What is the solution for xy^'-y= 1/2 xln(x) ?

The solution for xy^'-y= 1/2 xln(x) is y= 1/4 ln^2(x)x+c_{1}x

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xy^'-y= 1/2 xln(x)

x y^{^{'}} - y = \frac{1}{2} x ln (x)

y = \frac{1}{4} ln^{2} (x) x + c_{1} x

Solution steps

x y^{'} (x) - y = \frac{1}{2} x ln (x)

Solve linear ODE:

y = \frac{1}{4} ln^{2} (x) x + c_{1} x

y = \frac{1}{4} ln^{2} (x) x + c_{1} x

\int \frac{e ^{a x} + b}{e ^{a x} - b} d x

\int 10 x^{4} - 2 x - 6 d x

\int x x + 6 d x

\frac{d}{d x} (\frac{x ^{2}}{1 + x ^{3}})

(tan^{3} (x))^{^{'}}

What is the solution for xy^'-y= 1/2 xln(x) ?
The solution for xy^'-y= 1/2 xln(x) is y= 1/4 ln^2(x)x+c_{1}x