HW 5 solutions

§2.2 pp. 149-150
Math 4430 HW 5
Selected Solutions
2. Find the general solution of the differential equation
Solution
4y ′′ − 12y ′ + 9y = 0.
We compute b 2 − 4ac = (−12) 2 − 4(4)(9) = 0. Then the characteristic equation has a double
root
r = −b/(2a) = 12/8 = 3/2.
Then the general solution is
y(t) = c1e 3t/2 + c2te 3t/2 .
4. Solve the IVP
Solution
4y ′′ − 4y ′ + y = 0; y(0) = 0, y ′ (0) = 3.
We compute b 2 − 4ac = (−4) 2 − 4(4)(1) = 0. Then the characteristic equation has a double
root
r = −b/(2a) = 4/8 = 1/2.
Then the general solution is
y(t) = c1e t/2 + c2te t/2 .
Plugging in our first initial condition we get
0 = c1e 0 + c2 · 0 · e 0 = c1.
So c1 = 0. To apply the second initial condition we need to compute y ′ (x). Since we know
c1 = 0, we will just differentiate y(x) = c2te t/2 to get
y ′ (x) = c2[e t/2 + (t/2)e t/2 ].
Now plugging in the second initial condition we get
Thus the solution to the IVP is
3 = c2[e 0 + 0 · e 0 ] = c2.
y(x) = 3te t/2 .

§2.3 pp. 152-153
2. Three solutions of a certain second-order linear nonhomogeneous equation are
ψ1(t) = 1 + e t2
ψ2(t) = 1 + te t2
ψ2(t) = (t + 1)e t2
+ 1.
Find the general solution to this equation.
Solution
We know that the general solution is of the form
y(x) = ψ + c1y1 + c2y2
where ψ is any solution to the nonhomogeneous equation (so we could take any of the
above solutions to be ψ) and where y1 and y2 are two linearly independent solutions to the
homogeneous equation.
We know that the difference between any two solutions of the nonhomogeneous equation is
a solution to the homogeneous equation, so we may take
and
y1 = ψ3 − ψ2 = e t2
y2 = ψ3 − ψ1 = te t2
are both solutions to the homogeneous equation. Clearly y1 and y2 are not constant multiples
of each other, so they are linearly independent. Taking ψ = ψ1, we get the general solution
or more simply
y(x) = 1 + e t2
+ c1e t2
+ c2te t2
y(x) = 1 + c1e t2
+ c2te t2
.

§2.4 pp. 156-157
2. Find the general solution of the equation
Solution In the homogeneous equation
y ′′ − 4y ′ + 4y = te 2t .
y ′′ − 4y ′ + 4y = 0
we have b 2 − 4ac = (−4) 2 − 4(1)(4) = 0 so the characteristic equation has the double root
r = −b/(2a) = 4/2 = 2.
Then the general solution to the homogeneous equation is c1y1 + c2y2 where we have
and
y1 = e 2t
y2 = te 2t .
By the variation of parameters method, there is a particular solution of the form
ψ = u1y1 + u2y2
Since there formulas for u1 and u2 involve the Wronskian, we compute
W [y1, y2] = e 2t (e 2t + 2te 2t ) − 2e 2t (te 2t ) = e 4t
Now we apply the formulas for u1 and u2 to get
and also
Then we have the particular solution
−gy2
W dt
2t 2t −(te )te
=
e4t dt
= −t 2 dt
u1 =
= −t 3 /3
gy1
W dt
2t 2t (te )e
=
e4t dt
= t dt
u2 =
= t 2 /2
ψ = −t 3 /3 e 2t + t 2 /2 te 2t = 1
6 t3 e 2t
so the general solution to the nonhomogeneous equation is
y(x) = 1
6 t3 e 2t + c1e 2t + c2te 2t .

9. Find two linearly independent solutions of
t 2 y ′′ − 2y = 0
of the form y = t r and using these find the general solution of
t 2 y ′′ − 2y = t 2 .
Solution The nonhomogeneous equation above is Euler’s equation (see problem 10 on pg
140 from hw 4) since it is of the form
t 2 y ′′ + αty ′ + βy = 0
where we have α = 0 and β = −2 in this case. In problem 10 you showed that y = t r would
be a solution if r was a root of
so in this case we must find the roots of
r 2 + (α − 1)r + β = 0
r 2 − r − 2 = 0
which are r1 = −1 and r2 = 2. Thus we have solutions
y1 = 1
t
and
y2 = t 2
which are clearly linearly independent. A particular solution is given by
ψ = u1y1 + u2y2
and to find u1 and u2 we will need to compute the Wronskian:
W [y1, y2] = 1 −1
(2t) −
t t2 t2 = 3.
We also need to know the RHS g(t), and here it is important to first put our equation in the
form
y ′′ + py ′ + qy = g(t)
so after doing so we in fact see that g(t) = 1. Then applying the formulas for u1 and u2 we
get
2
−gy2 −t
u1 = dt =
W 3 dt = −t3 /9
and also
gy1
u2 = dt = 1/(3t) dt = (1/3) ln |t|.
W
Then we have the particular solution
ψ = (−t 3 /9)(1/t) + (1/3)(ln |t|)(t 2 ) = −t2
9 + t2 ln |t|
.
3
Then the general solution to the nonhomogeneous equation is
or more simply
y(x) = −t2
9 + t2 ln |t| c1 2
+ + c2t
3 t
y(x) = t2 ln |t|
3
+ c1
t + c2t 2 .