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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\int (cos (x)) e^{s i n (x)} d x

\frac{\partial}{\partial x} (4 x y - 5 x^{2} y)

\frac{d}{d x} (3 sec (x))

y = 3 x^{2} + 2

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