What is the solution for xy^'-y= 5/4 xln(x) ?

The solution for xy^'-y= 5/4 xln(x) is y= 5/8 ln^2(x)x+c_{1}x

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xy^'-y= 5/4 xln(x)

x y^{^{'}} - y = \frac{5}{4} x ln (x)

y = \frac{5}{8} ln^{2} (x) x + c_{1} x

Solution steps

x y^{'} (x) - y = \frac{5}{4} x ln (x)

Solve linear ODE:

y = \frac{5}{8} ln^{2} (x) x + c_{1} x

y = \frac{5}{8} ln^{2} (x) x + c_{1} x

y = 8 x arccos (x - 1), at x = 1

(t^{2} + 5)^{2} - 7 (t^{2} + 5)

2 x \frac{d y}{d x} - 8 y = x^{- 8}

\frac{2}{x}, \frac{1}{2} x, 1, 2

\frac{d}{d x} (ln (\frac{2 x - 1}{x - 1}))

What is the solution for xy^'-y= 5/4 xln(x) ?
The solution for xy^'-y= 5/4 xln(x) is y= 5/8 ln^2(x)x+c_{1}x