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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + y^{^{'}} = 2 x + 3

y^{^{''}} - 4 y^{^{'}} + 4 y = 14 x + 5

y^{^{''}} + 10 y^{^{'}} + 25 y = e^{\frac{- 10 x}{x ^{2}}}

d (r) = \frac{( ln ( 2 ) )}{ln ( 1 + 0.04 )}

y^{^{''}} + y^{^{'}} = 2 x - 2

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