What is the solution for y^{''}+y^'-2y=sin(x) ?

The solution for y^{''}+y^'-2y=sin(x) is y=c_{1}e^x+c_{2}e^{-2x}-3/10 sin(x)-1/10 cos(x)

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

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

y = c_{1} e^{x} + c_{2} e^{- 2 x} - \frac{3}{10} sin (x) - \frac{1}{10} cos (x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- 2 x} - \frac{3}{10} sin (x) - \frac{1}{10} cos (x)

y = c_{1} e^{x} + c_{2} e^{- 2 x} - \frac{3}{10} sin (x) - \frac{1}{10} cos (x)

\frac{d y}{d x} - y = e^{2 x} y^{3}

\frac{d ^{2} y}{d x ^{2}} + y = x^{2} + 2

\int \frac{1}{( t + 4 ) ( t - 1 )} d t

(x y - y^{2}) d x - x^{2} d y = 0

\frac{\partial}{\partial x} (- 5 y^{2} sin (x y^{2}))

What is the solution for y^{''}+y^'-2y=sin(x) ?
The solution for y^{''}+y^'-2y=sin(x) is y=c_{1}e^x+c_{2}e^{-2x}-3/10 sin(x)-1/10 cos(x)