MAT 142 - COLLEGE MATHEMATICS - Differential
Equations Practice Material - Set 4
1. Find the particular solution to the differential equation dy
dx =3x2−4
2ygiven that y(1) = 2.
Ans. Solution: 1. Rearrange the differential equation to separate the variables:
dy
2y=3x2−4
dx
2. Integrate both sides with respect to their respective variables:
Z1
2ydy =Z(3x2−4)dx
1
2ln |y|=x3−4x+C
where Cis the constant of integration.
3. Solve for yby eliminating the natural logarithm:
ln |y|= 2x3−8x+C1
|y|=e2x3−8x+C1
y=±e2x3−8x+C1
4. Apply the initial condition y(1) = 2 to find the particular solution:
2 = ±e2(1)3−8(1)+C1
2 = ±e−6+C1
C1= 6 + ln(2)
Thus, the particular solution to the differential equation is:
y=e2x3−8x+6+ln(2)
y= 2e2x3−8x+6
2. Find the general solution to the differential equation:
y′′ + 4y= 0
Ans. Let’s solve the differential equation step by step:
1. First, we assume the solution is in the form y=emx, where mis a constant to be
determined.
Substituting y=emx into the differential equation, we get:
(m2+ 4)emx = 0
2. Since emx is never zero, we must have m2+ 4 = 0.
Solving this equation, we find that m=±2i.
3. Therefore, the general solution to the differential equation is:
y(x) = c1cos(2x) + c2sin(2x)
where c1and c2are arbitrary constants.
3. Question: Solve the following differential equation:
dy
dx =2x+ 3y
x+ 2y
Ans. Step-by-step solution: 1. Rearrange the given differential equation to get it in a more
recognizable form:
(x+ 2y)dy
dx = 2x+ 3y
2. Notice that the left-hand side of the equation resembles the derivative of a product rule. Let
u=x+ 2y. Then, du
dx = 1 + 2dy
dx . 3. Substitute u=x+ 2yinto the differential equation and
simplify the expression:
u1+2dy
dx= 2x+ 3y
u+ 2udy
dx = 2x+ 3y
4. Substitute dy
dx =u−x
2uback into the equation and solve for u:
u+ 2uu−x
2u= 2x+ 3 u−x
2
u+u−x= 2x+3u−3x
2
5. Simplify the equation further to obtain:
2u−x= 2x+3u−3x
2
6. Solve for uin terms of x:
4u−2x= 4x+ 3u−3x
u= 5x(x+ 1)
7. Now substitute u=x+ 2y= 5x(x+ 1) back into the equation and solve for y:
x+ 2y= 5x(x+ 1)
2y= 5x(x+ 1) −x
y=5x2+ 5x−x
2=5x2+ 4x
2
y=5
2x2+ 2x
Therefore, the solution to the differential equation is y=5
2x2+ 2x.
4. Find the general solution to the differential equation: dy
dx =x2−y2
2xy .
Ans. Solution: 1. Let’s rewrite the given differential equation in the form dy
dx =P(x) + Q(x)y.
We have: dy
dx =x2−y2
2xy Rearranging terms, we get: 2xy dy = (x2−y2)dx
2. Let’s split the variables and integrate both sides. R2xy dy =R(x2−y2)dx
3. Integrating both sides gives us: x2y=x3
3−y3
3+C
4. To simplify further, we multiply through by 3 to get: 3x2y=x3−y3+C
5. This is the general solution to the differential equation dy
dx =x2−y2
2xy .
5. Question:
Solve the following first-order differential equation:
dy
dx +y=1
x
Ans. Solution:
1. This is a first-order linear differential equation of the form dy
dx +P(x)y=Q(x), where
P(x) = 1 and Q(x) = 1
x.
2. To solve this, we use an integrating factor which is given by e∫P(x)dx.
3. Calculating the integrating factor, e∫1dx, we get ex.
4. Multiply both sides of the differential equation by the integrating factor:
exdy
dx +exy=ex1
x
5. Notice that the left side can be rewritten using the product rule for derivatives:
d
dx(exy) = exdy
dx +exy
6. Therefore, we have:
d
dx(exy) = ex1
x
7. Integrating both sides with respect to xgives:
exy=Zex1
xdx
8. The integral on the right-hand side does not have an elementary form, so we express it in
terms of the exponential integral function Ei(x):
exy=Ei(x) + C
where Cis the constant of integration.
9. Finally, we can solve for y:
y=e−x(Ei(x) + C)
So, the general solution to the differential equation is y=e−x(Ei(x) + C), where Cis an
arbitrary constant.
6. Question: Solve the following initial value problem:
y′=2x
y, y(0) = 1
Ans. Step-by-step solution: 1. Rewrite the given differential equation in terms of separable
variables: dy
dx =2x
y
2. Separate variables by multiplying both sides by yand dividing both sides by dx:
ydy = 2xdx
3. Integrate both sides with respect to their respective variables:
Zydy =Z2xdx
4. Perform the integrations: 1
2y2=x2+C
5. To find C, use the initial condition y(0) = 1:
1
2(1)2= 0 + C
C=1
2
6. Substitute C=1
2back into the equation:
1
2y2=x2+1
2
7. Solve for yby taking the square root of both sides:
y=±√2x+ 1
8. However, since y(0) = 1, we choose the positive square root:
y=√2x+ 1
Therefore, the solution to the initial value problem is y=√2x+ 1.
7. Question: Find the general solution to the differential equation y′′ + 4y= 0.
Ans. Step-by-step solution: 1. We first write the characteristic equation for the given differential
equation:
r2+ 4 = 0
2. Solve the characteristic equation by setting the discriminant equal to zero:
r=±2i
3. The general solution to the differential equation is:
y(t) = c1cos(2t) + c2sin(2t)
where c1and c2are arbitrary constants.
8. Question: Solve the following first-order linear differential equation:
dy
dx + 2y= 4x
Ans. Step-by-step solution: 1. First, we rewrite the differential equation in standard form:
dy
dx + 2y= 4x
This equation is in the form of dy
dx +P(x)y=Q(x).
2. Let’s find the integrating factor, denoted by I(x), which is defined as:
I(x) = e∫P(x)dx
In this case, P(x) = 2, so:
I(x) = e∫2dx =e2x
3. We multiply both sides of the differential equation by the integrating factor I(x):
e2xdy
dx + 2e2xy= 4xe2x
4. Recognizing that the left-hand side is the derivative of ye2xwith respect to x, we can
rewrite the equation as:
d
dx(ye2x) = 4xe2x
5. We now integrate both sides with respect to xto solve for y:
Zd
dx(ye2x)dx =Z4xe2xdx
ye2x=Z4xe2xdx
6. We can integrate the right-hand side using integration by parts: Let u=xand dv =
4e2xdx. Then, du =dx and v= 2e2x.
Z4xe2xdx = 2xe2x−Z2e2xdx
= 2xe2x−e2x
7. Substituting this back into our equation, we get:
ye2x= 2xe2x−e2x+C
where Cis the constant of integration.
8. Finally, we solve for yby dividing both sides by e2x:
y= 2x−1 + Ce−2x
where Cis an arbitrary constant. Therefore, the general solution to the differential equation is
y= 2x−1 + Ce−2x.
9. Question: Find the general solution to the differential equation: dy
dx −2y=x2.
Ans. Step-by-step solution: 1. First, we identify the linear differential equation in standard
form: dy
dx −2y=x2. 2. Next, we find the integrating factor, denoted by I(x), using the
formula I(x) = e∫−2dx. 3. Calculating the integral gives us I(x) = e−2x. 4. We multiply
both sides of the differential equation by the integrating factor I(x):e−2xdy
dx −2e−2xy=e−2xx2.
5. By the product rule, the left-hand side can be rewritten as d
dx (e−2xy) = e−2xx2. 6. We
integrate both sides with respect to x:Rd
dx (e−2xy)dx =Re−2xx2dx. 7. This simplifies to
e−2xy=−1
2x2e−2x−1
2Rx2(−2e−2x)dx. 8. Solving the integral on the right-hand side, we
get e−2xy=−1
2x2e−2x+1
2R2x2e−2xdx. 9. Applying integration by parts to the integral gives
us e−2xy=−1
2x2e−2x+1
2−x2e−2x+ 2 Rxe−2xdx. 10. Solving the remaining integral, we
obtain e−2xy=−1
2x2e−2x−1
2x2e−2x+R−xe−2xdx. 11. Finally, integrating the last term on
the right-hand side gives e−2xy=−x2e−2x+1
4e−2x+C, where Cis the constant of integration.
