution is X(x) = c3+c4x.
Case 2: λ > 0
The resulting solution is X(x) = c5e2x+c6e−2x.
Case 3: λ < 0
The solution is X(x) = c7cos(√|λ|x)+c8sin(√|λ|x).
7. Combining X(x)and Y(y), we obtain the general solution
u(x, y) = (c3+c4x)(c1e2y+c2e2y)+(c5e2x+c6e−2x)(c1e2y+c2e2y)+(c7cos(√|λ|x)+c8sin(√|λ|x))(c1e2y+c2e2y)
16. Question 16: Solve the following partial differential equation using the method of separation
of variables:
∂u
∂t =k∂2u
∂x2
subject to the boundary conditions u(0, t) = 0 and u(L, t) = 0, and the initial condition
u(x, 0) = f(x).
Ans. Solution:
We assume the solution to be of the form u(x, t) = X(x)T(t).
Substitute this into the given PDE:
T′(t)X(x) = kX′′(x)T(t)
Divide by kX(x)T(t)to separate variables:
T′(t)
kT (t)=X′′(x)
X(x)=−λ
This gives us two ordinary differential equations:
1. T′(t)
kT (t)=−λ(1) 2. X′′(x) + λX(x) = 0 (2)
From the boundary conditions u(0, t) = 0 and u(L, t) = 0, we have X(0) = X(L) = 0.
Solving Equation (2) with these boundary conditions gives the eigenfunctions Xn(x) =
sin (nπx
L)and the corresponding eigenvalues λn=(nπ
L)2for n= 1,2,3, ....
Now we solve Equation (1) with λ=(nπ
L)2:
T′(t)
kT (t)=−(nπ
L)2
This is a simple first-order ODE, whose solution is Tn(t) = Ae−k(nπ
L)2t.
Thus, the general solution to the PDE is:
u(x, t) = ∞
∑
n=1
Bnsin (nπx
L)e−k(nπ
L)2t
where Bncan be determined by the initial condition u(x, 0) = f(x)using Fourier series.
17. Question 17:
Consider the partial differential equation given by
uxx + 4uxy + 4uyy = 0
with initial conditions
u(x, 0) = x2
and
uy(x, 0) = ex
.
Find the solution
u(x, y)
to the partial differential equation.
Ans. Solution:
We first recognize that the given partial differential equation can be expressed in terms of the
differential operator
∆ = ∂2
∂x2+ 2 ∂2
∂x∂y +∂2
∂y2
. Therefore, we have
∆u= 0
.
To solve this equation, we will use the method of characteristics. Let
p(x, y) = ux(x, y)
and
q(x, y) = uy(x, y)
. We will introduce new variables
ξ=x+y
and
η=x−y
in order to simplify the partial differential equation.
1. Compute the characteristic equations:
The characteristic equations are given by
dx
ds = 1
,dy
ds = 1
,du
ds =p
,dp
ds = 4q
,dq
ds = 4p
.
2. Solve the characteristic equations:
Integrating the first two equations with initial conditions
x(0) = x0
and
y(0) = y0
, we get
x=s+x0
and
y=s+y0
.
3. Express
u, p, q
in terms of characteristic variables:
Using the method of characteristics, we have
u(x, y) = F(ξ) = F(x+y)
,
p(x, y) = ux(x, y) = F′(ξ) = F′(x+y)
, and
q(x, y) = uy(x, y) = F′(η) = F′(ξ−2y) = F′(x−y)
.
4. Use the initial conditions to find F(s):
Applying the initial conditions
u(x, 0) = x2=F(x)
, we get
F(x) = x2
. Similarly, using
uy(x, 0) = ex=F′(x−0)
, we find
F′(x) = ex
.
5. Final solution:
Therefore, the solution to the partial differential equation is
u(x, y) = F(x+y) = (x+y)2
.
Hence, the solution to the partial differential equation
uxx + 4uxy + 4uyy = 0
with initial conditions
u(x, 0) = x2
and
uy(x, 0) = ex
is
u(x, y) = (x+y)2
.
18. Question 18: Solve the partial differential equation uxx −2uxy +uyy = 0 subject to the
initial conditions u(x, 0) = sin(x)and uy(x, 0) = x.
