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

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

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

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

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} + \frac{e ^{2 x} x}{4} + \frac{x}{2}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} + \frac{e ^{2 x} x}{4} + \frac{x}{2}

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} + \frac{e ^{2 x} x}{4} + \frac{x}{2}

y^{^{''}} + 4 y^{^{'}} + 5 y = 4 e^{- t} cos (2 t)

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

y^{^{''}} - 2 y^{^{'}} + y = \frac{e ^{x}}{( 1 + x ) ( 2 + x )}

\frac{d ^{2} y}{d x ^{2}} - \frac{d y}{d x} - 45 y = e^{5 x}

\frac{d ^{2} x}{d t ^{2}} + 36 x = 10 sin (4 t)

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