12. Multiplying through by e2x, we find the general solution to the differential equation: y=
−x2+1
4+Ce2x, where Cis an arbitrary constant.
10. Question: Solve the following first-order linear differential equation:
dy
dx +2y
x=xln(x)
Ans. Step-by-step solution: 1. This is a first-order linear differential equation of the form
dy
dx +P(x)y=Q(x)where P(x) = 2
xand Q(x) = xln(x). 2. The integrating factor is given
by µ(x) = e∫P(x)dx. So, in this case, µ(x) = e∫2
xdx. 3. Calculating the integral, we find
µ(x) = e2ln(x)=x2. 4. Multiply both sides of the differential equation by the integrating factor
µ(x):
x2dy
dx + 2xy =x3ln(x)
5. Rewrite the left-hand side as d
dx (x2y):
d
dx(x2y) = x3ln(x)
6. Integrate both sides with respect to x:
Zd
dx(x2y)dx =Zx3ln(x)dx
x2y=1
4x4(ln(x)−1
4) + C
where Cis the constant of integration. 7. Finally, solving for y:
y=1
4x2(ln(x)−1
4) + C
x2
11. Question: Solve the following differential equation: dy
dx =x2+y2
xy .
Ans. Step-by-step solution: 1. Let’s rearrange the given differential equation in terms of yand
x:dy
dx =x2+y2
xy .⇒ydy
dx =x2+y2
x.⇒ydy =x2+y2
xdx. 2. Now, let’s rewrite the equation in a more
manageable form: ydy −y2
xdx =xdx. 3. We notice that the left side of the equation can be
written in the form of an exact differential: d(y2)−2ydy =xdx. 4. Integrating both sides, we
get: Rd(y2)−R2ydy =Rxdx.y2−y2=1
2x2+C.0 = 1
2x2+C.⇒C=−1
2x2. 5. Therefore,
the solution to the differential equation is y2=1
2x2−1
2x2.⇒y2=1
2x2.⇒y=±1
√2x. Hence,
the general solution to the given differential equation is y=±1
√2x.
12. Question: Solve the following first-order differential equation:
dy
dx = 2xy2
Ans. Step-by-step solution:
1. We start by separating variables. Rearranging the equation, we have:
dy
y2= 2xdx
2. Next, we integrate both sides with respect to their respective variables. Integrating the
left side gives: Z1
y2dy =Z2xdx
−1
y=x2+C
where C is the constant of integration.
3. Solving for y, we get:
y=−1
x2+C
Therefore, the solution to the given differential equation is y=−1
x2+C, where C is a constant
of integration.
13. Question: Solve the differential equation dy
dx =2x+3y−4
3x−2y+5 .
Ans. Solution: 1. Rewrite the differential equation in a more standard form by multiplying
both sides by dx:
dy
dx =2x+ 3y−4
3x−2y+ 5
2. Rearrange terms to get all terms involving yon the left side and all terms involving xon
the right side:
2x+ 3y−4 = (3x−2y+ 5) dy
dx
3. Multiply through by dx and rearrange to separate xand yterms:
2x+ 3y−4 = (3x−2y+ 5)dy
4. Expand the terms on the right side:
2x+ 3y−4 = 3xdy −2ydy + 5dy
5. Rearrange to collect the terms with dy on one side:
3xdy −2ydy −5dy = 2x−3y+ 4
6. Factor out dy on the left side:
(3x−2y−5)dy = 2x−3y+ 4
7. Integrate both sides with respect to y:
Z(3x−2y−5)dy =Z(2x−3y+ 4)dx
8. This simplifies to:
3xy −y2−5y= 2xy −3
2y2+ 4x+C
9. Rearrange the terms to get an equation involving only y:
−3
2y2−y+ 4x−3xy + 2xy −5y=C
10. Simplify further to get:
−3
2y2−6y+ 4x=C
11. This is the general solution to the differential equation.
14. Question: Solve the following differential equation:
y′′ −4y′+ 4y= 4e2x
Ans. Step-by-step solution:
1. First, we find the complementary function by solving the homogeneous differential equation:
y′′ −4y′+ 4y= 0
The characteristic equation is r2−4r+ 4 = 0, which simplifies to (r−2)2= 0. This gives us a
repeated root r= 2. Therefore, the complementary function is:
yc=c1e2x+c2xe2x
2. Next, we find the particular integral for the non-homogeneous equation:
yp=Ae2x
We substitute ypback into the differential equation and solve for A:
4Ae2x−4(2Ae2x)+4Ae2x= 4e2x
4Ae2x−8Ae2x+ 4Ae2x= 4e2x
0 = 4e2x
This means that A= 1. So the particular integral is yp=e2x.
3. Combining the complementary function and the particular integral, we get the general
solution:
y=yc+yp=c1e2x+c2xe2x+e2x
Therefore, the general solution to the given differential equation is:
y=c1e2x+c2xe2x+e2x
15. Find the general solution to the differential equation:
y′′ −4y′+ 4y= 0
Ans. Solution: 1. First, we write the characteristic equation for the given differential equation:
r2−4r+ 4 = 0
2. Solve for rby factoring the quadratic equation:
(r−2)2= 0
r= 2
3. Since the characteristic equation has a repeated root, the general solution to the differential
equation is:
y(t) = c1e2t+c2te2t
where c1, c2are arbitrary constants.
16. Question 16: Solve the following initial value problem:
y′+ 2y=sin(2x), y(0) = 1
Ans. Step 1. First, find the integrating factor µ(x)by solving the differential equation
y′+ 2y= 0: Given y′+ 2y=sin(2x), the integrating factor is given by:
µ(x) = e∫2dx =e2x
Step 2. Multiply the given differential equation by the integrating factor µ(x):
e2xy′+ 2e2xy=e2xsin(2x)
Step 3. Rewrite the left-hand side as the derivative of the product y(x)e2x:
(ye2x)′=e2xsin(2x)
Step 4. Integrate both sides with respect to x:
Z(ye2x)′dx =Ze2xsin(2x)dx
ye2x=−1
2e2xcos(2x) + C
Step 5. Solve for yby dividing both sides by e2x:
y=−1
2cos(2x) + Ce−2x
Step 6. Apply the initial condition y(0) = 1 to find the value of C:
1 = −1
2cos(0) + Ce0
1 = −1
2+C
C=3
2
Step 7. Substitute the value of Cback into the solution to find the final answer:
y=−1
2cos(2x) + 3
2e−2x
Therefore, the solution to the initial value problem is y=−1
2cos(2x) + 3
2e−2x.
17. Question: Solve the initial value problem: dy
dx =2x+3y−1
3x+2y+2 ,y(0) = 1.
Ans. Step-by-step solution: 1. Rearrange the given differential equation to the form dy
dx =
A(x) + B(y), where A(x)depends only on xand B(y)depends only on y.
dy
dx =2x+ 3y−1
3x+ 2y+ 2 ⇒(3x+ 2y+ 2) dy = (2x+ 3y−1) dx
3x dy + 2y dy + 2 dy = 2x dx + 3y dx −dx
3x dy −3y dx =−2x dx −2dy
Hence, we have
3x dy + 2 dy −3y dx + 2x dx = 0
2dy + 3x dy −3y dx + 2x dx = 0
2. We notice that the equation is exact as
∂M
∂y = 2 = ∂N
∂x = 2
Therefore, we can write the equation as
2dy + 3 dx −3dx + 2 dx = 0
d(2y+ 3x) = d(c)
2y+ 3x=c
where cis the constant of integration. 3. To find the value of c, we use the initial condition
y(0) = 1.
2(1) + 3(0) = c
c= 2
Thus, the solution to the initial value problem is 2y+ 3x= 2.
18. Question: Solve the following differential equation: y′′ −5y′+ 6y= 0.
Ans. Step-by-step solution: 1. Write down the characteristic equation: r2−5r+ 6 = 0. 2.
Solve the characteristic equation: (r−2)(r−3) = 0, so r= 2,3. 3. The general solution is
then y(t) = c1e2t+c2e3t, where c1and c2are arbitrary constants. Thus, the solution to the
differential equation is y(t) = c1e2t+c2e3t, where c1and c2are constants determined by initial
conditions.
19. Question: Solve the following initial value problem:
dy
dx =2x
1 + y2, y(0) = 1
Ans. Step-by-step solution: 1. Rewrite the given differential equation as:
(1 + y2)dy = 2xdx
2. Integrate both sides of the equation:
Z(1 + y2)dy =Z2xdx
3. Integrating the left side gives:
y+y3
3=x2+C
where C is the constant of integration.
