What is the solution for ydx-xdy=x(dy-ydx) ?

The solution for ydx-xdy=x(dy-ydx) is y=e^{1/2 (x+ln(x))+c_{1}}

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ydx-xdy=x(dy-ydx)

y d x - x d y = x (d y - y d x)

y = e^{\frac{1}{2} (x + l n (x)) + c_{1}}

Solution steps

y d x - x d y = x (d y - y d x)

Solve separable ODE:

y = e^{\frac{1}{2} (x + l n (x)) + c_{1}}

y = e^{\frac{1}{2} (x + l n (x)) + c_{1}}

x^{2} y^{^{'}} - 2 - 3 y = 0

y^{^{'}} + x^{2} y = x^{2} y^{- 3}

(1 - 3 y^{2}) d y - x sin (x) d x = 0, y (\frac{π}{2}) = 2

\frac{d y}{d x} = 6 x 25 - y^{2}

3 x y y^{^{'}} + 4 y^{1} - 2 = 0

What is the solution for ydx-xdy=x(dy-ydx) ?
The solution for ydx-xdy=x(dy-ydx) is y=e^{1/2 (x+ln(x))+c_{1}}