Diff equations
Zenkii
HW_4-solve.pdf
HW#4 - Solutions
Section 1.7
3. dy dt
= y2 −ay + 1
Let fa(y) = y 2 −ay + 1. We first need to find equilibrium points, so we need to solve
fa(y) = y 2 −ay + 1 = 0 (1)
which has solutions
y = a± √ a2 − 4 2
. (2)
If a < 2, there are no real solutions. If a > 2, there are 2 real solutions , and if a = 2, there is only 1 real solution, meaning there is only one equilibrium solution. Thus a = 2 is the bifurcation value. We need to sketch a phase line for a = 2, and slightly below the bifurcation value (I’ll choose a = 1) and slightly above the bifurcation value (I’ll choose a=3).
For a = 1, there is no equilibrium solution. For a = 1, the equilibrium solution is y = 1. For a = 3, the equilibria are 3+ √ 5
2 and 3−
√ 5
2 .
Noting that the arrows on the phase lines in between the equilibra corresponds to when fa(y) is positive (hence y is in- creasing and the arrow points upward) and when fa(y) is negative (hence y is decreasing and the arrow points downward).
4. dy dt
= y3 + αy2
Let fα = y 3 + αy2. First we need to find the equilibra by solving
y3 + αy2 = 0 (3)
y2(y + α) = 0 (4)
which has solutions y = 0 and y = −α. There are two equilibria if α 6= 0, and one equilibria if α = 0. Hence α = 0 is the bifurcation value.
1
From the graphs of fα(y) for different α, we note when fα(y) is positive or negative, and add arrows between the equilibra on the phase lines to match.
Section 1.8
7. The associated homogeneous equation is dy/dt + 2y = 0 or equivalently, dy/dt = −2y and its general solution (i.e., the complementary solution) is yh(t) = ke
−2t, where k is an arbitrary constant.
For a particular solution to the original nonhomogeneous equation
dy
dt + 2y = et/3, (5)
we guess yp(t) = αe t/3. Substituting this yp into (1) and simplifying gives
d
dt (αet/3) + 2(αet/3) = et/3
1
3 αet/3 + 2αet/3 = et/3
7
3 αet/3 = 1et/3
Comparing the coefficient of et/3 on both sides of the last equation, we must have
7
3 α = 1.
Therefore, α = 3/7 and yp(t) = 3 7 et/3. Hence by the Extended Linearity Principle, the general solution to the original
nonhomogeneous equation is the sum of yh and yp, that is,
y(t) = ke−2t + 3
7 et/3.
Finally, we use the given initial condition y(0) = 1 to determine the constant k. To this end,
y(0) = 1
k e−2(0)︸ ︷︷ ︸ = 1
+ 3
7 e(0)/3︸ ︷︷ ︸ = 1
= 1
k + 3
7 = 1
k = 4
7 .
Putting all these together, the solution to the given IVP is
y(t) = 4
7 e−2t +
3
7 et/3 .
2
8. The associated homogeneous equation is dy/dt− 2y = 0 or dy/dt = 2y and its general solution is yh(t) = ke2t.
For a particular solution yp(t) to the original nonhomogeneous equation
dy
dt − 2y = 3e−2t, (6)
we guess yp(t) = αe −2t. Substituting this yp into (2) and simplifying gives
d
dt (αe−2t) − 2(αe−2t) = 3e−2t
−2αe−2t − 2αe−2t = 3e−2t
−4αe−2t = 3e−2t
Comparing the coefficient of e−2t on both sides of the last equation, we must have
−4α = 3.
Therefore, α = −3/4 and yp(t) = −34e −2t. Hence by the Extended Linearity Principle, the general solution to the
original nonhomogeneous equation is the sum of yh and yp, that is,
y(t) = ke2t − 3
4 e−2t.
Finally, we use the given initial condition y(0) = 10 to determine the constant k. To this end,
y(0) = 10
k e2(0)︸︷︷︸ = 1
− 3
4 e−2(0)︸ ︷︷ ︸
= 1
= 10
k − 3
4 = 10
k = 43
4 .
Putting all these together, the solution to the given IVP is
y(t) = 43
4 e2t −
3
4 e−2t .
