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

The solution for y^{''}-4y=xe^x is y=c_{1}e^{2x}+c_{2}e^{-2x}-(e^xx)/3-2/9 e^x

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y^{''}-4y=xe^x

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

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} - \frac{e ^{x} x}{3} - \frac{2}{9} e^{x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} - \frac{e ^{x} x}{3} - \frac{2}{9} e^{x}

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} - \frac{e ^{x} x}{3} - \frac{2}{9} e^{x}

y^{^{''}} - y^{^{'}} - 2 y = cosh (2 t)

y^{^{''}} - y = sin (x)

y^{^{''}} - 4 y = (x^{2} - 3) sin (2 x)

y^{^{''}} + 9 y = 9 sec^{2} (3 t)

y^{^{''}} + 25 y = 40 cos (5 t)

What is the solution for y^{''}-4y=xe^x ?
The solution for y^{''}-4y=xe^x is y=c_{1}e^{2x}+c_{2}e^{-2x}-(e^xx)/3-2/9 e^x