If the solution curve y=y(x) of (1+y^2)(1+_ex),dx+x,dy=0, x0, passes through (1,1) and y(…

Question

If the solution curve $y=y(x)$ of $(1+y^2)(1+\log_ex)\,dx+x\,dy=0$, $x>0$, passes through $(1,1)$ and $y(e)=\dfrac{\alpha-	an3}{\beta+	an3}$, then $\alpha+2\beta$ is \_\_\_\_\_\_.

Answer: 3

Accepted Answer

Separating variables: $\dfrac{dy}{1+y^2} = -\dfrac{(1+\ln x)}{x}\,dx$. Integrating:
$$\tan^{-1}y = -\left(\ln x+\frac{(\ln x)^2}{2}\right)+C.$$

Using $(1,1)$: $\dfrac{\pi}{4}=C$. At $x=e$: $\tan^{-1}y = \dfrac{\pi}{4}-\dfrac32$, leading (via tangent subtraction formula) to the stated form with $\alpha=1,\beta=1$, so $\alpha+2\beta=3$.

Solutions · 1