Ans. Solution: 1. Let’s first find the characteristics of the given PDE by solving the charac-
teristic equation dy/dx = (2 ±√(−2)2−4)/2 = 1. So the characteristics curves are given by
y−x=Cwhere Cis a constant.
2. We can rewrite the given PDE using the characteristics as follows: let u(x, y) = F(y−x),
where Fis an arbitrary function. Then we have the following:
ux=−F′(y−x), uy=F′(y−x)
uxx =F′′(y−x), uyy =F′′(y−x), uxy =−F′′(y−x)
3. Substituting these expressions into the original PDE, we get:
F′′(y−x)−2(−F′′(y−x)) + F′′(y−x) = 0
⇒F′′(y−x) = 0
4. Integrating twice with respect to y−x, we have F′(y−x) = Aand F(y−x) = Ay +B,
where Aand Bare constants.
5. Now, using the initial conditions u(x, 0) = sin(x)and uy(x, 0) = x, we find:
F(−x) = A·0 + B=sin(x)
F′(−x) = A=x
6. Thus, we have B=sin(x)and A=x. Therefore, the solution to the PDE with the initial
conditions is u(x, y) = x·y+sin(x−y).
19. Question: Solve the partial differential equation uxx −2uxy +uyy = 0 subject to the
boundary conditions u(0, y) = sin(y)and ux(x, 0) = ex.
Ans. Solution: To solve the partial differential equation uxx −2uxy +uyy = 0, we will first
find the characteristic equation corresponding to the given equation. Then we will use the given
boundary conditions to find the solution.
1. Find the characteristic equation: The characteristic equation corresponding to the
given PDE is given by dr2−2sdr +ds2= 0, where r=dx
dt and s=dy
dt . Solving the characteristic
equation, we get (r−s)2= 0. Thus, the characteristics are defined by the equation r−s= 0,
or equivalently, by x−y=Cwhere Cis a constant.
2. Express the solution in terms of characteristic variables: Let v(x, y) = u(x, x −y)
be a new function defined in terms of the characteristic variables. Thus, we have vx=ux−uy
and vy=−uy. Differentiating with respect to xand y, we get
vxx =uxx −2uxy +uyy and vyy =uyy.
Substitute the above expressions into the given PDE to get
vxx −2vxy +vyy = 0.
3. Solve the transformed PDE: The transformed PDE simplifies to vyy = 0, which has
the general solution v(A, B) = f(A) + g(B), where A=x−yand B=xare the characteristic
variables. Thus, the solution to the transformed PDE is of the form v(x, y) = f(x−y) + g(x).
4. Apply the boundary conditions: Using the first boundary condition u(0, y) = sin(y),
we have
u(0, x −x) = f(0) + g(0) = sin(x).
This implies that g(0) = sin(x)since f(0) is a constant. Using the second boundary condition
ux(x, 0) = ex, we have
vy(x, x) = ux(x, 0) −uy(x, 0) = ex.
Differentiating v(x, y) = f(x−y) + g(x)with respect to yand setting y= 0, we find
f′(x) = ex.
Integrating both sides, we get f(x) = ex+C1.
5. Determine the final solution: Substitute f(x) = ex+C1and g(0) = sin(x)back into
the general solution form v(x, y) = f(x−y) + g(x)to get
v(x, y) = ex−y+C1+sin(x).
Since we defined v(x, y) = u(x, x−y), the final solution to the given PDE subject to the boundary
conditions is
u(x, y) = ex−(x−y)+C1+sin(x) = ey+C1+sin(x).
where C1is a constant of integration.
Therefore, the solution to the partial differential equation uxx −2uxy +uyy = 0 subject to
the boundary conditions u(0, y) = sin(y)and ux(x, 0) = exis u(x, y) = ey+C1+sin(x).
20. Question: Solve the following partial differential equation using the method of separation
of variables:
ut= 4uxx, u(x, 0) = x(1 −x), u(0, t) = 0, u(π, t) = 0
Ans. Step-by-step solution:
1. Assume a separation of variables: Let u(x, t) = X(x)T(t). Substituting into the
PDE, we have:
X(x)T′(t) = 4X′′(x)T(t)
Dividing by u, we get:
T′(t)
4T(t)=X′′(x)
X(x)
2. Solve the time ODE: Solving the time ODE T′(t)
4T(t)=λ, we get T(t) = c1e4λt, where λ
is a constant.