10. The associated homogeneous equation is dy/dt + 3y = 0 or dy/dt = −3y and its general solution is yh(t) = ke−3t.
For a particular solution yp(t) to the original nonhomogeneous equation
dy
dt + 3y = cos(2t), (7)
we guess yp(t) = α cos(2t) + β sin(2t), where both α and β are undetermined coefficients. (Note: You may also complexify the problem to find a particular solution. I will leave this up to you! In any case, you should be able to obtain the same result.) Substituting this yp into (3) and collecting the terms involving ‘cos(2t)’ and ‘sin(2t)’, we get
d
dt
( α cos(2t) + β sin(2t)
) + 3 ( α cos(2t) + β sin(2t)
) = cos(2t)
−2α sin(2t) + 2β cos(2t) + 3α cos(2t) + 3β sin(2t) = cos(2t)
(3α + 2β) cos(2t) + (−2α + 3β) sin(2t) = cos(2t)
Notice that the right-hand side of the last equation can be written as ‘1 cos(2t) + 0 sin(2t)’. (Since we are missing the sine term sin(2t), its coefficient must be 0. Obviously, the coefficient of cos(2t) is 1!) Hence we have
(3α + 2β) cos(2t) + (−2α + 3β) sin(2t) = 1 cos(2t) + 0 sin(2t). (8)
3
Comparing the coefficients of both cos(2t) and sin(2t) on both sides of (4), we must have{ 3α + 2β = 1 −2α + 3β = 0
Solving for α and β simultaneously, we obtain α = 3/13 and β = 2/13 (Check this on your own!) and therefore, yp(t) =
3 13
cos(2t) + 2 13
sin(2t). Hence by the Extended Linearity Principle, the general solution to the original nonhomogeneous equation is the sum of yh and yp, that is,
y(t) = ke−3t + 3
13 cos(2t) +
2
13 sin(2t).
Finally, we use the given initial condition y(0) = −1 to determine the constant k. To this end,
y(0) = −1
k e−2(0)︸ ︷︷ ︸ = 1
+ 3
13 cos(2 · 0)︸ ︷︷ ︸
= 1
+ 2
13 sin(2 · 0)︸ ︷︷ ︸
= 0
= −1
k + 3
13 = −1
k = − 16
13 .
Putting all these together, the solution to the given IVP is
y(t) = − 16
13 e−3t +
3
13 cos(2t) +
2
13 sin(2t) .
12. The associated homogeneous equation is the same as in Exercise 8 and thus the general solution is yh(t) = ke 2t.
For a particular solution yp(t) to the original nonhomogeneous equation
dy
dt − 2y = 7e2t, (9)
we guess yp(t) = αte 2t. (We need an extra factor of ‘t’ for yp.) Substituting this yp into (5) and simplifying gives
d
dt (αte2t)︸ ︷︷ ︸
ProductRule
− 2(αte2t) = 7e2t
d
dt (αt) ·e2t + αt ·
d
dt (e2t) − 2αte2t = 7e2t
α ·e2t + αt · 2e2t − 2αte2t = 7e2t
αe2t + 2αte2t − 2αte2t = 7e2t
αe2t = 7e2t
Comparing the coefficient of e2t on both sides of the last equation, we must have
α = 7.
Therefore, yp(t) = 7te 2t. Hence by the Extended Linearity Principle, the general solution to the original nonhomo-
geneous equation is the sum of yh and yp, that is,
y(t) = ke2t + 7te2t.
Finally, we use the given initial condition y(0) = 3 to determine the constant k. To this end,
y(0) = 3
k e2(0)︸︷︷︸ = 1
+ 7(0)e2(0)︸ ︷︷ ︸ = 0
= 3
k = 3.
Putting all these together, the solution to the given IVP is
y(t) = 3e2t + 7te2t .
4
20. Substituting the suggested particular solution yp(t) = at 2 + bt + c into the given linear ODE and simplifying yields
d
dt (at2 + bt + c) + 2(at2 + bt + c) = 3t2 + 2t− 1
2at + b + 2at2 + 2bt + 2c = 3t2 + 2t− 1 2at2 + (2a + 2b)t + (b + 2c) = 3t2 + 2t−1.
Comparing the coefficients of t2, t, and the constant terms on both sides of the last equation, we must have
2a = 3
2a + 2b = 2
b + 2c = −1.
Solving the first equation for a gives a = 3/2. Substituting this into the second equation and solving for b, we get
2
( 3
2
) + 2b = 2
3 + 2b = 2
2b = −1
b = − 1
2 .
Finally, substituting the last result (i.e., b = −1/2) into the third equation and solving for c yields
− 1
2 + 2c = −1
2c = − 1
2
c = − 1
4 .
Therefore, in order for yp(t) = at 2 + bt + c to be a particular solution to the given linear ODE, we must have
a = 3
2 , b = −
1
2 , c = −
1
4
5
dy
dt +
1
1 + t y = t2.
µ(t) = e ∫ g(t) dt
= e ∫
1 1+t
dt
= eln |1+t|
= 1 + t.
Therefore, the general solution is (justify each step)
y(t) = 1
µ(t)
[∫ µ(t)b(t) dt
] =
1
1 + t
[∫ (1 + t)t2 dt
] =
1
1 + t
[∫ (t2 + t3) dt
] =
1
1 + t
[ t3
3 + t4
4 + C
] or
t3
3(1 + t) +
t4
4(1 + t) +
C
1 + t .
