What is the solution for y^{''}+3y^'+2y=4e^x ?

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

2 y^{^{''}} + 4 y^{^{'}} + 2 y = e^{2 x} cos (2 x)

x^{^{''}} - x = 1

2 y^{^{''}} - 4 y^{^{'}} + 4 y = e^{x} tan (x) sec (x)

x^{^{''}} + 49 x = 99 cos (4 t)

y^{^{''}} - 8 y^{^{'}} + 20 y = 100 x^{2} - 65

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