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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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