What is the solution for y^{''}+2y^'-3y=8e^x ?

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

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

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

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

Solution steps

y^{''} (x) + 2 y^{'} (x) - 3 y = 8 e^{x}

Solve linear ODE:

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

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

y^{^{''}} + 16 y = 4 cos (4 x) + 6 sin (4 x)

y^{^{''}} + 3 y^{^{'}} + 2 y = e - t, y (4) = 1, y^{^{'}} (4) = 3

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

y^{^{''}} - 4 y^{^{'}} + 4 y = 4 x + 6

y^{^{''}} - 4 y = 28

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