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}

y^{^{'''}} + y^{^{'}} = tan (t)

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

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

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

y^{^{''}} - 9 y = \frac{9 x}{e ^{3 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