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

The solution for y^{''}-4y^'+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^{''}-4y^'+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^{^{''}} - 2 y^{^{'}} + 2 y = 1, y^{^{'}} 0 = 0

y^{^{''}} - 2 y^{^{'}} + y = \frac{7 e ^{t}}{1 + t ^{2}}

y^{^{''}} - 9 y = 9 e^{3 x} - 6 x^{4}

y^{^{''}} + y^{^{'}} - 12 y = t, y (0) = 1, y^{^{'}} (0) = 1

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

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