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

The solution for y^{''}+4y=sin(t),y(0)=0,y^'(0)=0 is y=-1/6 sin(2t)+1/3 sin(t)

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

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

y = - \frac{1}{6} sin (2 t) + \frac{1}{3} sin (t)

Solution steps

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

Solve linear ODE:

y = - \frac{1}{6} sin (2 t) + \frac{1}{3} sin (t)

y = - \frac{1}{6} sin (2 t) + \frac{1}{3} sin (t)

y^{^{''}} - y = 8 x e^{x} + 2 e^{x}

y^{^{''}} + 5 y^{^{'}} = 4 x + 8

y^{^{''}} - 2 y^{^{'}} + y = \frac{e ^{x}}{x ^{3}}

y^{^{''}} + 4 y = 4 csc^{2} (2 t)

\frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} + 4 y = e^{x}

What is the solution for y^{''}+4y=sin(t),y(0)=0,y^'(0)=0 ?
The solution for y^{''}+4y=sin(t),y(0)=0,y^'(0)=0 is y=-1/6 sin(2t)+1/3 sin(t)