What is the solution for xy^'-y= 2/3 xln(x) ?

The solution for xy^'-y= 2/3 xln(x) is y= 1/3 ln^2(x)x+c_{1}x

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xy^'-y= 2/3 xln(x)

x y^{^{'}} - y = \frac{2}{3} x ln (x)

y = \frac{1}{3} ln^{2} (x) x + c_{1} x

Solution steps

x y^{'} (x) - y = \frac{2}{3} x ln (x)

Solve linear ODE:

y = \frac{1}{3} ln^{2} (x) x + c_{1} x

y = \frac{1}{3} ln^{2} (x) x + c_{1} x

x^{2.4} + e^{2.4}

x \to 4 - lim (\frac{15}{x - 4})

y = 20 + x - x^{2}, x^{2} - 5 x

\int (3 x^{5} - 2 x^{3}) d x

cos (tan (x))

What is the solution for xy^'-y= 2/3 xln(x) ?
The solution for xy^'-y= 2/3 xln(x) is y= 1/3 ln^2(x)x+c_{1}x