What is the solution for xy^'-y= 5/2 xln(x) ?

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

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xy^'-y= 5/2 xln(x)

x y^{^{'}} - y = \frac{5}{2} x ln (x)

y = \frac{5}{4} ln^{2} (x) x + c_{1} x

Solution steps

x y^{'} (x) - y = \frac{5}{2} x ln (x)

Solve linear ODE:

y = \frac{5}{4} ln^{2} (x) x + c_{1} x

y = \frac{5}{4} ln^{2} (x) x + c_{1} x

\frac{d y}{d x}, 3 e^{x y} - x = 0

\int z^{2} d z

\frac{\partial}{\partial t} ((x - c t)^{4} + (x + c t)^{4})

\int \frac{( 4 x + 2 )}{x ^{3} - x ^{2} + 2 x} d x

y^{^{'}} = 2 t y^{2}

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