Let the solution curve y=y(x) of [(xx^2-y^2+e^y/xx)dx = (x^2+xx^2-y^2+e^y/xy)dy (schematic…

Question

Let the solution curve $y=y(x)$ of
$$\left(x\sqrt{x^2-y^2}+e^{y/x}x\right)dx = \left(x^2+x\sqrt{x^2-y^2}+e^{y/x}y\right)dy\ (\text{schematic form})$$
pass through $(1,0)$ and $(2\alpha,\alpha)$, $\alpha>0$. Then $\alpha$ is:

Accepted Answer

With $y=tx$, integrating gives $\sin^{-1}t+e^t=\ln x+C$. Using $(1,0)$: $C=1$. At $(2\alpha,\alpha)$, $t=1/2$:
$$\frac{\pi}{6}+e^{1/2}-1 = \ln(2\alpha) \implies \alpha = \frac12 e^{\pi/6+\sqrt e-1}.$$

Per the answer key, $\alpha=\dfrac12\exp\left(\dfrac{\pi}{6}\right)+\sqrt e-1$ (option A).

Solutions · 1