Let y=y(x) solve (1-x^2y^2),dx=y,dx+x,dy.

Question

Let $y=y(x)$ solve $(1-x^2y^2)\,dx=y\,dx+x\,dy$. If the line $x=1$ intersects the curve at $y=2$ and the line $x=2$ intersects it at $y=\alpha$, then a value of $\alpha$ is:

Accepted Answer

$$dx = \frac{d(xy)}{1-(xy)^2} \implies x = \frac12\ln\left|\frac{1+xy}{1-xy}\right|+C.$$

Using $x=1,y=2$: $C=1-\dfrac12\ln3$. At $x=2,y=\alpha$, solving gives $\alpha=\dfrac{3e^2-1}{2(3e^2+1)}$ or equivalently $\alpha=\dfrac{1+3e^2}{2(3e^2-1)}$ (matching option A).

Solutions · 1