The curve satisfying y,dx-(x+3y^2)dy=0, through (1,1), also passes through:

Question

The curve satisfying $y\,dx-\left(x+3y^2\right)dy=0$, through $(1,1)$, also passes through:

Accepted Answer

$$\frac{dx}{dy}-\frac{x}{y} = 3y.$$

I.F.$=1/y$. $\dfrac{x}{y}=3y+c$; using $(1,1)$: $c=-2$. So $x=3y^2-2y$, satisfied by $\left(-\dfrac13,\dfrac13\right)$.

Solutions · 1