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

The solution for y^{''}+4y=g(t),y(0)=0,y^'(0)=0 is y=(gt)/4-(gsin(2t))/8

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

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

y = \frac{g t}{4} - \frac{g sin ( 2 t )}{8}

Solution steps

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

Solve linear ODE:

y = \frac{g t}{4} - \frac{g sin ( 2 t )}{8}

y = \frac{g t}{4} - \frac{g sin ( 2 t )}{8}

\frac{d ^{2} y}{d x ^{2}} - 8 \frac{d y}{d x} + 6 y = 3 x^{2} - 8

y^{^{'''}} + 8 y = 2 x - 5 + 8 e^{- 2} x

y^{^{'''}} + y^{^{''}} + y^{^{'}} + y = x e^{x}

y^{^{''}} - 4 y = 4 e^{3 x}

y^{^{''}} + 2 y^{^{'}} + y = g (t)

What is the solution for y^{''}+4y=g(t),y(0)=0,y^'(0)=0 ?
The solution for y^{''}+4y=g(t),y(0)=0,y^'(0)=0 is y=(gt)/4-(gsin(2t))/8