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

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

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y^{''}-y=sin(t),y(0)=0,y^'(0)=1

y^{^{''}} - y = sin (t), y (0) = 0, y^{^{'}} (0) = 1

y = \frac{3}{4} e^{t} - \frac{3}{4} e^{- t} - \frac{1}{2} sin (t)

Solution steps

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

Solve linear ODE:

y = \frac{3}{4} e^{t} - \frac{3}{4} e^{- t} - \frac{1}{2} sin (t)

y = \frac{3}{4} e^{t} - \frac{3}{4} e^{- t} - \frac{1}{2} sin (t)

y^{^{''}} + 6 y^{^{'}} + 9 y = 2 sin (x)

y^{^{''}} - 8 y^{^{'}} + 16 y = 8 e^{4 x}

y^{^{''}} + 2 y^{^{'}} + y = \frac{e ^{- x}}{x ^{2} + 1}

y^{^{''}} - 2 y^{^{'}} + 17 y = e^{x} cos (4 x)

y^{^{'''}} - 4 y^{^{'}} = x e^{2 x} + x^{2} + sin (x)

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