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

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

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y^{''}-4y^'+4y= 1/x e^{2x}

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

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

y^{^{''}} - 2 y^{^{'}} + y = - 6 e^{t}

4 \frac{d ^{2} y}{d x ^{2}} + 36 y = csc (3 x)

4 y^{^{''}} - 6 y^{^{'}} + 7 y = 8

y^{^{''}} + 8 y^{^{'}} + 16 y = 8 sin (4 t)

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