Lecture Notes in Differential Equations - Bruce E. Shapiro

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Example 19.4. Solve
⎫
y ′′ + 4y = 3 sin 2t⎪⎬
y(0) = 0
⎪⎭
y ′ (0) = −1
(19.46)
The characteristic equation is
r 2 + 4 = 0 (19.47)
So the roots are ±2i and the homogeneous solution is
y H = C 1 cos 2t + C 2 sin 2t (19.48)
For a particular solution we use
y = t(A cos 2t + B sin 2t) (19.49)
y ′ = Aa cos 2t + B sin 2t + t(2B cos 2t − 2A sin 2t) (19.50)
y ′′ = 4B cos 2t − 4A sin 2t + t(−4A cos 2t − 4B sin 2t) (19.51)
Plugging into the differential equation,
4B cos 2t − 4A sin 2t + t(−4A cos 2t − 4B sin 2t)
+ 4t(A cos 2t + B sin 2t) = 3 sin 2t (19.52)
Canceling like terms,
4B cos 2t − 4A sin 2t = 3 sin 2t (19.53)
equating coefficients of like trigonometric functions, A = 0, B = 3 4 , hence
y = C 1 cos 2t + C 2 sin 2t + 3 t sin 2t (19.54)
4
From the first initial condition, y(0) = 0,
0 = C 1 (19.55)
hence
y = C 2 sin 2t + 3 t sin 2t (19.56)
4
Differentiating,
y ′ = 2C 2 cos 2t + 3 4 sin 2t + 3 t cos 2t (19.57)
4