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

The solution for y^{''}-4y=(e^{2x}-e^{-2x})/2 is y=c_{1}e^{2x}+c_{2}e^{-2x}+(e^{2x}x+e^{-2x}x)/8

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

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

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} + \frac{e ^{2 x} x + e ^{- 2 x} x}{8}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} + \frac{e ^{2 x} x + e ^{- 2 x} x}{8}

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} + \frac{e ^{2 x} x + e ^{- 2 x} x}{8}

(D^{2} + D + 1) y = e^{x} sin (x)

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

y^{^{''}} + 9 y = 5 sin (3 x), y (0) = 2, y^{^{'}} (0) = 0

y^{^{''}} + 25 y = \frac{1}{100} cos (5 x)

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

What is the solution for y^{''}-4y=(e^{2x}-e^{-2x})/2 ?
The solution for y^{''}-4y=(e^{2x}-e^{-2x})/2 is y=c_{1}e^{2x}+c_{2}e^{-2x}+(e^{2x}x+e^{-2x}x)/8