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

The solution for y^{''}+3y^'-4y=e^x+x is y=c_{1}e^x+c_{2}e^{-4x}+(e^xx)/5-x/4-3/16

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

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

y = c_{1} e^{x} + c_{2} e^{- 4 x} + \frac{e ^{x} x}{5} - \frac{x}{4} - \frac{3}{16}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- 4 x} + \frac{e ^{x} x}{5} - \frac{x}{4} - \frac{3}{16}

y = c_{1} e^{x} + c_{2} e^{- 4 x} + \frac{e ^{x} x}{5} - \frac{x}{4} - \frac{3}{16}

What is the solution for y^{''}+3y^'-4y=e^x+x ?
The solution for y^{''}+3y^'-4y=e^x+x is y=c_{1}e^x+c_{2}e^{-4x}+(e^xx)/5-x/4-3/16