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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d y}{d x} = \frac{x - 4}{y - 6}

\int_{1}^{2} \frac{5}{x ^{3} + 2 x} d x

\int_{0}^{1} (5^{x} - 3^{x}) d x

y = 2 x^{3}, (1, 2)

f (x) = 6 x + 3

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