What is the solution for y^'=y+y/(xln(x)),y(e)=1 ?

The solution for y^'=y+y/(xln(x)),y(e)=1 is y=e^{x-e}ln(x)

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y^'=y+y/(xln(x)),y(e)=1

y^{^{'}} = y + \frac{y}{x ln ( x )}, y (e) = 1

y = e^{x - e} ln (x)

Solution steps

y^{'} (x) = y + \frac{y}{x ln ( x )}

Solve separable ODE:

y = e^{x - e} ln (x)

y = e^{x - e} ln (x)

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

(x - 1)^{2} y^{^{'}} + 8 (x - 1) y^{^{'}} + 12 y = 0

\frac{d y}{d x} = \frac{4 x}{y ^{2}}

y^{^{'}} = \frac{2 y ^{4} + 2}{4 x y ^{3}}

y^{^{'}} = y + 5, y (0) = 11 e^{x} - 5

What is the solution for y^'=y+y/(xln(x)),y(e)=1 ?
The solution for y^'=y+y/(xln(x)),y(e)=1 is y=e^{x-e}ln(x)