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

The solution for y^{''}+2y^'-3y=e^{2x}-2x is y=c_{1}e^x+c_{2}e^{-3x}+1/5 e^{2x}+(2x)/3+4/9

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

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

y = c_{1} e^{x} + c_{2} e^{- 3 x} + \frac{1}{5} e^{2 x} + \frac{2 x}{3} + \frac{4}{9}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- 3 x} + \frac{1}{5} e^{2 x} + \frac{2 x}{3} + \frac{4}{9}

y = c_{1} e^{x} + c_{2} e^{- 3 x} + \frac{1}{5} e^{2 x} + \frac{2 x}{3} + \frac{4}{9}

\frac{d ^{3} y}{d x ^{3}} + y = cos (x)

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

y^{^{'''}} - 5 y^{^{''}} + 9 y = t cos (2 t)

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

y^{^{''}} + 6 y^{^{'}} + 5 y = sin (x)

What is the solution for y^{''}+2y^'-3y=e^{2x}-2x ?
The solution for y^{''}+2y^'-3y=e^{2x}-2x is y=c_{1}e^x+c_{2}e^{-3x}+1/5 e^{2x}+(2x)/3+4/9