What is the solution for y^{''}+y^'-27=xe^x ?

The solution for y^{''}+y^'-27=xe^x is y=c_{1}+c_{2}e^{-x}+(e^xx)/2-3/4 e^x+27x

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y^{''}+y^'-27=xe^x

y^{^{''}} + y^{^{'}} - 27 = x e^{x}

y = c_{1} + c_{2} e^{- x} + \frac{e ^{x} x}{2} - \frac{3}{4} e^{x} + 27 x

Solution steps

y^{''} (x) + y^{'} (x) - 27 = x e^{x}

Solve linear ODE:

y = c_{1} + c_{2} e^{- x} + \frac{e ^{x} x}{2} - \frac{3}{4} e^{x} + 27 x

y = c_{1} + c_{2} e^{- x} + \frac{e ^{x} x}{2} - \frac{3}{4} e^{x} + 27 x

y^{^{''}} + 5 y^{^{'}} - 9 y = e^{4 x}

D (x) = (x - \frac{11}{2})^{2} + x

y^{^{''}} - 7 y^{^{'}} - 18 y = - x + cos (3 x)

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

y^{^{''}} - y^{^{'}} + 12 y = e^{2 x}

What is the solution for y^{''}+y^'-27=xe^x ?
The solution for y^{''}+y^'-27=xe^x is y=c_{1}+c_{2}e^{-x}+(e^xx)/2-3/4 e^x+27x