3. Solve the spatial ODE: Solving the spatial ODE X′′(x) = λX(x), we have:
X(x) = c2sin(√λx)+c3cos(√λx)
Applying the boundary conditions u(0, t) = 0 and u(π, t) = 0, we have:
X(0)T(t) = 0 =⇒c3= 0
and
X(π)T(t) = 0 =⇒√λπ =nπ =⇒λ=n2, n = 1,2,3, . . .
4. Form the general solution: The general solution is given by:
u(x, t) = ∞
∑
n=1 (cnsin(nx)e−4n2t)
5. Apply the initial condition to find the coefficients: Applying the initial condition
u(x, 0) = x(1 −x), we have:
x(1 −x) = ∞
∑
n=1
cnsin(nx)
Multiplying both sides by sin(mx)and integrating over [0, π], we get:
cm=2
π∫π
0
x(1 −x)sin(mx)dx
6. Calculate the coefficients: Calculating the coefficients cn, we get:
cn={8
πn3if nis odd
0if nis even
7. Final solution: The solution to the given partial differential equation is:
u(x, t) = ∞
∑
n=1 (8
πn3sin(nx)e−4n2t)
21. Question 21: Solve the partial differential equation ∂u
∂t = 4∂2u
∂x2subject to the initial condition
u(x, 0) = 2x2and the boundary conditions u(0, t) = 0 and u(1, t) = 0.
Ans. Solution: 1. We first assume a solution of the form u(x, t) = X(x)T(t)and substitute
it into the given PDE.
2. This gives us T′(t)
T(t)= 4X′′ (x)
X(x).
3. Rearranging terms, we obtain T′(t)
T(t)= 4X′′ (x)
X(x)=−λ, where λis a constant (separation
constant).
4. So we have two ordinary differential equations to solve: T′(t) + λT (t)=0and X′′(x) +
λ
4X(x) = 0.
5. The general solution to T′(t) + λT (t) = 0 is T(t) = Acos(√λt)+Bsin(√λt).
6. For the boundary conditions, u(0, t) = 0 gives X(0)T(t) = 0, which implies X(0) = 0.
7. Similarly, u(1, t) = 0 gives X(1)T(t) = 0, leading to X(1) = 0.
8. This gives us a countable set of eigenvalues λn= (2n−1)2π2with corresponding eigen-
vectors Xn(x) = sin((2n−1)πx).
9. The general solution for X′′(x) + λ
4X(x) = 0 is then X(x) = ∑∞
n=1 Cnsin((2n−1)πx),
where Cnare constants to be determined.
10. Now, using the initial condition u(x, 0) = 2x2, we have X(x)T(0) = 2x2. This gives us
X(x) = 2x2.
11. Substituting X(x) = 2x2into the general solution above, we find the particular solution
u(x, t) = ∑∞
n=1 Cnsin((2n−1)πx)(2 cos((2n−1)πt)).
12. Applying the boundary conditions, we solve for the coefficients Cnand obtain the final
solution u(x, t) = 8 ∑∞
n=1
1
(2n−1)3π3sin((2n−1)πx)cos((2n−1)πt).
22. Question: Solve the following partial differential equation using the method of characteris-
tics:
ux+xyuy=y
Ans. Step 1. We can rewrite the given partial differential equation as a system of ordinary
differential equations (ODEs) using the method of characteristics:
Let x(t),y(t), and u(t)be functions of tsuch that
dx
dt = 1, x(0) = x0,
dy
dt =xy, y(0) = y0,
du
dt =y, u(0) = u0.
Step 2. Solve the ODE dx
dt = 1:
x(t) = t+x0.
Step 3. Solve the ODE dy
dt =xy:
y(t) = y0et2/2.
Step 4. Solve the ODE du
dt =y:
u(t) = u0+∫t
0
y(τ)dτ =u0+∫t
0
y0eτ2/2dτ.
Step 5. Substitute the initial conditions to find u(t):
u(t) = u0+y0∫t
0
eτ2/2dτ =u0+y0et2/2 ∫t
0
e−τ2/2dτ.
Step 6. Using the error function, we have:
u(t) = u0+y0et2/2√π
2erf (t
√2).