4. Apply the initial condition y(0) = 1 to find the value of C:
1 + 1
3= 0 + C
C=4
3
5. Substitute the value of C back into the equation:
y+y3
3=x2+4
3
6. The solution to the initial value problem is:
y(x) = √3x2+ 4 −1
20. Question: Solve the differential equation dy
dx =2x+1
2y.
Ans. Step-by-step solution: 1. Rewrite the differential equation in the form 2y dy = (2x+1) dx.
2. Integrate both sides to solve for y:
Z2y dy =Z(2x+ 1) dx
3. This gives us y2=x2+x+C, where Cis the constant of integration. 4. Therefore, the
general solution to the differential equation is y=±√x2+x+C.
21. Question: Solve the initial value problem y′+y=sin(x)with initial condition y(0) = 2.
Ans. Step-by-step solution:
1. First, we write the differential equation in standard form, y′+y=sin(x), where y′=dy
dx .
This is a first-order linear non-homogeneous ordinary differential equation.
2. To solve the differential equation, we first find the integrating factor I(x), given by
I(x) = e∫1dx =ex.
3. We then multiply both sides of the differential equation by the integrating factor:
ex(y′+y) = exsin(x)
4. The left side can be rewritten using the product rule:
d
dx(exy) = exsin(x)
5. Integrating both sides with respect to x, we get:
exy=−cos(x) + C
, where Cis an arbitrary constant of integration.
6. Now, we divide both sides by exto solve for y:
y=−e−xcos(x) + Ce−x
7. Finally, we use the initial condition y(0) = 2 to find the value of C:
2 = −e0cos(0) + Ce0=−1 + C
C= 3
8. Therefore, the solution to the initial value problem is:
y=−e−xcos(x)+3e−x
22. Question: Solve the following differential equation:
dy
dx + 2y= 3e−2x
Ans. Step-by-step solution: 1. First, we identify the differential equation as a first-order linear
differential equation in the standard form:
dy
dx +P(x)y=Q(x)
where P(x) = 2 and Q(x) = 3e−2x.
2. To solve this linear differential equation, we begin by finding the integrating factor µ(x),
defined by:
µ(x) = e∫P(x)dx
3. In this case, P(x) = 2, so we have:
ZP(x)dx =Z2dx = 2x
µ(x) = e2x
4. Next, we multiply both sides of the differential equation by the integrating factor µ(x):
e2xdy
dx + 2e2xy= 3e0
5. Recognizing that the left side is the derivative of the product (e2xy)with respect to x, we
can rewrite the equation as:
d
dx e2xy= 3
6. Integrating both sides with respect to x, we get:
e2xy= 3x+C
where Cis the constant of integration.
7. Solving for y, we have:
y=3x+C
e2x
y=3x
e2x+C
e2x
Therefore, the general solution to the differential equation is:
y=3x
e2x+C
e2x
where Cis an arbitrary constant.
23. Question:
Solve the following differential equation:
dy
dx =2x+y
x+ 2y
Ans. Step-by-step solution:
1. Rewrite the given differential equation in the form dy
dx =f(x, y):
dy
dx =2x+y
x+ 2y
2. Notice that the given differential equation is not exact. To check for exactness, calculate
the partial derivatives ∂f
∂y and ∂f
∂x :
∂f
∂y =1
x+ 2y−(2x+y)(2)
(x+ 2y)2
∂f
∂x =2
x+ 2y−(x+ 2y)(2)
(x+ 2y)2
3. Determine if the given equation is exact by verifying if ∂f
∂y =∂f
∂x . In this case, we see that
they are not equal.
4. To find an integrating factor µ(x, y)such that µ(x, y)Mdx +µ(x, y)Ndy = 0 is exact, we
set the left-hand side equation of the given differential equation as the total derivative of some
function U(x, y)with respect to x.
5. Let’s assume U(x, y) = µ(x, y)M(x, y)dx +N(x, y)dy, whereM = 2x+yandN =
x+2y.T hen, find∂U ∂y and ∂U
∂x :
∂U
∂y =µ(x, y)N+µ(x, y)∂N
∂y
∂U
∂x =µ(x, y)M+µ(x, y)∂M
∂x
6. Equate ∂U
∂y to ∂U
∂x and solve for µ(x, y)to find the integrating factor.
7. After finding the integrating factor µ(x, y), rewrite the differential equation in the form
d
dx (µ(x, y)y) = µ(x, y)M−∂µ
∂y .
8. Finally, integrate the above expression to obtain the general solution to the differential
equation.
24. Question: Solve the following initial value problem:
y′=2y
x+x2, y(1) = 0
Ans. Step-by-step solution:
1. The given differential equation is a first-order linear ordinary differential equation in the
form of y′+P(x)y=Q(x). To solve this differential equation, we will use the integrating factor
method. First, we identify P(x) = 2
xand Q(x) = x2.
2. The integrating factor, denoted by µ(x), is given by µ(x) = e∫P(x)dx. Therefore, in this
case, µ(x) = e∫2
xdx =e2ln x=x2.
3. Multiply the differential equation by the integrating factor µ(x):
x2y′−2xy =x4
4. Rewrite the equation using the product rule for differentiation:
(x2y)′=x4
5. Integrate both sides with respect to x:
Z(x2y)′dx =Zx4dx
x2y=1
5x5+C
6. Solve for y:
y=1
5x3+C
x2
7. Apply the initial condition y(1) = 0 to find the value of the constant C:
0 = 1
5(1)3+C
(1)2
0 = 1
5+C
C=−1
5
8. Substitute the value of Cback into the general solution to obtain the particular solution:
y=1
5x3−1
5x2
Therefore, the solution to the initial value problem is y=1
5x3−1
5x2.
25. Question: Solve the following initial value problem:
dy
dx =y(1 −x), y(0) = 2
Ans. Solution: 1. We first write the given differential equation in the standard form for solving:
dy
dx =y(1 −x)=⇒dy
y= (1 −x)dx
2. Next, we integrate both sides of the equation:
Z1
ydy =Z(1 −x)dx
3. Integrating both sides gives us:
ln |y|=x−x2
2+C
where Cis the constant of integration.
4. Exponentiating both sides, we get:
|y|=ex−x2
2+C
5. Since the initial condition is y(0) = 2, we substitute x= 0 and y= 2 into our equation
to find the value of the constant C:
|2|=e0−02
2+C=⇒2 = eC
6. Therefore, C=ln 2, and the solution to the differential equation is:
y=±2ex−x2
2+ln 2
7. Simplifying, we get:
y= 2ex−x2
2+ln 2
Hence, the solution to the initial value problem is y= 2ex−x2
2+ln 2.
26. Question: Solve the differential equation y′′ −2y′+y= 2et.
Ans. Step-by-step solution:
1. First, we solve the homogeneous part of the differential equation by setting y′′−2y′+y= 0.
The characteristic equation is r2−2r+1 = (r−1)2= 0. This gives us a repeated root r1=r2= 1.
So, the general solution to the homogeneous equation is yh(t) = c1et+c2tet, where c1and c2
are constants.
2. Next, we find a particular solution to the non-homogeneous part of the differential equation,
2et. Since the right-hand side is of the form Cet, a good guess for a particular solution would
be of the form yp(t) = Aet, where Ais a constant to be determined. Substitute yp(t)and its
derivatives into the differential equation to find A. We get: 2Aet−2Aet+Aet= 2etAet= 2et
A= 2
So, a particular solution is yp(t) = 2et.
3. The general solution to the entire differential equation is the sum of the homogeneous
and particular solutions: y(t) = yh(t) + yp(t) = c1et+c2tet+ 2et, where c1and c2are arbitrary
constants.
27. Question 27: Solve the following first-order linear differential equation:
dy
dx + 2xy =ex2
Ans. Solution: 1. We can solve this first-order linear differential equation using the integrating
factor method. First, we need to rewrite the equation in the standard form:
dy
dx + 2xy =ex2
2. The standard form is: dy
dx +P(x)y=Q(x), where P(x) = 2xand Q(x) = ex2.
3. Next, we find the integrating factor I(x), which is given by I(x) = e∫P(x)dx.
4. In this case, P(x) = 2x, so RP(x)dx =R2xdx =x2.
5. Therefore, the integrating factor I(x) = ex2.
6. Multiply the original differential equation by the integrating factor:
ex2dy
dx + 2xex2y=e2x2
7. The left side now becomes the derivative of the product of yand the integrating factor:
d
dx(ex2y) = e2x2
8. Integrate both sides with respect to xto solve for y:
Zd
dx(ex2y)dx =Ze2x2dx
9. By integrating the left side, we get ex2y=1
2e2x2+C, where Cis the constant of
integration.
