What is the solution for y^{''}+3y^'+2y=1,+4xe^{-3x}y(0)=0 ?

The solution for y^{''}+3y^'+2y=1,+4xe^{-3x}y(0)=0 is y=c_{1}e^{-x}+c_{2}e^{-2x}+1/2

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y^{''}+3y^'+2y=1,+4xe^{-3x}y(0)=0

y^{^{''}} + 3 y^{^{'}} + 2 y = 1, + 4 x e^{- 3 x} y (0) = 0

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

Solution steps

y^{''} (x) + 3 y^{'} (x) + 2 y = 1

Solve linear ODE:

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

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

What is the solution for y^{''}+3y^'+2y=1,+4xe^{-3x}y(0)=0 ?
The solution for y^{''}+3y^'+2y=1,+4xe^{-3x}y(0)=0 is y=c_{1}e^{-x}+c_{2}e^{-2x}+1/2