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}

y^{^{''}} + 2 y^{^{'}} - 3 y = (x + 1) e^{x}

\frac{d ^{2} y}{d x ^{2}} - 14 \frac{d y}{d x} + 49 y = (x + 1) e^{x}

y^{^{''}} + y^{^{'}} = 4 cos (2 x)

6 y^{^{''}} - 5 y^{^{'}} + y = 3 x^{2} + x

\frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} + 6 y = e^{2 x}

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)