10. Finally, we solve for yby dividing by ex2:
y=1
2ex2+Ce−x2
Hence, the solution to the given differential equation is y=1
2ex2+Ce−x2, where Cis an
arbitrary constant.
28. Question 28: Solve the initial value problem:
dy
dx = (x2+y2)e−y, y(0) = 1
Ans. Solution: 1. First, rewrite the differential equation as:
dy
dx = (x2+y2)e−y
2. Notice that this is a separable differential equation. We can write it as:
dy
dx = (x2+y2)e−y=⇒eydy = (x2+y2)dx
3. Integrate both sides with respect to their respective variables:
Zeydy =Z(x2+y2)dx
4. This gives us:
ey=x3
3+y3
3+C
where C is the constant of integration.
5. To determine the value of the constant C, we use the initial condition y(0) = 1:
e1=03
3+13
3+C=⇒e=1
3+C=⇒C=e−1
3
6. Therefore, the particular solution to the initial value problem is:
ey=x3
3+y3
3+e−1
3
29. Question 29: Solve the following differential equation using the method of undetermined
coefficients: y′′ −3y′+ 2y= 4ex+ 3 sin(x).
Ans. Solution: 1. First, find the complementary solution ycby solving the characteristic
equation r2−3r+ 2 = 0. The characteristic roots are r1= 1 and r2= 2, so the complementary
solution is yc=c1ex+c2e2x.
2. Next, find the particular solution yp. Since the right-hand side of the equation includes both
4exand sin(x), assume yp=Aex+Bsin(x). Then, calculate the derivatives: y′
p=Aex+Bcos(x)
and y′′
p=Aex−Bsin(x).
3. Substitute yp,y′
p, and y′′
pinto the differential equation to get: (Aex−Bsin(x)) −3(Aex+
Bcos(x)) + 2(Aex+Bsin(x)) = 4ex+ 3 sin(x).
4. Simplify the equation and group like terms to solve for Aand B. You should find A= 2
and B=−1.
5. Therefore, the particular solution is yp= 2ex−sin(x).
6. The general solution is the sum of the complementary and particular solutions: y=
yc+yp=c1ex+c2e2x+ 2ex−sin(x).
30. Question: Solve the following differential equation:
dy
dx =2x2y+y2
2xy +x2.
Ans. Step-by-step solution: 1. Rewrite the given differential equation in terms of differentials:
dy
dx =2x2y+y2
2xy +x2.
2. Multiply both sides by dx to separate variables:
dy
2x2y+y2=dx
2xy +x2.
3. Perform partial fraction decomposition on the left side:
dy
y(2x+y)=dx
x(2y+x).
4. Decompose the left side into partial fractions:
1
y(2x+y)=A
y+B
2x+y.
5. Clear the denominators and solve for A and B:
1 = A(2x+y) + By.
6. Setting y = 0 gives A=1
2x, setting x = -y/2 gives B=−1
4y.
7. Substitute the values of A and B back into the partial fractions decomposition:
1
y(2x+y)=1
2x·y−1
4y·(2x+y).
8. Similarly, decompose the right side into partial fractions:
1
x(2y+x)=C
x+D
2y+x.
9. Clear the denominators and solve for C and D:
1 = C(2y+x) + Dx.
10. Setting x = 0 gives C=1
2y, setting y = -x/2 gives D=−1
4x.
11. Substitute the values of C and D back into the partial fractions decomposition:
1
x(2y+x)=1
2y·x−1
4x·(2y+x).
12. Now, integrate both sides:
Z1
y(2x+y)dy =Z1
x(2y+x)dx.
13. Integrating the left side gives 1
2ln |y| − 1
2ln |2y+x|=ln |x|+1
2ln |2y+x|+C.
14. Rearrange the equation and simplify it to obtain the solution to the differential equation:
ln |y| − ln |2y+x|= 2 ln |x|+ln |2y+x|+ 2C.
15. Combine the logarithmic terms to get:
ln
y
2y+x
=ln |x|2+C.
16. Exponentiate both sides to solve for y:
y=cx2
1−c,
where c=±eC.
31. Question 31: Solve the initial value problem dy
dx =y3
x2,y(1) = −1.
Ans. Solution: 1. First, notice that the given differential equation is separable, so we can
rewrite it as dy
y3=dx
x2.
2. Integrate both sides with respect to their variables:
Z1
y3dy =Z1
x2dx
3. This gives us −1
2y2=−1
x+C, where Cis the constant of integration.
4. Now, we need to find the value of the constant C. Substituting the initial condition
y(1) = −1into the equation, we get
−1
2(−1)2=−1
1+C
1
2=−1 + C
C=1
2−1
C=−1
2
5. Substituting C=−1
2back into our equation, we have
−1
2y2=−1
x−1
2
6. Solving for y, we get 1
2y2=1
x+1
2
2y2=1
1
x+1
2
y2=1
2
x+ 1
y=±s1
2
x+ 1
7. Since y(1) = −1, we have y=−q1
2
1+1 =−1, which satisfies the initial condition.
So, the solution to the initial value problem is y=−q1
2
x+1 .
32. Question 32: Solve the following differential equation using an integrating factor: dy
dx +2y
x=
x2.
Ans. Solution: 1. The given differential equation is of the form dy
dx +P(x)y=Q(x), where
P(x) = 2
xand Q(x) = x2. 2. First, we need to find the integrating factor. The integrating factor,
denoted by µ(x), is given by µ(x) = e∫P(x)dx. 3. In this case, P(x) = 2
x, so the integrating
factor becomes µ(x) = e∫2
xdx. 4. Integrating 2
xwith respect to xgives us ln |x|2= 2 ln |x|.
5. Therefore, the integrating factor is µ(x) = e2ln |x|=eln |x|2=x2. 6. Multiply both sides of
the original differential equation by the integrating factor x2:x2dy
dx + 2xy =x4. 7. Notice that
the left side of the equation is now the result of applying the product rule to x2y. So, we can
rewrite the equation as d
dx (x2y) = x4. 8. Integrate both sides with respect to xto solve for x2y:
Rd
dx (x2y)dx =Rx4dx. 9. This yields x2y=1
5x5+C, where Cis the constant of integration.
10. Finally, solving for y, we get y=1
5x3+C
x2.
33. Question: Solve the following differential equation using the method of integrating factors:
dy
dx −2y= 3x.
Ans. Step-by-step solution: 1. The given differential equation can be written in the form
dy
dx +P(x)y=Q(x), where P(x) = −2and Q(x) = 3x. 2. To solve this differential equation,
we first find the integrating factor (IF ) by integrating e∫P(x)dx.
Find IF :IF =e∫−2dx =e−2x
3. Multiply both sides of the differential equation by the integrating factor:
e−2xdy
dx −2e−2xy= 3xe−2x
4. Rewrite the left side as the derivative of the product of yand the integrating factor:
d
dx ye−2x= 3xe−2x
5. Integrate both sides with respect to x:
Zd
dx ye−2xdx =Z3xe−2xdx
ye−2x=−3
2xe−2x−3
4e−2x+Cwhere Cis the constant of integration
6. Solve for yby multiplying through by e2xand simplifying:
y=−3
2x−3
4+Ce2x
Therefore, the solution to the differential equation is y=−3
2x−3
4+Ce2x, where Cis the
constant of integration.
34. Question: Find the particular solution of the differential equation dy
dx = 4x+1
ythat passes
through the point (1, 2).
Ans. Step-by-step solution: 1. Rearrange the differential equation:
dy
dx = 4x+1
y
ydy
dx = 4xy + 1
2. Integrate both sides with respect to x:
Zy dy =Z(4xy + 1) dx
y2
2= 2x2y+x+C
where Cis the constant of integration.
3. Plug in the coordinates of the point (1, 2) to find the particular solution:
22
2= 2(1)2(2) + 1 + C
2 = 4 + 1 + C
C=−3
4. Substitute C=−3back into the general solution:
y2
2= 2x2y+x−3
5. Simplify the particular solution:
y2= 4x2y+ 2x−6
y2−4x2y= 2x−6
y2−4x2y= 2(x−3)
y(y−4x2) = 2(x−3)
y=2(x−3)
y−4x2
Therefore, the particular solution passing through the point (1, 2) is y=2(x−3)
y−4x2.
