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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

10 y^{^{''}} + 90 y^{^{'}} + \frac{1}{0.005} y = 500 sin (x)

y^{^{''}} + 3 y = cos (3 x)

y^{^{''}} + 3 y^{^{'}} = 9

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

y^{^{''}} - 11 y^{^{'}} + 24 y = 280 e^{t}

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