What is the solution for xln(x)y^'+y=xe^x,y(2)=-5 ?

The solution for xln(x)y^'+y=xe^x,y(2)=-5 is y=(e^x)/(ln(x))+(-5ln(2)-e^2)/(ln(x))

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xln(x)y^'+y=xe^x,y(2)=-5

x ln (x) y^{^{'}} + y = x e^{x}, y (2) = - 5

y = \frac{e ^{x}}{ln ( x )} + \frac{- 5 ln ( 2 ) - e ^{2}}{ln ( x )}

Solution steps

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

Solve linear ODE:

y = \frac{e ^{x}}{ln ( x )} + \frac{- 5 ln ( 2 ) - e ^{2}}{ln ( x )}

y = \frac{e ^{x}}{ln ( x )} + \frac{- 5 ln ( 2 ) - e ^{2}}{ln ( x )}

\frac{d}{d x} (5 x x)

\frac{d}{d x} (2 cos (\frac{1}{2} x))

\int_{0}^{\frac{π}{4}} cos (x) d x

\frac{d}{d x} (20 x^{- 8} + 5 x^{- 3} - 7 x^{- 7})

\frac{d}{d x} (\frac{1}{x ^{2} - 9 y ^{2}})

What is the solution for xln(x)y^'+y=xe^x,y(2)=-5 ?
The solution for xln(x)y^'+y=xe^x,y(2)=-5 is y=(e^x)/(ln(x))+(-5ln(2)-e^2)/(ln(x))