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

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

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

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

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

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

4 y^{^{''}} + 8 y^{^{'}} + 5 y = e^{- x} sec (\frac{x}{2})

y^{^{''}} + 2 y^{^{'}} = 24 x + e^{- 2 x}

x^{^{''}} + 16 x = sin (t)

\frac{d ^{2} y}{d t ^{2}} + 8 \frac{d y}{d t} - 20 y = 5 e^{- 3 t}

y^{^{''}} + y^{^{'}} + 4 = - 20 x

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