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

The solution for y^{''}+2y^'+y=xe^{-x}+cos(x) is y=c_{1}e^{-x}+c_{2}xe^{-x}+(e^{-x}x^3)/6+1/2 sin(x)

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

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

y = c_{1} e^{- x} + c_{2} x e^{- x} + \frac{e ^{- x} x ^{3}}{6} + \frac{1}{2} sin (x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} x e^{- x} + \frac{e ^{- x} x ^{3}}{6} + \frac{1}{2} sin (x)

y = c_{1} e^{- x} + c_{2} x e^{- x} + \frac{e ^{- x} x ^{3}}{6} + \frac{1}{2} sin (x)

y^{^{''}} + 4 y = x^{2} + 5 \cdot cos (2 x)

y^{^{''}} + 4 y = 16 sin (2 t), y (0) = 0, y^{^{'}} (0) = 0

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

\frac{d ^{2} x}{d t ^{2}} - \frac{d x}{d t} - 2 x = 4 t

x^{^{''}} + 2 x + 5 x = 1, x^{^{'}} (0) = 0, x (0) = 0

What is the solution for y^{''}+2y^'+y=xe^{-x}+cos(x) ?
The solution for y^{''}+2y^'+y=xe^{-x}+cos(x) is y=c_{1}e^{-x}+c_{2}xe^{-x}+(e^{-x}x^3)/6+1/2 sin(x)