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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

x^{^{''}} + 4 x = 4 t

y^{^{''}} + 16 y = g (t), y (0) = 3, y^{^{'}} (0) = 0

\frac{d ^{2} y}{d x ^{2}} - 3 \frac{d y}{d x} + 2 y = x^{2} + 8 x + 9

y^{^{''}} + 4 y^{^{'}} + 4 y = 1

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

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