35. Question: Solve the following differential equation:
dy
dx −2y= 4e2x
Ans. Solution: 1. First, we identify the integrating factor. The integrating factor is given by:
IF =e∫−2dx =e−2x
y′′ + 4y= 0
Ans. Let’s solve the differential equation step by step:
1. First, we assume the solution is in the form y=emx, where mis a constant to be
determined.
Substituting y=emx into the differential equation, we get:
(m2+ 4)emx = 0
2. Since emx is never zero, we must have m2+ 4 = 0.
Solving this equation, we find that m=±2i.
3. Therefore, the general solution to the differential equation is:
y(x) = c1cos(2x) + c2sin(2x)
where c1and c2are arbitrary constants.
3. Question: Solve the following differential equation:
dy
dx =2x+ 3y
x+ 2y
Ans. Step-by-step solution: 1. Rearrange the given differential equation to get it in a more
recognizable form:
(x+ 2y)dy
dx = 2x+ 3y
2. Notice that the left-hand side of the equation resembles the derivative of a product rule. Let
u=x+ 2y. Then, du
dx = 1 + 2dy
dx . 3. Substitute u=x+ 2yinto the differential equation and
simplify the expression:
u1+2dy
dx= 2x+ 3y
u+ 2udy
dx = 2x+ 3y
4. Substitute dy
dx =u−x
2uback into the equation and solve for u:
u+ 2uu−x
2u= 2x+ 3 u−x
2
u+u−x= 2x+3u−3x
2
5. Simplify the equation further to obtain:
2u−x= 2x+3u−3x
2
6. Solve for uin terms of x:
4u−2x= 4x+ 3u−3x
u= 5x(x+ 1)
7. Now substitute u=x+ 2y= 5x(x+ 1) back into the equation and solve for y:
x+ 2y= 5x(x+ 1)
2y= 5x(x+ 1) −x
y=5x2+ 5x−x
2=5x2+ 4x
2
y=5
2x2+ 2x
Therefore, the solution to the differential equation is y=5
2x2+ 2x.
4. Find the general solution to the differential equation: dy
dx =x2−y2
2xy .
Ans. Solution: 1. Let’s rewrite the given differential equation in the form dy
dx =P(x) + Q(x)y.
We have: dy
dx =x2−y2
2xy Rearranging terms, we get: 2xy dy = (x2−y2)dx
2. Let’s split the variables and integrate both sides. R2xy dy =R(x2−y2)dx
3. Integrating both sides gives us: x2y=x3
3−y3
3+C
4. To simplify further, we multiply through by 3 to get: 3x2y=x3−y3+C
5. This is the general solution to the differential equation dy
dx =x2−y2
2xy .
5. Question:
Solve the following first-order differential equation:
dy
dx +y=1
x
Ans. Solution:
1. This is a first-order linear differential equation of the form dy
dx +P(x)y=Q(x), where
P(x) = 1 and Q(x) = 1
x.
2. To solve this, we use an integrating factor which is given by e∫P(x)dx.
3. Calculating the integrating factor, e∫1dx, we get ex.
4. Multiply both sides of the differential equation by the integrating factor:
exdy
dx +exy=ex1
x
5. Notice that the left side can be rewritten using the product rule for derivatives:
d
dx(exy) = exdy
dx +exy
6. Therefore, we have:
d
dx(exy) = ex1
x
7. Integrating both sides with respect to xgives:
exy=Zex1
xdx
8. The integral on the right-hand side does not have an elementary form, so we express it in
terms of the exponential integral function Ei(x):
exy=Ei(x) + C
where Cis the constant of integration.
9. Finally, we can solve for y:
y=e−x(Ei(x) + C)
So, the general solution to the differential equation is y=e−x(Ei(x) + C), where Cis an
arbitrary constant.
6. Question: Solve the following initial value problem:
y′=2x
y, y(0) = 1
Ans. Step-by-step solution: 1. Rewrite the given differential equation in terms of separable
variables: dy
dx =2x
y
2. Separate variables by multiplying both sides by yand dividing both sides by dx:
ydy = 2xdx
3. Integrate both sides with respect to their respective variables:
Zydy =Z2xdx
4. Perform the integrations: 1
2y2=x2+C
5. To find C, use the initial condition y(0) = 1:
1
2(1)2= 0 + C
C=1
2
6. Substitute C=1
2back into the equation:
1
2y2=x2+1
2
7. Solve for yby taking the square root of both sides:
y=±√2x+ 1
8. However, since y(0) = 1, we choose the positive square root:
y=√2x+ 1
Therefore, the solution to the initial value problem is y=√2x+ 1.
7. Question: Find the general solution to the differential equation y′′ + 4y= 0.
Ans. Step-by-step solution: 1. We first write the characteristic equation for the given differential
equation:
r2+ 4 = 0
2. Solve the characteristic equation by setting the discriminant equal to zero:
r=±2i
3. The general solution to the differential equation is:
y(t) = c1cos(2t) + c2sin(2t)
where c1and c2are arbitrary constants.
8. Question: Solve the following first-order linear differential equation:
dy
dx + 2y= 4x
Ans. Step-by-step solution: 1. First, we rewrite the differential equation in standard form:
dy
dx + 2y= 4x
This equation is in the form of dy
dx +P(x)y=Q(x).
2. Let’s find the integrating factor, denoted by I(x), which is defined as:
I(x) = e∫P(x)dx
In this case, P(x) = 2, so:
I(x) = e∫2dx =e2x
3. We multiply both sides of the differential equation by the integrating factor I(x):
e2xdy
dx + 2e2xy= 4xe2x
4. Recognizing that the left-hand side is the derivative of ye2xwith respect to x, we can
rewrite the equation as:
d
dx(ye2x) = 4xe2x
5. We now integrate both sides with respect to xto solve for y:
Zd
dx(ye2x)dx =Z4xe2xdx
ye2x=Z4xe2xdx
6. We can integrate the right-hand side using integration by parts: Let u=xand dv =
4e2xdx. Then, du =dx and v= 2e2x.
Z4xe2xdx = 2xe2x−Z2e2xdx
= 2xe2x−e2x
7. Substituting this back into our equation, we get:
ye2x= 2xe2x−e2x+C
where Cis the constant of integration.
8. Finally, we solve for yby dividing both sides by e2x:
y= 2x−1 + Ce−2x
where Cis an arbitrary constant. Therefore, the general solution to the differential equation is
y= 2x−1 + Ce−2x.
9. Question: Find the general solution to the differential equation: dy
dx −2y=x2.
Ans. Step-by-step solution: 1. First, we identify the linear differential equation in standard
form: dy
dx −2y=x2. 2. Next, we find the integrating factor, denoted by I(x), using the
formula I(x) = e∫−2dx. 3. Calculating the integral gives us I(x) = e−2x. 4. We multiply
both sides of the differential equation by the integrating factor I(x):e−2xdy
dx −2e−2xy=e−2xx2.
5. By the product rule, the left-hand side can be rewritten as d
dx (e−2xy) = e−2xx2. 6. We
integrate both sides with respect to x:Rd
dx (e−2xy)dx =Re−2xx2dx. 7. This simplifies to
e−2xy=−1
2x2e−2x−1
2Rx2(−2e−2x)dx. 8. Solving the integral on the right-hand side, we
get e−2xy=−1
2x2e−2x+1
2R2x2e−2xdx. 9. Applying integration by parts to the integral gives
us e−2xy=−1
2x2e−2x+1
2−x2e−2x+ 2 Rxe−2xdx. 10. Solving the remaining integral, we
obtain e−2xy=−1
2x2e−2x−1
2x2e−2x+R−xe−2xdx. 11. Finally, integrating the last term on
the right-hand side gives e−2xy=−x2e−2x+1
4e−2x+C, where Cis the constant of integration.
12. Multiplying through by e2x, we find the general solution to the differential equation: y=
−x2+1
4+Ce2x, where Cis an arbitrary constant.
10. Question: Solve the following first-order linear differential equation:
dy
dx +2y
x=xln(x)
Ans. Step-by-step solution: 1. This is a first-order linear differential equation of the form
dy
dx +P(x)y=Q(x)where P(x) = 2
xand Q(x) = xln(x). 2. The integrating factor is given
by µ(x) = e∫P(x)dx. So, in this case, µ(x) = e∫2
xdx. 3. Calculating the integral, we find
µ(x) = e2ln(x)=x2. 4. Multiply both sides of the differential equation by the integrating factor
µ(x):
x2dy
dx + 2xy =x3ln(x)
5. Rewrite the left-hand side as d
dx (x2y):
d
dx(x2y) = x3ln(x)
6. Integrate both sides with respect to x:
Zd
dx(x2y)dx =Zx3ln(x)dx
x2y=1
4x4(ln(x)−1
4) + C
where Cis the constant of integration. 7. Finally, solving for y:
y=1
4x2(ln(x)−1
4) + C
x2
11. Question: Solve the following differential equation: dy
dx =x2+y2
xy .
