What is the solution for y^{''}+4y^'+3y=32xe^x ?

The solution for y^{''}+4y^'+3y=32xe^x is y=c_{1}e^{-x}+c_{2}e^{-3x}+4xe^x-3e^x

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y^{''}+4y^'+3y=32xe^x

y^{^{''}} + 4 y^{^{'}} + 3 y = 32 x e^{x}

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

Solution steps

y^{''} (x) + 4 y^{'} (x) + 3 y = 32 x e^{x}

Solve linear ODE:

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

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

\int x^{5} cos (x) d x

f (x) = \frac{6 x}{x ^{2} + 1}

\frac{d}{d x} (3 x (18 x^{4} + \frac{13}{x + 1}))

\frac{d}{d x} (ln (sin (x^{3})))

x y = - 121

What is the solution for y^{''}+4y^'+3y=32xe^x ?
The solution for y^{''}+4y^'+3y=32xe^x is y=c_{1}e^{-x}+c_{2}e^{-3x}+4xe^x-3e^x