What is the solution for y^{''}+2y^'-3y=2x ?

The solution for y^{''}+2y^'-3y=2x is y=c_{1}e^x+c_{2}e^{-3x}-(2x)/3-4/9

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

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

y = c_{1} e^{x} + c_{2} e^{- 3 x} - \frac{2 x}{3} - \frac{4}{9}

Solution steps

y^{''} (x) + 2 y^{'} (x) - 3 y = 2 x

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- 3 x} - \frac{2 x}{3} - \frac{4}{9}

y = c_{1} e^{x} + c_{2} e^{- 3 x} - \frac{2 x}{3} - \frac{4}{9}

y^{^{''}} + 2 y^{^{'}} + y = \frac{ln ( x )}{e ^{x}}

y^{^{''}} - y^{^{'}} - 2 y = \frac{( e ^{2 t} + e ^{- 2 t} )}{2}

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

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

(D^{3} + 10 D^{2} + 25 D) y = e^{x}

What is the solution for y^{''}+2y^'-3y=2x ?
The solution for y^{''}+2y^'-3y=2x is y=c_{1}e^x+c_{2}e^{-3x}-(2x)/3-4/9