Therefore, the general solution to the given partial differential equation is:
u(x, y) = u0+y0e(x−x0)2/2√π
2erf (x−x0
√2).
23. Question: Solve the partial differential equation
∂u
∂t −4∂u
∂x = 0
subject to the initial condition u(x, 0) = sin(x).
Ans. Solution:
We begin by writing the given partial differential equation in the standard form of a first-order
linear partial differential equation:
∂u
∂t −4∂u
∂x = 0
This is a first-order linear partial differential equation, so we can solve it by the method of
characteristics. Let x=x(t)be the path of a characteristic curve.
1. Finding the characteristic equations:
The characteristic equations are given by the system of ordinary differential equations
dx
dt =−4,du
dt = 0
Solving the first equation gives x(t) = −4t+C1for some constant C1. We will determine
C1using the initial condition u(x, 0) = sin(x).
2. Applying the initial condition:
From the initial condition u(x, 0) = sin(x), we have u(x(0),0) = u(C1,0) = sin(C1). Since
x(0) = C1, we can substitute into the characteristic equation to find that C1=x(0) = x.
3. Solving for u(x, t):
Now we know that x=−4t+x, so we have x=x−4t, which implies t= 0.
Therefore, the solution to the partial differential equation is
u(x, t) = sin(x−4t)
24. Find the general solution to the partial differential equation
∂2u
∂x2−∂u
∂t = 0
subject to the initial conditions u(0, t) = f(t)and ux(0, t) = g(t).
Ans. To solve the given partial differential equation, we can first assume a separable solution
of the form u(x, t) = X(x)T(t).
1. Plug this into the PDE to get:
X′′(x)T(t)−X(x)T′(t) = 0
Divide by X(x)T(t)to separate variables:
X′′(x)
X(x)=T′(t)
T(t)=λ
where λis a constant. This gives two ordinary differential equations:
X′′(x)−λX(x) = 0 and T′(t)−λT (t) = 0
2. Solve the ODE for X(x):
The characteristic equation is r2−λ= 0, which has solutions r=±√λ. So the general
solution for X(x)is:
X(x) = c1e√λx +c2e−√λx
3. Solve the ODE for T(t):
The ODE for T(t)has the general solution:
T(t) = c3eλt
4. Combine the solutions for X(x)and T(t):
The general solution to the PDE with separable variables is given by the product of the
solutions for X(x)and T(t), so:
u(x, t) = X(x)T(t) = (c1e√λx +c2e−√λx)(c3eλt)
5. Apply the initial conditions:
From u(0, t) = f(t), we have:
f(t) = (c1+c2)(c3eλt)
And from ux(0, t) = g(t), we have:
g(t) = √λ(c1−c2)(c3eλt)
6. Solve the system of equations to find the constants:
Solving the system of equations formed by the initial conditions will give the specific values
of the constants c1,c2, and c3in the general solution.
25. Question:
Consider the partial differential equation
ut= 4uxx
subject to the boundary conditions
u(0, t) = 0
and
u(π, t) = 0,
and the initial condition
u(x, 0) = x(π−x).
Find the solution to this partial differential equation using the method of separation of vari-
ables.
Ans. Step-by-step solution:
1. We begin by assuming that the solution to the partial differential equation can be expressed
as a product of functions of xand t, i.e., u(x, t) = X(x)T(t).
2. Substituting u(x, t) = X(x)T(t)into the partial differential equation ut= 4uxx, we get
X(x)T′(t) = 4X′′(x)T(t).
3. Dividing both sides by X(x)T(t), we obtain
T′(t)
T(t)= 4X′′(x)
X(x).
4. This equation can be separated into two ordinary differential equations:
T′(t)
T(t)=−λ
and
X′′(x) + λX(x) = 0,
where λis a constant (separation constant).
5. Solving the first ODE, we have
T′(t) + λT (t) = 0,
which gives the solution
T(t) = c1e−λt,
where c1is a constant.
6. Solving the second ODE, we have
X′′(x) + λX(x) = 0,
which has solutions depending on the sign of λ:
For λ= 0, the solution is X(x) = c2x+c3.
For λ < 0, the solution is X(x) = c4sin(√−λx)+c5cos(√−λx).
