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

The solution for y^{''}+4y^'+3y=3e^x+x is y=c_{1}e^{-x}+c_{2}e^{-3x}+3/8 e^x+x/3-4/9

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

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

y = c_{1} e^{- x} + c_{2} e^{- 3 x} + \frac{3}{8} e^{x} + \frac{x}{3} - \frac{4}{9}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} e^{- 3 x} + \frac{3}{8} e^{x} + \frac{x}{3} - \frac{4}{9}

y = c_{1} e^{- x} + c_{2} e^{- 3 x} + \frac{3}{8} e^{x} + \frac{x}{3} - \frac{4}{9}

What is the solution for y^{''}+4y^'+3y=3e^x+x ?
The solution for y^{''}+4y^'+3y=3e^x+x is y=c_{1}e^{-x}+c_{2}e^{-3x}+3/8 e^x+x/3-4/9