What is the solution for y^{''}-4y=x^{-1}e^{2x} ?

The solution for y^{''}-4y=x^{-1}e^{2x} 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=x^{-1}e^{2x}

y^{^{''}} - 4 y = x^{- 1} e^{2 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 = x^{- 1} e^{2 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^{^{''}} + 2 y^{^{'}} + 4 y = cos (4 x)

\frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} 4 + 8 y = 8 x^{2} + 8 x + 1

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

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

\frac{d ^{2} y}{d x ^{2}} + 8 \frac{d y}{d x} + 16 y = x^{2} + x + 1

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