For λ > 0, the solution is X(x) = c6e√λx +c7e−√λx.
7. Applying the boundary condition u(0, t) = 0, we find X(0)T(t) = 0, which implies
X(0) = 0.
8. Applying the boundary condition u(π, t) = 0, we find X(π)T(t) = 0, which implies
X(π) = 0.
9. Using the initial condition u(x, 0) = x(π−x), we substitute u(x, 0) = X(x)T(0) =
x(π−x)into the general solution u(x, t) = X(x)T(t)to determine the constants.
10. Finally, we express the solution u(x, t)as a series by combining the solutions for different
λvalues and the appropriate constants.
26. Question 26:
Solve the partial differential equation:
uxx −4uxy + 4uyy = 0
subject to the boundary condition:
u(0, y) = ey
∂u
∂x (1, y) = ey
Ans. Step-by-step solution:
1. Let’s first find the characteristic equation for the given partial differential equation. The
characteristic equation is given by:
dr2−4rs + 4s2= (r−2s)2= 0
So, we have a repeated characteristic root: r= 2s.
2. Next, we express the general solution in terms of s. The general solution is of the form:
u(x, y) = A(y)e2x+B(y)xe2x
where A(y)and B(y)are arbitrary functions to be determined.
3. To determine A(y)and B(y), we substitute the general solution into the boundary condi-
tions.
4. Applying the first boundary condition u(0, y) = eygives:
A(y) + 0 = ey⇒A(y) = ey
5. Applying the second boundary condition ∂u
∂x (1, y) = eygives:
2A(y)+2B(y) = ey⇒B(y) = 0
6. Therefore, the solution to the partial differential equation subject to the given boundary
conditions is:
u(x, y) = eye2x=ey+2x
27. Question 27: Solve the following partial differential equation using the method of separation
of variables:
∂u
∂t =k∂2u
∂x2
subject to the initial condition u(x, 0) = x(1−x)and boundary conditions u(0, t) = u(1, t) =
0, where kis a constant.
Ans. Solution:
Step 1. Assume a separated solution of the form u(x, t) = X(x)T(t), substitute into the
PDE, and rearrange:
T′(t)X(x) = kX′′(x)T(t)
Divide by kX(x)T(t)to separate the variables:
T′(t)
kT (t)=X′′(x)
X(x)=−λ
This leads to two ordinary differential equations:
T′(t) + kλT (t) = 0, X′′(x) + λX(x) = 0
Step 2. Solve the ODE for T(t):
The solution to T′(t) + kλT (t) = 0 is T(t) = Ae−kλt, where Ais a constant.
Step 3. Solve the ODE for X(x):
The solution to X′′(x) + λX(x) = 0 subject to the boundary conditions X(0) = X(1) = 0 is
Xn(x) = sin(nπx),where λ=n2π2
Step 4. Write the general solution as a sum of the separated solutions:
u(x, t) = ∞
∑
n=1
Bne−kn2π2tsin(nπx)
Step 5. Apply the initial condition u(x, 0) = x(1 −x)to find the coefficients Bn:
x(1 −x) = ∞
∑
n=1
Bnsin(nπx)
Multiplying both sides by sin(mπx)and integrating from 0 to 1 gives:
Bm= 4 ∫1
0
x(1 −x)sin(mπx)dx
Step 6. Calculate the coefficients Bn:
Bn= 4 ∫1
0
x(1 −x)sin(nπx)dx =4
n3π3((−1)n−1)
Step 7. Substitute the coefficients back into the general solution to obtain the final solution
for u(x, t):
u(x, t) = ∞
∑
n=1
4
n3π3((−1)n−1)e−kn2π2tsin(nπx)
28. Question: Consider the partial differential equation (PDE) defined by ∂u
∂t + 4∂u
∂x = 0. Find
the general solution to this PDE.
