What is the solution for y^{''}+4y=32sin(4t),y(0)=1,y^'(0)=0 ?

The solution for y^{''}+4y=32sin(4t),y(0)=1,y^'(0)=0 is y=cos(2t)+16/3 sin(2t)-8/3 sin(4t)

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y^{''}+4y=32sin(4t),y(0)=1,y^'(0)=0

y^{^{''}} + 4 y = 32 sin (4 t), y (0) = 1, y^{^{'}} (0) = 0

y = cos (2 t) + \frac{16}{3} sin (2 t) - \frac{8}{3} sin (4 t)

Solution steps

y^{''} (t) + 4 y = 32 sin (4 t)

Solve linear ODE:

y = cos (2 t) + \frac{16}{3} sin (2 t) - \frac{8}{3} sin (4 t)

y = cos (2 t) + \frac{16}{3} sin (2 t) - \frac{8}{3} sin (4 t)

y^{^{''}} + y^{^{'}} - 6 = 4 x^{3} + 3 x

x^{^{''}} - 25 x = 6 t

x^{^{''}} - 4 x^{^{'}} + 4 x = e^{t} + e^{2 t} + 1

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} + 10 y = 2 cos (6 x)

y^{^{''}} + 25 y = sec^{3} (5 x)

What is the solution for y^{''}+4y=32sin(4t),y(0)=1,y^'(0)=0 ?
The solution for y^{''}+4y=32sin(4t),y(0)=1,y^'(0)=0 is y=cos(2t)+16/3 sin(2t)-8/3 sin(4t)