What is the solution for y^{''}+4y^'=2x-3e^x ?

The solution for y^{''}+4y^'=2x-3e^x is y=c_{1}+c_{2}e^{-4x}+(x^2)/4-x/8-3/5 e^x

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

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

y = c_{1} + c_{2} e^{- 4 x} + \frac{x ^{2}}{4} - \frac{x}{8} - \frac{3}{5} e^{x}

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} e^{- 4 x} + \frac{x ^{2}}{4} - \frac{x}{8} - \frac{3}{5} e^{x}

y = c_{1} + c_{2} e^{- 4 x} + \frac{x ^{2}}{4} - \frac{x}{8} - \frac{3}{5} e^{x}

\int 5 cos^{3} (x) d x

\frac{d}{d x} (\frac{3}{( 2 x - 1 ) ^{3}})

\frac{d}{d x} (\frac{1 + x}{1 - x})

(t^{3} - 3 t + 2) e^{- 2 t}

\int_{0}^{5} \frac{1}{1 + x ^{2}} d x

What is the solution for y^{''}+4y^'=2x-3e^x ?
The solution for y^{''}+4y^'=2x-3e^x is y=c_{1}+c_{2}e^{-4x}+(x^2)/4-x/8-3/5 e^x