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)

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)