What is the solution for (20y^2-xy)dx+x^2dy=0 ?

The solution for (20y^2-xy)dx+x^2dy=0 is y= x/(20ln(x)+c_{1)}

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(20y^2-xy)dx+x^2dy=0

(20 y^{2} - x y) d x + x^{2} d y = 0

y = \frac{x}{20 ln ( x ) + c _{1}}

Solution steps

(20 y^{2} - x y) d x + x^{2} d y = 0

Solve Bernoulli ODE:

y = \frac{x}{20 ln ( x ) + c _{1}}

y = \frac{x}{20 ln ( x ) + c _{1}}

(12 x^{2} - 24 x + 24) e^{- 8 x}

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What is the solution for (20y^2-xy)dx+x^2dy=0 ?
The solution for (20y^2-xy)dx+x^2dy=0 is y= x/(20ln(x)+c_{1)}