What is the solution for (19y^2-xy)dx+x^2dy=0 ?

The solution for (19y^2-xy)dx+x^2dy=0 is y= x/(19ln(x)+c_{1)}

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(19y^2-xy)dx+x^2dy=0

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

y = \frac{x}{19 ln ( x ) + c _{1}}

Solution steps

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

Solve Bernoulli ODE:

y = \frac{x}{19 ln ( x ) + c _{1}}

y = \frac{x}{19 ln ( x ) + c _{1}}

\int cos (π) t d t

\frac{\partial}{\partial x} (a^{3} - a + x)

\frac{d}{d x} (sin (cos^{2} (x)))

n = 0 \sum \infty (\frac{4}{9})^{n}

x^{2} - 8, 1

What is the solution for (19y^2-xy)dx+x^2dy=0 ?
The solution for (19y^2-xy)dx+x^2dy=0 is y= x/(19ln(x)+c_{1)}