What is the solution for y^{''}+16y=sin(t),y(0)=0,y^'(0)=0 ?

The solution for y^{''}+16y=sin(t),y(0)=0,y^'(0)=0 is y=-1/60 sin(4t)+1/15 sin(t)

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y^{''}+16y=sin(t),y(0)=0,y^'(0)=0

y^{^{''}} + 16 y = sin (t), y (0) = 0, y^{^{'}} (0) = 0

y = - \frac{1}{60} sin (4 t) + \frac{1}{15} sin (t)

Solution steps

y^{''} (t) + 16 y = sin (t)

Solve linear ODE:

y = - \frac{1}{60} sin (4 t) + \frac{1}{15} sin (t)

y = - \frac{1}{60} sin (4 t) + \frac{1}{15} sin (t)

9 y^{^{''}} - 6 y^{^{'}} + y = 50 sin (x)

4 y^{^{''}} + 4 y^{^{'}} + y = x e^{\cdot \frac{x}{2}} sin (x)

y^{^{''}} - y^{^{'}} = 2 x^{2} - 3

y^{^{''}} + 2 y^{^{'}} + y = e^{- x l n (x)}

y^{^{''}} - 8 y^{^{'}} + 80 y = - 3 sin (3 t)

What is the solution for y^{''}+16y=sin(t),y(0)=0,y^'(0)=0 ?
The solution for y^{''}+16y=sin(t),y(0)=0,y^'(0)=0 is y=-1/60 sin(4t)+1/15 sin(t)