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} (9 x^{2} - 2 x + 1)

\frac{d}{d x} (x a^{2} + x^{2})

a re a x = y^{2} - 1, x = 1 - y^{2}

in v erse l a pl a ce \frac{2}{s ^{2} ( s + 2 )}

\int \frac{16 x}{8 x ^{2} + 1} d x

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