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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 9 y = 20 sin (x) + 3 x^{2}

y^{^{''}} + y^{^{'}} + y = k x

y^{^{''}} - y^{^{'}} - 2 y = 2 x^{3} - 3 x^{2} + 8 x + 1

y^{^{''}} + 16 y = 8 e^{3 x}

y^{^{''}} + 4 y = x - \frac{x ^{2}}{20}

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