Ans. Step-by-step solution: 1. Let’s rearrange the given differential equation in terms of yand
x:dy
dx =x2+y2
xy .⇒ydy
dx =x2+y2
x.⇒ydy =x2+y2
xdx. 2. Now, let’s rewrite the equation in a more
manageable form: ydy −y2
xdx =xdx. 3. We notice that the left side of the equation can be
written in the form of an exact differential: d(y2)−2ydy =xdx. 4. Integrating both sides, we
get: Rd(y2)−R2ydy =Rxdx.y2−y2=1
2x2+C.0 = 1
2x2+C.⇒C=−1
2x2. 5. Therefore,
the solution to the differential equation is y2=1
2x2−1
2x2.⇒y2=1
2x2.⇒y=±1
√2x. Hence,
the general solution to the given differential equation is y=±1
√2x.
12. Question: Solve the following first-order differential equation:
dy
dx = 2xy2
Ans. Step-by-step solution:
1. We start by separating variables. Rearranging the equation, we have:
dy
y2= 2xdx
2. Next, we integrate both sides with respect to their respective variables. Integrating the
left side gives: Z1
y2dy =Z2xdx
−1
y=x2+C
where C is the constant of integration.
3. Solving for y, we get:
y=−1
x2+C
Therefore, the solution to the given differential equation is y=−1
x2+C, where C is a constant
of integration.
13. Question: Solve the differential equation dy
dx =2x+3y−4
3x−2y+5 .
Ans. Solution: 1. Rewrite the differential equation in a more standard form by multiplying
both sides by dx:
dy
dx =2x+ 3y−4
3x−2y+ 5
2. Rearrange terms to get all terms involving yon the left side and all terms involving xon
the right side:
2x+ 3y−4 = (3x−2y+ 5) dy
dx
3. Multiply through by dx and rearrange to separate xand yterms:
2x+ 3y−4 = (3x−2y+ 5)dy
4. Expand the terms on the right side:
2x+ 3y−4 = 3xdy −2ydy + 5dy
5. Rearrange to collect the terms with dy on one side:
3xdy −2ydy −5dy = 2x−3y+ 4
6. Factor out dy on the left side:
(3x−2y−5)dy = 2x−3y+ 4
7. Integrate both sides with respect to y:
Z(3x−2y−5)dy =Z(2x−3y+ 4)dx
8. This simplifies to:
3xy −y2−5y= 2xy −3
2y2+ 4x+C
9. Rearrange the terms to get an equation involving only y:
−3
2y2−y+ 4x−3xy + 2xy −5y=C
10. Simplify further to get:
−3
2y2−6y+ 4x=C
11. This is the general solution to the differential equation.
14. Question: Solve the following differential equation:
y′′ −4y′+ 4y= 4e2x
Ans. Step-by-step solution:
1. First, we find the complementary function by solving the homogeneous differential equation:
y′′ −4y′+ 4y= 0
The characteristic equation is r2−4r+ 4 = 0, which simplifies to (r−2)2= 0. This gives us a
repeated root r= 2. Therefore, the complementary function is:
yc=c1e2x+c2xe2x
2. Next, we find the particular integral for the non-homogeneous equation:
yp=Ae2x
We substitute ypback into the differential equation and solve for A:
4Ae2x−4(2Ae2x)+4Ae2x= 4e2x
4Ae2x−8Ae2x+ 4Ae2x= 4e2x
0 = 4e2x
This means that A= 1. So the particular integral is yp=e2x.
3. Combining the complementary function and the particular integral, we get the general
solution:
y=yc+yp=c1e2x+c2xe2x+e2x
Therefore, the general solution to the given differential equation is:
y=c1e2x+c2xe2x+e2x
15. Find the general solution to the differential equation:
y′′ −4y′+ 4y= 0
Ans. Solution: 1. First, we write the characteristic equation for the given differential equation:
r2−4r+ 4 = 0
2. Solve for rby factoring the quadratic equation:
(r−2)2= 0
r= 2
3. Since the characteristic equation has a repeated root, the general solution to the differential
equation is:
y(t) = c1e2t+c2te2t
where c1, c2are arbitrary constants.
16. Question 16: Solve the following initial value problem:
y′+ 2y=sin(2x), y(0) = 1
Ans. Step 1. First, find the integrating factor µ(x)by solving the differential equation
y′+ 2y= 0: Given y′+ 2y=sin(2x), the integrating factor is given by:
µ(x) = e∫2dx =e2x
Step 2. Multiply the given differential equation by the integrating factor µ(x):
e2xy′+ 2e2xy=e2xsin(2x)
Step 3. Rewrite the left-hand side as the derivative of the product y(x)e2x:
(ye2x)′=e2xsin(2x)
Step 4. Integrate both sides with respect to x:
Z(ye2x)′dx =Ze2xsin(2x)dx
ye2x=−1
2e2xcos(2x) + C
Step 5. Solve for yby dividing both sides by e2x:
y=−1
2cos(2x) + Ce−2x
Step 6. Apply the initial condition y(0) = 1 to find the value of C:
1 = −1
2cos(0) + Ce0
1 = −1
2+C
C=3
2
Step 7. Substitute the value of Cback into the solution to find the final answer:
y=−1
2cos(2x) + 3
2e−2x
Therefore, the solution to the initial value problem is y=−1
2cos(2x) + 3
2e−2x.
17. Question: Solve the initial value problem: dy
dx =2x+3y−1
3x+2y+2 ,y(0) = 1.
Ans. Step-by-step solution: 1. Rearrange the given differential equation to the form dy
dx =
A(x) + B(y), where A(x)depends only on xand B(y)depends only on y.
dy
dx =2x+ 3y−1
3x+ 2y+ 2 ⇒(3x+ 2y+ 2) dy = (2x+ 3y−1) dx
3x dy + 2y dy + 2 dy = 2x dx + 3y dx −dx
3x dy −3y dx =−2x dx −2dy
Hence, we have
3x dy + 2 dy −3y dx + 2x dx = 0
2dy + 3x dy −3y dx + 2x dx = 0
2. We notice that the equation is exact as
∂M
∂y = 2 = ∂N
∂x = 2
Therefore, we can write the equation as
2dy + 3 dx −3dx + 2 dx = 0
d(2y+ 3x) = d(c)
2y+ 3x=c
where cis the constant of integration. 3. To find the value of c, we use the initial condition
y(0) = 1.
2(1) + 3(0) = c
c= 2
Thus, the solution to the initial value problem is 2y+ 3x= 2.
18. Question: Solve the following differential equation: y′′ −5y′+ 6y= 0.
Ans. Step-by-step solution: 1. Write down the characteristic equation: r2−5r+ 6 = 0. 2.
Solve the characteristic equation: (r−2)(r−3) = 0, so r= 2,3. 3. The general solution is
then y(t) = c1e2t+c2e3t, where c1and c2are arbitrary constants. Thus, the solution to the
differential equation is y(t) = c1e2t+c2e3t, where c1and c2are constants determined by initial
conditions.
19. Question: Solve the following initial value problem:
dy
dx =2x
1 + y2, y(0) = 1
Ans. Step-by-step solution: 1. Rewrite the given differential equation as:
(1 + y2)dy = 2xdx
2. Integrate both sides of the equation:
Z(1 + y2)dy =Z2xdx
3. Integrating the left side gives:
y+y3
3=x2+C
where C is the constant of integration.
4. Apply the initial condition y(0) = 1 to find the value of C:
1 + 1
3= 0 + C
C=4
3
5. Substitute the value of C back into the equation:
y+y3
3=x2+4
3
6. The solution to the initial value problem is:
y(x) = √3x2+ 4 −1
20. Question: Solve the differential equation dy
dx =2x+1
2y.
Ans. Step-by-step solution: 1. Rewrite the differential equation in the form 2y dy = (2x+1) dx.
2. Integrate both sides to solve for y:
Z2y dy =Z(2x+ 1) dx
3. This gives us y2=x2+x+C, where Cis the constant of integration. 4. Therefore, the
general solution to the differential equation is y=±√x2+x+C.
