What is the solution for (16y^2-xy)dx+x^2dy=0 ?

The solution for (16y^2-xy)dx+x^2dy=0 is y= x/(16ln(x)+c_{1)}

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(16y^2-xy)dx+x^2dy=0

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

y = \frac{x}{16 ln ( x ) + c _{1}}

Solution steps

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

Solve Bernoulli ODE:

y = \frac{x}{16 ln ( x ) + c _{1}}

y = \frac{x}{16 ln ( x ) + c _{1}}

\frac{d}{d x} (2^{x} tan (3 x))

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

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\frac{d}{d x} ((7 x + 8) (8 x - 9))

What is the solution for (16y^2-xy)dx+x^2dy=0 ?
The solution for (16y^2-xy)dx+x^2dy=0 is y= x/(16ln(x)+c_{1)}