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

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

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

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

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

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

y^{^{''}} + y^{^{'}} - 2 y = 4 e^{3 t}

y^{^{''}} - 12 y^{^{'}} + 5 y = - 8 sin (t)

4 y^{^{''}} + y = t - cos (\frac{t}{2})

y^{^{''}} + 4 y = 2 e^{2 t}, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} - y^{^{'}} - 6 y = 2 sin (t)

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