21. Question: Solve the initial value problem y′+y=sin(x)with initial condition y(0) = 2.
Ans. Step-by-step solution:
1. First, we write the differential equation in standard form, y′+y=sin(x), where y′=dy
dx .
This is a first-order linear non-homogeneous ordinary differential equation.
2. To solve the differential equation, we first find the integrating factor I(x), given by
I(x) = e∫1dx =ex.
3. We then multiply both sides of the differential equation by the integrating factor:
ex(y′+y) = exsin(x)
4. The left side can be rewritten using the product rule:
d
dx(exy) = exsin(x)
5. Integrating both sides with respect to x, we get:
exy=−cos(x) + C
, where Cis an arbitrary constant of integration.
6. Now, we divide both sides by exto solve for y:
y=−e−xcos(x) + Ce−x
7. Finally, we use the initial condition y(0) = 2 to find the value of C:
2 = −e0cos(0) + Ce0=−1 + C
C= 3
8. Therefore, the solution to the initial value problem is:
y=−e−xcos(x)+3e−x
22. Question: Solve the following differential equation:
dy
dx + 2y= 3e−2x
Ans. Step-by-step solution: 1. First, we identify the differential equation as a first-order linear
differential equation in the standard form:
dy
dx +P(x)y=Q(x)
where P(x) = 2 and Q(x) = 3e−2x.
2. To solve this linear differential equation, we begin by finding the integrating factor µ(x),
defined by:
µ(x) = e∫P(x)dx
3. In this case, P(x) = 2, so we have:
ZP(x)dx =Z2dx = 2x
µ(x) = e2x
4. Next, we multiply both sides of the differential equation by the integrating factor µ(x):
e2xdy
dx + 2e2xy= 3e0
5. Recognizing that the left side is the derivative of the product (e2xy)with respect to x, we
can rewrite the equation as:
d
dx e2xy= 3
6. Integrating both sides with respect to x, we get:
e2xy= 3x+C
where Cis the constant of integration.
7. Solving for y, we have:
y=3x+C
e2x
y=3x
e2x+C
e2x
Therefore, the general solution to the differential equation is:
y=3x
e2x+C
e2x
where Cis an arbitrary constant.
23. Question:
Solve the following differential equation:
dy
dx =2x+y
x+ 2y
Ans. Step-by-step solution:
1. Rewrite the given differential equation in the form dy
dx =f(x, y):
dy
dx =2x+y
x+ 2y
2. Notice that the given differential equation is not exact. To check for exactness, calculate
the partial derivatives ∂f
∂y and ∂f
∂x :
∂f
∂y =1
x+ 2y−(2x+y)(2)
(x+ 2y)2
∂f
∂x =2
x+ 2y−(x+ 2y)(2)
(x+ 2y)2
3. Determine if the given equation is exact by verifying if ∂f
∂y =∂f
∂x . In this case, we see that
they are not equal.
4. To find an integrating factor µ(x, y)such that µ(x, y)Mdx +µ(x, y)Ndy = 0 is exact, we
set the left-hand side equation of the given differential equation as the total derivative of some
function U(x, y)with respect to x.
5. Let’s assume U(x, y) = µ(x, y)M(x, y)dx +N(x, y)dy, whereM = 2x+yandN =
x+2y.T hen, find∂U ∂y and ∂U
∂x :
∂U
∂y =µ(x, y)N+µ(x, y)∂N
∂y
∂U
∂x =µ(x, y)M+µ(x, y)∂M
∂x
6. Equate ∂U
∂y to ∂U
∂x and solve for µ(x, y)to find the integrating factor.
7. After finding the integrating factor µ(x, y), rewrite the differential equation in the form
d
dx (µ(x, y)y) = µ(x, y)M−∂µ
∂y .
8. Finally, integrate the above expression to obtain the general solution to the differential
equation.
24. Question: Solve the following initial value problem:
y′=2y
x+x2, y(1) = 0
Ans. Step-by-step solution:
1. The given differential equation is a first-order linear ordinary differential equation in the
form of y′+P(x)y=Q(x). To solve this differential equation, we will use the integrating factor
method. First, we identify P(x) = 2
xand Q(x) = x2.
2. The integrating factor, denoted by µ(x), is given by µ(x) = e∫P(x)dx. Therefore, in this
case, µ(x) = e∫2
xdx =e2ln x=x2.
3. Multiply the differential equation by the integrating factor µ(x):
x2y′−2xy =x4
4. Rewrite the equation using the product rule for differentiation:
(x2y)′=x4
5. Integrate both sides with respect to x:
Z(x2y)′dx =Zx4dx
x2y=1
5x5+C
6. Solve for y:
y=1
5x3+C
x2
7. Apply the initial condition y(1) = 0 to find the value of the constant C:
0 = 1
5(1)3+C
(1)2
0 = 1
5+C
C=−1
5
8. Substitute the value of Cback into the general solution to obtain the particular solution:
y=1
5x3−1
5x2
Therefore, the solution to the initial value problem is y=1
5x3−1
5x2.
25. Question: Solve the following initial value problem:
dy
dx =y(1 −x), y(0) = 2
Ans. Solution: 1. We first write the given differential equation in the standard form for solving:
dy
dx =y(1 −x)=⇒dy
y= (1 −x)dx
2. Next, we integrate both sides of the equation:
Z1
ydy =Z(1 −x)dx
3. Integrating both sides gives us:
ln |y|=x−x2
2+C
where Cis the constant of integration.
4. Exponentiating both sides, we get:
|y|=ex−x2
2+C
5. Since the initial condition is y(0) = 2, we substitute x= 0 and y= 2 into our equation
to find the value of the constant C:
|2|=e0−02
2+C=⇒2 = eC
6. Therefore, C=ln 2, and the solution to the differential equation is:
y=±2ex−x2
2+ln 2
7. Simplifying, we get:
y= 2ex−x2
2+ln 2
Hence, the solution to the initial value problem is y= 2ex−x2
2+ln 2.
26. Question: Solve the differential equation y′′ −2y′+y= 2et.
Ans. Step-by-step solution:
1. First, we solve the homogeneous part of the differential equation by setting y′′−2y′+y= 0.
The characteristic equation is r2−2r+1 = (r−1)2= 0. This gives us a repeated root r1=r2= 1.
So, the general solution to the homogeneous equation is yh(t) = c1et+c2tet, where c1and c2
are constants.
2. Next, we find a particular solution to the non-homogeneous part of the differential equation,
2et. Since the right-hand side is of the form Cet, a good guess for a particular solution would
be of the form yp(t) = Aet, where Ais a constant to be determined. Substitute yp(t)and its
derivatives into the differential equation to find A. We get: 2Aet−2Aet+Aet= 2etAet= 2et
A= 2
So, a particular solution is yp(t) = 2et.
3. The general solution to the entire differential equation is the sum of the homogeneous
and particular solutions: y(t) = yh(t) + yp(t) = c1et+c2tet+ 2et, where c1and c2are arbitrary
constants.
27. Question 27: Solve the following first-order linear differential equation:
dy
dx + 2xy =ex2
Ans. Solution: 1. We can solve this first-order linear differential equation using the integrating
factor method. First, we need to rewrite the equation in the standard form:
dy
dx + 2xy =ex2
2. The standard form is: dy
dx +P(x)y=Q(x), where P(x) = 2xand Q(x) = ex2.
3. Next, we find the integrating factor I(x), which is given by I(x) = e∫P(x)dx.
4. In this case, P(x) = 2x, so RP(x)dx =R2xdx =x2.
5. Therefore, the integrating factor I(x) = ex2.
6. Multiply the original differential equation by the integrating factor:
ex2dy
dx + 2xex2y=e2x2
7. The left side now becomes the derivative of the product of yand the integrating factor:
d
dx(ex2y) = e2x2
8. Integrate both sides with respect to xto solve for y:
Zd
dx(ex2y)dx =Ze2x2dx
9. By integrating the left side, we get ex2y=1
2e2x2+C, where Cis the constant of
integration.
10. Finally, we solve for yby dividing by ex2:
y=1
2ex2+Ce−x2
Hence, the solution to the given differential equation is y=1
2ex2+Ce−x2, where Cis an
arbitrary constant.
