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

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

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y^{''}-4y^'+4y=(e^{2x})/x

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

y = c_{1} e^{2 x} + c_{2} x e^{2 x} - e^{2 x} x + e^{2 x} x ln (x)

Solution steps

y^{''} (x) - 4 y^{'} (x) + 4 y = \frac{e ^{2 x}}{x}

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} x e^{2 x} - e^{2 x} x + e^{2 x} x ln (x)

y = c_{1} e^{2 x} + c_{2} x e^{2 x} - e^{2 x} x + e^{2 x} x ln (x)

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