Ans. Step-by-step solution: 1. We are given the PDE ∂u
∂t + 4∂u
∂x = 0. 2. To solve this first-order
linear PDE, we can use the method of characteristics. Let x=x(s)and t=t(s)be the equations
of the characteristic curves. 3. The characteristic equations are given by:
dx
ds = 4 and dt
ds = 1
Integrating these equations with respect to s, we get:
x= 4s+C1and t=s+C2
where C1and C2are constants of integration. 4. Rewriting the equations in terms of s, we have
u(x(s), t(s)) = f(s), where fis an arbitrary function. 5. Using the chain rule, we can express
∂u
∂t and ∂u
∂x as:
∂u
∂t =df
ds and ∂u
∂x =df
ds ·4
6. Substituting these expressions into the PDE, we get:
df
ds + 4 ·df
ds = 0 =⇒5·df
ds = 0
7. Integrating this equation with respect to s, we find f(s) = C3, where C3is another constant
of integration. 8. Therefore, the general solution to the given PDE is u(x, t) = C3, where C3is
an arbitrary constant.
29. Question:
Consider the partial differential equation ut=k(uxx +uyy)with initial condition u(x, 0) =
xe−x2−y2for x, y ∈Rand t > 0. Solve the partial differential equation subject to the initial
condition.
Ans. Solution:
We are given the partial differential equation ut=k(uxx +uyy). To solve this equation, we
will use separation of variables and Fourier series.
1. Find the solution using separation of variables:
Let u(x, y, t) = X(x)Y(y)T(t). Then, the partial differential equation becomes
XT ′=k(X′′Y+XY ′′).
Dividing by kXY T gives
T′
kT =X′′Y+XY ′′
XY =λ.
This gives us three ordinary differential equations:
T′=kλT, X′′ −λX = 0, Y ′′ −λY = 0.
2. Solve the first ODE:
Solving T′=kλT gives T(t) = Cekλt.
3. Solve the second ODE:
The solution to X′′ −λX = 0 is X(x) = Acos(√λx)+Bsin(√λx).
4. Solve the third ODE:
The solution to Y′′ −λY = 0 is Y(y) = Ccos(√λy)+Dsin(√λy).
5. Combine the solutions:
The general solution to the PDE is
u(x, y, t) = ∞
∑
n=1
∞
∑
m=1
(anm cos(√λnmx)+bnm sin(√λnmx))(cnm cos(√λnmy)+dnm sin(√λnmy))ekλnmt.
6. Apply the initial condition:
Given initial condition u(x, 0) = xe−x2−y2, we can solve for the coefficients and obtain the
final solution.
30. Consider the partial differential equation (PDE) given by:
ut= 4uxx
subject to the boundary conditions:
u(0, t) = 0
u(π, t) = 0
and the initial condition:
u(x, 0) = sin(x)
Determine the solution to the PDE using the method of separation of variables.
Ans. Solution:
1. Write the general solution using separation of variables
Let’s assume a solution of the form u(x, t) = X(x)·T(t).
Substitute this into the PDE ut= 4uxx:
X(x)·T′(t) = 4X′′(x)·T(t)
Divide both sides by X(x)·T(t)to separate variables:
T′(t)
T(t)= 4X′′(x)
X(x)=−λ
where λis a constant.
2. Solve the ODE for X(x)
Solve the ODE 4X′′(x) + λX(x) = 0 with the boundary conditions X(0) = 0 and X(π) = 0.
The general solution to the ODE is given by X(x) = Asin(√λ
4x)+Bcos(√λ
4x).
Applying the boundary conditions gives λ=n2for n∈Z+, and the corresponding eigenfunc-
tions are Xn(x) = sin(nx).
3. Solve the ODE for T(t)
Now solve the ODE T′(t)
T(t)=−λ:
T′(t) + n24T(t) = 0
T′(t)+4n2T(t) = 0
T(t) = Cne−4n2t
where Cnis a constant.
4. Form the general solution
The general solution to the PDE is the sum of the separable solutions:
u(x, t) = ∞
∑
n=1
Cnsin(nx)e−4n2t
5. Find the coefficients using the initial condition
Given the initial condition u(x, 0) = sin(x), we need to find Cnsuch that:
∞
∑
n=1
Cnsin(nx) = sin(x)
Using the orthogonality property of sine functions, we find that Cn=2
π∫π
0sin(x)sin(nx)dx.
Calculating the integral gives Cn=2
π·2(−1)n
n2−1=4(−1)n
π(n2−1) .
Therefore, the solution to the PDE is:
u(x, t) = ∞
∑
n=1
4(−1)n
π(n2−1) sin(nx)e−4n2t
