If the curve y=y(x) satisfying (2xy^2-y),dx+x,dy=0 passes through the intersection of 2x-3…

Question

If the curve $y=y(x)$ satisfying $(2xy^2-y)\,dx+x\,dy=0$ passes through the intersection of $2x-3y=1$ and $3x+2y=8$, then $|y(1)|$ is \_\_\_\_\_\_.

Answer: 1

Accepted Answer

Bernoulli: $\dfrac{dx}{dy}$-form with $v=1/y$: $x^2+c=\dfrac{x}{y}$. Intersection of the two lines: $(2,1)$. Using this: $c=-2$, so $\dfrac{x}{y}=x^2-2$, i.e. $y=\dfrac{x}{x^2-2}$.

At $x=1$: $y=\dfrac{1}{-1}=-1$, so $|y(1)|=1$.

Solutions · 1