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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

x \to a lim (sin (x))

\int_{6}^{8} \frac{1}{x ^{2} - 1} d x

x \to 0 lim (x sec (x) csc (x))

\frac{d y}{d x} + y + 3 sin (x) = 0

\int_{0}^{\frac{π}{3}} (2 sec^{2} (x)) d x

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