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}+1/8 e^x

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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{1}{8} e^{x}

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{1}{8} e^{x}

y = c_{1} e^{- x} + c_{2} e^{- 3 x} + \frac{1}{8} e^{x}

\frac{d}{d x} (x^{l n (x)})

\frac{d}{d x} (9 x^{2} - 2 x + 1)

x sin (\frac{4}{x})

\int_{0}^{1} 2 π (x) (x - x^{2}) d x

\frac{d}{d x} (lo g_{7} (\frac{x ^{4}}{5 x + 11}))

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}+1/8 e^x