What is the solution for y^{''}+4y^'+3y=6e^{-x} ?

The solution for y^{''}+4y^'+3y=6e^{-x} is y=c_{1}e^{-x}+c_{2}e^{-3x}+3xe^{-x}

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y^{''}+4y^'+3y=6e^{-x}

y^{^{''}} + 4 y^{^{'}} + 3 y = 6 e^{- x}

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

Solution steps

y^{''} (x) + 4 y^{'} (x) + 3 y = 6 e^{- x}

Solve linear ODE:

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

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

\int \frac{1}{a x + b} d x

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

y = ln (x^{2} + y^{2})

\frac{\partial}{\partial y} (e^{x^{2} - y^{2}})

\frac{d}{d x} (1.5 (\frac{1}{3} + \frac{x}{2} - \frac{x ^{3}}{6}))

What is the solution for y^{''}+4y^'+3y=6e^{-x} ?
The solution for y^{''}+4y^'+3y=6e^{-x} is y=c_{1}e^{-x}+c_{2}e^{-3x}+3xe^{-x}