What is the solution for y^{''}+2y^'+y=4e^{-x}ln(x) ?

The solution for y^{''}+2y^'+y=4e^{-x}ln(x) is y=c_{1}e^{-x}+c_{2}xe^{-x}+2e^{-x}x^2ln(x)-3e^{-x}x^2

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y^{''}+2y^'+y=4e^{-x}ln(x)

y^{^{''}} + 2 y^{^{'}} + y = 4 e^{- x} ln (x)

y = c_{1} e^{- x} + c_{2} x e^{- x} + 2 e^{- x} x^{2} ln (x) - 3 e^{- x} x^{2}

Solution steps

y^{''} (x) + 2 y^{'} (x) + y = 4 e^{- x} ln (x)

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} x e^{- x} + 2 e^{- x} x^{2} ln (x) - 3 e^{- x} x^{2}

y = c_{1} e^{- x} + c_{2} x e^{- x} + 2 e^{- x} x^{2} ln (x) - 3 e^{- x} x^{2}

3 y^{^{''}} + 8 y^{^{'}} - 3 y = 8 - 3 t

2 y^{^{''}} + 4 y = 6 t, y (0) = 4.9, y^{^{'}} (0) = 0

y^{^{''}} - 5 y^{^{'}} + 6 y = 24

D (t) = - 200 t + 9000

\frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} + 4 y = 4

What is the solution for y^{''}+2y^'+y=4e^{-x}ln(x) ?
The solution for y^{''}+2y^'+y=4e^{-x}ln(x) is y=c_{1}e^{-x}+c_{2}xe^{-x}+2e^{-x}x^2ln(x)-3e^{-x}x^2