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

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

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

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

y = c_{1} e^{- x} + c_{2} x e^{- x} + \frac{1}{4} e^{x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} x e^{- x} + \frac{1}{4} e^{x}

y = c_{1} e^{- x} + c_{2} x e^{- x} + \frac{1}{4} e^{x}

(d^{4} + d^{2}) y = 18 x - 4 sin (x)

y^{^{''}} + 6 y^{^{'}} + 5 y = 12 e^{t}, y (0) = - 1, y^{^{'}} (0) = 7

y^{^{''}} - 6 y^{^{'}} + 9 y = x^{2} + e^{x}

D, D (t) = \frac{100 ln ( t + 2 )}{t + 2}

2 y^{^{''}} + 3 y^{^{'}} - 2 y = 6 x^{2} + 5 x - 8

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