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}

y^{^{''}} + y^{^{'}} - 2 y = x^{2} + cos (3 x)

(d^{2} + 1) y = csc (x)

2 y^{^{''}} + y^{^{'}} - y = 2 e^{x}

y^{^{''}} + 1.44 y = 24 cos (1.2 x)

y^{^{''}} + y = 5 e^{- 4 x}, y (0) = 0, y^{^{'}} (0) = 0

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