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General Solutions
.

Therefore,
1 y (0)
′
y = t sin λt + cos λt + sin λt. (5.2)
2λ λ

Finally, from (5.2)
′
y (0) π
= 1 − ,
λ 4λ 2
gives
y (0) π
′
= 1 − ,
λ 4λ 2
and thus
1 ( π )
y = t sin λt + cos λt + 1 − sin λt.
2λ 4λ 2
Similarly, if the boundary data had been, say

π
( )
y(0) = 1, y ′ = 1,
4
then differentiating in (5.2)

1
′
′
y = (sin λt + λt cos λt) − λ sin λt + y (0) cos λt.
2λ
Thus,
π −π
( )
1 = y ′ = − y (0)
′
λ 2λ
and
( π )
′
y (0) = − 1 +
2λ
to yield
1 1 ( π )
y = t sin λt + cos λt − 1 + sin λt.
2λ λ 2λ
. . . .
.

Example 5.7, (Distinct real roots, but one matches the source
term.) Solve the initial value problem by Laplace transform

2t
′
′′
y − y − 2y = e , y(0) = 0, y (0) = 1.
′
. . . . .
.
Solution. Take Laplace transform on both sides of the equation, we get
1
2t
2
′
′′
L[y ] − L[y ] − 2L[y] = L[e ] ⇒ s Y (s) − 1 − sY (s) − 2Y (s) = .
s − 2
Solve it for Y (s) :
1 s − 1
2
(s − s − 2)Y (s) = + 1 = ⇒
. . . . s − 2 s − 2

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