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

The solution for y^{''}+4y^'=x is y=c_{1}+c_{2}e^{-4x}+(x^2)/8-x/(16)

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y^{''}+4y^'=x

y^{^{''}} + 4 y^{^{'}} = x

y = c_{1} + c_{2} e^{- 4 x} + \frac{x ^{2}}{8} - \frac{x}{16}

Solution steps

y^{''} (x) + 4 y^{'} (x) = x

Solve linear ODE:

y = c_{1} + c_{2} e^{- 4 x} + \frac{x ^{2}}{8} - \frac{x}{16}

y = c_{1} + c_{2} e^{- 4 x} + \frac{x ^{2}}{8} - \frac{x}{16}

What is the solution for y^{''}+4y^'=x ?
The solution for y^{''}+4y^'=x is y=c_{1}+c_{2}e^{-4x}+(x^2)/8-x/(16)