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

The solution for y^{''}+2y^'+y=e^{2x}+cos(x) is y=c_{1}e^{-x}+c_{2}xe^{-x}+1/9 e^{2x}+1/2 sin(x)

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

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

y = c_{1} e^{- x} + c_{2} x e^{- x} + \frac{1}{9} e^{2 x} + \frac{1}{2} sin (x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} x e^{- x} + \frac{1}{9} e^{2 x} + \frac{1}{2} sin (x)

y = c_{1} e^{- x} + c_{2} x e^{- x} + \frac{1}{9} e^{2 x} + \frac{1}{2} sin (x)

What is the solution for y^{''}+2y^'+y=e^{2x}+cos(x) ?
The solution for y^{''}+2y^'+y=e^{2x}+cos(x) is y=c_{1}e^{-x}+c_{2}xe^{-x}+1/9 e^{2x}+1/2 sin(x)