What is the solution for y^{''}-y=e^tsin(t) ?

The solution for y^{''}-y=e^tsin(t) is y=c_{1}e^t+c_{2}e^{-t}-1/5 e^tsin(t)-2/5 e^tcos(t)

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y^{''}-y=e^tsin(t)

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

y = c_{1} e^{t} + c_{2} e^{- t} - \frac{1}{5} e^{t} sin (t) - \frac{2}{5} e^{t} cos (t)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{t} + c_{2} e^{- t} - \frac{1}{5} e^{t} sin (t) - \frac{2}{5} e^{t} cos (t)

y = c_{1} e^{t} + c_{2} e^{- t} - \frac{1}{5} e^{t} sin (t) - \frac{2}{5} e^{t} cos (t)

y^{^{''}} + 3 y^{^{'}} - 4 y = 3 e^{2} x

y^{^{''}} - 2 y^{^{'}} - 3 y = e^{x}

y^{^{''}} - 2 y^{^{'}} + y = e^{x} (x^{2} - x + 1)

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

x^{^{''}} + 4 x^{^{'}} + 40 x = 4 e^{- 2 t} cos (6 t)

What is the solution for y^{''}-y=e^tsin(t) ?
The solution for y^{''}-y=e^tsin(t) is y=c_{1}e^t+c_{2}e^{-t}-1/5 e^tsin(t)-2/5 e^tcos(t)