28. Question 28: Solve the initial value problem:
dy
dx = (x2+y2)e−y, y(0) = 1
Ans. Solution: 1. First, rewrite the differential equation as:
dy
dx = (x2+y2)e−y
2. Notice that this is a separable differential equation. We can write it as:
dy
dx = (x2+y2)e−y=⇒eydy = (x2+y2)dx
3. Integrate both sides with respect to their respective variables:
Zeydy =Z(x2+y2)dx
4. This gives us:
ey=x3
3+y3
3+C
where C is the constant of integration.
5. To determine the value of the constant C, we use the initial condition y(0) = 1:
e1=03
3+13
3+C=⇒e=1
3+C=⇒C=e−1
3
6. Therefore, the particular solution to the initial value problem is:
ey=x3
3+y3
3+e−1
3
29. Question 29: Solve the following differential equation using the method of undetermined
coefficients: y′′ −3y′+ 2y= 4ex+ 3 sin(x).
Ans. Solution: 1. First, find the complementary solution ycby solving the characteristic
equation r2−3r+ 2 = 0. The characteristic roots are r1= 1 and r2= 2, so the complementary
solution is yc=c1ex+c2e2x.
2. Next, find the particular solution yp. Since the right-hand side of the equation includes both
4exand sin(x), assume yp=Aex+Bsin(x). Then, calculate the derivatives: y′
p=Aex+Bcos(x)
and y′′
p=Aex−Bsin(x).
3. Substitute yp,y′
p, and y′′
pinto the differential equation to get: (Aex−Bsin(x)) −3(Aex+
Bcos(x)) + 2(Aex+Bsin(x)) = 4ex+ 3 sin(x).
4. Simplify the equation and group like terms to solve for Aand B. You should find A= 2
and B=−1.
5. Therefore, the particular solution is yp= 2ex−sin(x).
6. The general solution is the sum of the complementary and particular solutions: y=
yc+yp=c1ex+c2e2x+ 2ex−sin(x).
30. Question: Solve the following differential equation:
dy
dx =2x2y+y2
2xy +x2.
Ans. Step-by-step solution: 1. Rewrite the given differential equation in terms of differentials:
dy
dx =2x2y+y2
2xy +x2.
2. Multiply both sides by dx to separate variables:
dy
2x2y+y2=dx
2xy +x2.
3. Perform partial fraction decomposition on the left side:
dy
y(2x+y)=dx
x(2y+x).
4. Decompose the left side into partial fractions:
1
y(2x+y)=A
y+B
2x+y.
5. Clear the denominators and solve for A and B:
1 = A(2x+y) + By.
6. Setting y = 0 gives A=1
2x, setting x = -y/2 gives B=−1
4y.
7. Substitute the values of A and B back into the partial fractions decomposition:
1
y(2x+y)=1
2x·y−1
4y·(2x+y).
8. Similarly, decompose the right side into partial fractions:
1
x(2y+x)=C
x+D
2y+x.
9. Clear the denominators and solve for C and D:
1 = C(2y+x) + Dx.
10. Setting x = 0 gives C=1
2y, setting y = -x/2 gives D=−1
4x.
11. Substitute the values of C and D back into the partial fractions decomposition:
1
x(2y+x)=1
2y·x−1
4x·(2y+x).
12. Now, integrate both sides:
Z1
y(2x+y)dy =Z1
x(2y+x)dx.
13. Integrating the left side gives 1
2ln |y| − 1
2ln |2y+x|=ln |x|+1
2ln |2y+x|+C.
14. Rearrange the equation and simplify it to obtain the solution to the differential equation:
ln |y| − ln |2y+x|= 2 ln |x|+ln |2y+x|+ 2C.
15. Combine the logarithmic terms to get:
ln
y
2y+x
=ln |x|2+C.
16. Exponentiate both sides to solve for y:
y=cx2
1−c,
where c=±eC.
31. Question 31: Solve the initial value problem dy
dx =y3
x2,y(1) = −1.
Ans. Solution: 1. First, notice that the given differential equation is separable, so we can
rewrite it as dy
y3=dx
x2.
2. Integrate both sides with respect to their variables:
Z1
y3dy =Z1
x2dx
3. This gives us −1
2y2=−1
x+C, where Cis the constant of integration.
4. Now, we need to find the value of the constant C. Substituting the initial condition
y(1) = −1into the equation, we get
−1
2(−1)2=−1
1+C
1
2=−1 + C
C=1
2−1
C=−1
2
5. Substituting C=−1
2back into our equation, we have
−1
2y2=−1
x−1
2
6. Solving for y, we get 1
2y2=1
x+1
2
2y2=1
1
x+1
2
y2=1
2
x+ 1
y=±s1
2
x+ 1
7. Since y(1) = −1, we have y=−q1
2
1+1 =−1, which satisfies the initial condition.
So, the solution to the initial value problem is y=−q1
2
x+1 .
32. Question 32: Solve the following differential equation using an integrating factor: dy
dx +2y
x=
x2.
Ans. Solution: 1. The given differential equation is of the form dy
dx +P(x)y=Q(x), where
P(x) = 2
xand Q(x) = x2. 2. First, we need to find the integrating factor. The integrating factor,
denoted by µ(x), is given by µ(x) = e∫P(x)dx. 3. In this case, P(x) = 2
x, so the integrating
factor becomes µ(x) = e∫2
xdx. 4. Integrating 2
xwith respect to xgives us ln |x|2= 2 ln |x|.
5. Therefore, the integrating factor is µ(x) = e2ln |x|=eln |x|2=x2. 6. Multiply both sides of
the original differential equation by the integrating factor x2:x2dy
dx + 2xy =x4. 7. Notice that
the left side of the equation is now the result of applying the product rule to x2y. So, we can
rewrite the equation as d
dx (x2y) = x4. 8. Integrate both sides with respect to xto solve for x2y:
Rd
dx (x2y)dx =Rx4dx. 9. This yields x2y=1
5x5+C, where Cis the constant of integration.
10. Finally, solving for y, we get y=1
5x3+C
x2.
33. Question: Solve the following differential equation using the method of integrating factors:
dy
dx −2y= 3x.
Ans. Step-by-step solution: 1. The given differential equation can be written in the form
dy
dx +P(x)y=Q(x), where P(x) = −2and Q(x) = 3x. 2. To solve this differential equation,
we first find the integrating factor (IF ) by integrating e∫P(x)dx.
Find IF :IF =e∫−2dx =e−2x
3. Multiply both sides of the differential equation by the integrating factor:
e−2xdy
dx −2e−2xy= 3xe−2x
4. Rewrite the left side as the derivative of the product of yand the integrating factor:
d
dx ye−2x= 3xe−2x
5. Integrate both sides with respect to x:
Zd
dx ye−2xdx =Z3xe−2xdx
ye−2x=−3
2xe−2x−3
4e−2x+Cwhere Cis the constant of integration
6. Solve for yby multiplying through by e2xand simplifying:
y=−3
2x−3
4+Ce2x
Therefore, the solution to the differential equation is y=−3
2x−3
4+Ce2x, where Cis the
constant of integration.
34. Question: Find the particular solution of the differential equation dy
dx = 4x+1
ythat passes
through the point (1, 2).
Ans. Step-by-step solution: 1. Rearrange the differential equation:
dy
dx = 4x+1
y
ydy
dx = 4xy + 1
2. Integrate both sides with respect to x:
Zy dy =Z(4xy + 1) dx
y2
2= 2x2y+x+C
where Cis the constant of integration.
3. Plug in the coordinates of the point (1, 2) to find the particular solution:
22
2= 2(1)2(2) + 1 + C
2 = 4 + 1 + C
C=−3
4. Substitute C=−3back into the general solution:
y2
2= 2x2y+x−3
5. Simplify the particular solution:
y2= 4x2y+ 2x−6
y2−4x2y= 2x−6
y2−4x2y= 2(x−3)
y(y−4x2) = 2(x−3)
y=2(x−3)
y−4x2
Therefore, the particular solution passing through the point (1, 2) is y=2(x−3)
y−4x2.
35. Question: Solve the following differential equation:
dy
dx −2y= 4e2x
Ans. Solution: 1. First, we identify the integrating factor. The integrating factor is given by:
IF =e∫−2dx =e−2x
2. Multiply both sides of the differential equation by the integrating factor:
e−2xdy
dx −2e−2xy= 4e−2xe2x
e−2xdy
dx −2e−2xy= 4
3. Rewrite the left side as a derivative applying the product rule:
d
dx(e−2xy) = 4
4. Integrate both sides with respect to x:
Zd
dx(e−2xy)dx =Z4dx
e−2xy= 4x+C
5. Solve for y:
y= 4xe2x+Ce2x
where C is the constant of integration.
