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{\partial}{\partial y} (e^{- 2 x^{2} - y^{2}})

x \to - \infty lim (b x)

\frac{d}{d x} (ln (sec (x) + tan (x)))

\int 1200 e^{0.3 t} d t

\int_{0}^{1} \frac{1}{9 + 16 x ^{2}} d x

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