6
5. Notice that the original equation IS in the required form so we don’t need to rewrite it at all! Observe that g(t) = − 2t 1+t2
and b(t) = 3. First we compute the integrating factor µ(t):
µ(t) = e ∫ g(t) dt
= e ∫ − 2t
1+t2 dt .
Observe that the integral in the exponent (in blue) can be evaluated by substitution with u = 1 + t2, yielding ln |1 + t2|−1 (check this on your own). Therefore,
µ(t) = e ∫ − 2t
1+t2 dt
= eln |1+t 2|−1
= (1 + t2)−1.
Therefore, the general solution is (justify each step)
y(t) = 1
µ(t)
[∫ µ(t)b(t) dt
] =
1
(1 + t2)−1
[∫ (1 + t2)−1 · 3 dt
] = (1 + t2)
[ 3
∫ 1
1 + t2 dt
] = (1 + t2)
[ 3tan−1(t) + C
] .
9. Notice that the original equation is not quite in the desired form so we are going to first rewrite it by adding the term y t
to both sides, which gives us dy
dt + y
t = 2
or equivalently, dy
dt +
1
t y = 2
Observe that g(t) = 1 t
and b(t) = 2. The integrating factor µ(t) is
µ(t) = e ∫ g(t) dt
= e ∫
1 t dt
= eln |t|
= t.
Therefore, the general solution is (justify each step)
y(t) = 1
µ(t)
[∫ µ(t)b(t) dt
] =
1
t
[∫ t · 2 dt
] =
1
t
[ 2
∫ tdt
] =
1
t
[ 2 ·
t2
2 + C
] =
1
t
[ t2 + C
] = t +
C
t .
7
Now we use the given IC y(1) = 3 to determine the constant C. To this end,
y(1) = 3
(1) + C
(1) = 3
1 + C = 3
C = 2.
Therefore, the solution to the given IVP is
y(t) = t + 2
t .
10. Notice that the original equation is not in the required form so we are going to first rewrite it by adding the term 2ty to both sides, yielding
dy
dt + 2ty = 4e−t
2
.
Observe that g(t) = 2t and b(t) = 4e−t 2
. The integrating factor µ(t) is
µ(t) = e ∫ g(t) dt
= e ∫ 2tdt
= et 2
.
Therefore, the general solution is (justify each step)
y(t) = 1
µ(t)
[∫ µ(t)b(t) dt
] =
1
et 2
[∫ et
2
· 4e−t 2
dt
] = e−t
2
[ 4
∫ dt
] = e−t
2
[4t + C]
= 4te−t 2
+ Ce−t 2
.
Using the given IC y(0) = 3, we can determine the constant C. To this end,
y(0) = 3
4(0)e−(0) 2︸ ︷︷ ︸
= 0
+C e−(0) 2︸ ︷︷ ︸
= 1
= 3
C = 3
Putting all these together, the solution to the given IVP is
y(t) = 4te−t 2
+ 3e−t 2
.
12. Notice that the original equation IS already in the desired form so there is nothing we need to do to rewrite it. Observe that g(t) = −3
t and b(t) = 2t3e2t. The integrating factor µ(t) is
µ(t) = e ∫ g(t) dt
= e ∫ −3
t dt
= e−3 ∫
1 t dt
= e−3 ln |t|
= eln |t| −3
= t−3.
8
Therefore, the general solution is (justify each step)
y(t) = 1
µ(t)
[∫ µ(t)b(t) dt
] =
1
t−3
[∫ t−3 · 2t3e2t dt
] = t3
[ 2
∫ e2t dt
] = t3
[ 2 ·
1
2 e2t + C
] = t3
[ e2t + C
] = t3e2t + Ct3.
Finally, we use the given IC y(1) = 0 to determine the constant C. To this end,
y(1) = 0
(1)3e2(1) + C(1)3 = 0
e2 + C = 0
C = −e2
Putting all these together, the solution to the given IVP is
y(t) = t3e2t −e2t3 .
21. (a) Notice that the given equation is already in the desired form. Observe that g(t) = 0.4 and b(t) = 3 cos(2t). Computing the integrating factor µ(t), we obtain:
µ(t) = e ∫ g(t) dt
= e ∫ 0.4 dt
= e0.4t.
Therefore, the general solution is (justify each step)
v(t) = 1
µ(t)
[∫ µ(t)b(t) dt + C
] =
1
e0.4t
[∫ e0.4t · 3 cos(2t) dt + C
] = e−0.4t
[ 3
∫ e0.4t cos(2t) dt + C
] .
Observe that the cyan-colored integral needs to be integrated by parts. (It’s going to be a bit messy so I’m going to do separate calculations for this integral below.)
