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

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

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

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

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} + \frac{e ^{2 x} ln ( x ) - e ^{- 2 x} Ei ( 4 x )}{4}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} + \frac{e ^{2 x} ln ( x ) - e ^{- 2 x} Ei ( 4 x )}{4}

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} + \frac{e ^{2 x} ln ( x ) - e ^{- 2 x} Ei ( 4 x )}{4}

5 y^{^{''}} + 180 y = 90 sin (6 t)

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

y^{^{''}} + 9 y = x^{2} sin (4 x)

y^{^{''}} - y^{^{'}} - 6 y = 12 x - e^{x}, y (0) = 1

\frac{d ^{2} y}{d x ^{2}} - 3 \frac{d y}{d x} + 2 y = 2 x

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