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

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

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

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

y = g t - g sin (t)

Solution steps

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

Solve linear ODE:

y = g t - g sin (t)

y = g t - g sin (t)

x^{^{''}} - x = 4 e^{- t}

y^{^{''}} + 6 y^{^{'}} + 18 y = 11 t e^{4 t}

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

y^{^{''}} - 10 y^{^{'}} + 25 y = 6 + 3 x

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

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