To this end, let u = cos(2t) and dv = e0.4t dt. Then du = −2 sin(2t) dt and v = 1 0.4 e0.4t = 5
2 e0.4t. Using the
integration by parts formula,∫ e0.4t cos(2t) dt = cos(2t)︸ ︷︷ ︸
u
· 5
2 e0.4t︸ ︷︷ ︸ v
− ∫
5
2 e0.4t︸ ︷︷ ︸ v
·−2 sin(2t) dt︸ ︷︷ ︸ du
= 5
2 e0.4t cos(2t) + 5
∫ e0.4t sin(2t) dt.
9
The underlined integral at the bottom of the previous page, unfortunately, needs to be integrated by parts again... This time, let u = sin(2t) and dv = e0.4t dt. Then du = 2 cos(2t) dt and v = 1
0.4 e0.4t = 5
2 e0.4t. So∫
e0.4t cos(2t) dt = 5
2 e0.4t cos(2t) + 5
∫ e0.4t sin(2t) dt
= 5
2 e0.4t cos(2t) + 5
sin(2t)︸ ︷︷ ︸
u
· 5
2 e0.4t︸ ︷︷ ︸ v
− ∫
5
2 e0.4t︸ ︷︷ ︸ v
·2 cos(2t) dt︸ ︷︷ ︸ du
= 5
2 e0.4t cos(2t) + 5
[ 5
2 e0.4t sin(2t) − 5
∫ e0.4t cos(2t) dt
] =
5
2 e0.4t cos(2t) +
25
2 e0.4t sin(2t) − 25
∫ e0.4t cos(2t) dt.
In summary, we have∫ e0.4t cos(2t) dt =
5
2 e0.4t cos(2t) +
25
2 e0.4t sin(2t) − 25
∫ e0.4t cos(2t) dt. (10)
Since integrating by parts twice has produced a similar integral to the original one (off by a constant), there is no point of integrating by parts anymore. Instead, we are going to treat (6) as an algebraic equation for the integral we are looking for and try to get it isolated:∫
e0.4t cos(2t) dt = 5
2 e0.4t cos(2t) +
25
2 e0.4t sin(2t) − 25
∫ e0.4t cos(2t) dt
1
∫ e0.4t cos(2t) dt + 25
∫ e0.4t cos(2t) dt =
5
2 e0.4t cos(2t) +
25
2 e0.4t sin(2t)
26
∫ e0.4t cos(2t) dt =
5
2 e0.4t cos(2t) +
25
2 e0.4t sin(2t)∫
e0.4t cos(2t) dt = 1
26
[ 5
2 e0.4t cos(2t) +
25
2 e0.4t sin(2t)
] =
5
52 e0.4t cos(2t) +
25
52 e0.4t sin(2t).
Therefore, the general solution is
v(t) = e−0.4t [ 3
( 5
52 e0.4t cos(2t) +
25
52 e0.4t sin(2t)
) + C
] = e−0.4t
[ 15
52 e0.4t cos(2t) +
75
52 e0.4t sin(2t) + C
] =
15
52 cos(2t) +
75
52 sin(2t) + Ce−0.4t .
Don’t worry! There won’t be a problem like this on an exam.,
(b) The associated homogeneous equation is dv/dt+0.4v = 0 or equivalently dv/dt = −0.4v and its general solution is vh(t) = Ce
−0.4t (C, constant).
For a particular solution vp(t) to the original nonhomogeneous equation
dv
dt + 0.4v = 3 cos(2t), (11)
we guess vp(t) = α cos(2t) + β sin(2t), where both α and β are undetermined coefficients. Substituting this vp into the left-hand side of the original equation and collecting the terms involving ‘cos(2t)’ and ‘sin(2t)’, we get
d
dt
( α cos(2t) + β sin(2t)
) + 0.4
( α cos(2t) + β sin(2t)
) = 3 cos(2t)
−2α sin(2t) + 2β cos(2t) + 0.4α cos(2t) + 0.4β sin(2t) = 3 cos(2t)
(0.4α + 2β) cos(2t) + (−2α + 0.4β) sin(2t) = 3 cos(2t).
10
Notice that the right-hand side of the last equation can be written as ‘3 cos(2t) + 0 sin(2t)’. Hence we have
(0.4α + 2β) cos(2t) + (−2α + 0.4β) sin(2t) = 3 cos(2t) + 0 sin(2t) (12)
Comparing the coefficients of both cos(2t) and sin(2t) on both sides of (8), we must have{ 0.4α + 2β = 3 −2α + 0.4β = 0
Solving for α and β simultaneously, we obtain α = 15/52 and β = 75/52 (Check this on your own!) and therefore, vp(t) =
15 52
cos(2t) + 75 52
sin(2t). Hence by the Extended Linearity Principle, the general solution to the original nonhomogeneous equation is the sum of yh and yp, that is,
v(t) = Ce−0.4t + 15
52 cos(2t) +
75
52 sin(2t) ,
which is essentially the same result as in part (a) above!
11
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