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

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

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

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

y = \frac{1}{4} e^{t} - \frac{1}{4} e^{- t} - \frac{1}{2} sin (t)

Solution steps

y^{''} (t) - y = sin (t)

Solve linear ODE:

y = \frac{1}{4} e^{t} - \frac{1}{4} e^{- t} - \frac{1}{2} sin (t)

y = \frac{1}{4} e^{t} - \frac{1}{4} e^{- t} - \frac{1}{2} sin (t)

y^{^{''}} + 5 y^{^{'}} + 6 y = 4 x + 5

3 x^{^{''}} + 30 x^{^{'}} + 125 x = 350

y^{^{''}} + y = t^{2} + 1

y^{^{''}} + 4 y^{^{'}} + 4 y = 5 e^{3 t}

y^{^{''}} + y^{^{'}} + \frac{5}{4} y = 3 cos (x)

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