What is the solution for y^{''}+3y^'+2y=e^x,y(0)=0,y^'(0)=3 ?

The solution for y^{''}+3y^'+2y=e^x,y(0)=0,y^'(0)=3 is y=(8/3-1/6)e^{-x}-8/3 e^{-2x}+1/6 e^x

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y^{''}+3y^'+2y=e^x,y(0)=0,y^'(0)=3

y^{^{''}} + 3 y^{^{'}} + 2 y = e^{x}, y (0) = 0, y^{^{'}} (0) = 3

y = (\frac{8}{3} - \frac{1}{6}) e^{- x} - \frac{8}{3} e^{- 2 x} + \frac{1}{6} e^{x}

Solution steps

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

Solve linear ODE:

y = (\frac{8}{3} - \frac{1}{6}) e^{- x} - \frac{8}{3} e^{- 2 x} + \frac{1}{6} e^{x}

y = (\frac{8}{3} - \frac{1}{6}) e^{- x} - \frac{8}{3} e^{- 2 x} + \frac{1}{6} e^{x}

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

y^{^{''}} + 10 y^{^{'}} + 16 y = 40 cos (4 x)

y^{^{''}} - 8 y^{^{'}} + 6 y = 3 x^{2} - 8

y^{^{''}} - 4 y^{^{'}} + 4 y = 4 e^{2 x}

x^{^{''}} - 25 x = e^{5 t}

What is the solution for y^{''}+3y^'+2y=e^x,y(0)=0,y^'(0)=3 ?
The solution for y^{''}+3y^'+2y=e^x,y(0)=0,y^'(0)=3 is y=(8/3-1/6)e^{-x}-8/3 e^{-2x}+1/6 e^x