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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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