HW 5 solutions

§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 .