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

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

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

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

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + e^{x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + e^{x}

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + e^{x}

x \to 2 lim (825)

y = e^{- \frac{x}{2}}

\frac{d}{d y} (arctan (\frac{y}{x}))

12 (x - \frac{1}{x})^{4}

\int \frac{( 37 x + 5 )}{( 6 x + 1 ) ( x - 1 )} d x

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