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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 3 y^{^{'}} - 4 y = 6 e^{x}

y^{^{''}} - 2 y^{^{'}} + 2 y = 3 e^{x} sec (x)

y^{^{''}} - 3 y^{^{'}} - 4 y = 10 e^{t} - 250 cos (3 t)

y^{^{''}} + 4 y = H (x - π) - H (x - 2 π)

y^{^{''}} + 25 y = 60 cos (5 t)

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