What is the solution for xy^'-y= 3/4 xln(x) ?

The solution for xy^'-y= 3/4 xln(x) is y= 3/8 ln^2(x)x+c_{1}x

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xy^'-y= 3/4 xln(x)

x y^{^{'}} - y = \frac{3}{4} x ln (x)

y = \frac{3}{8} ln^{2} (x) x + c_{1} x

Solution steps

x y^{'} (x) - y = \frac{3}{4} x ln (x)

Solve linear ODE:

y = \frac{3}{8} ln^{2} (x) x + c_{1} x

y = \frac{3}{8} ln^{2} (x) x + c_{1} x

\frac{\partial}{\partial y} (2 x y)

\frac{d y}{d x}, x y = 24

\frac{d}{d r} (4 π r^{2})

f (x) = [(x - 1)^{2} - x]^{2}, at x = 2

What is the solution for xy^'-y= 3/4 xln(x) ?
The solution for xy^'-y= 3/4 xln(x) is y= 3/8 ln^2(x)x+c_{1}x