If the solution curve y=y(x) of the differential equation (1+y^2)(1+ _ex)dx+xdy=0,x0 pass…

Question

If the solution curve $y=y(x)$ of the differential equation $(1+{y}^{2})(1+{\log }_{e}x)dx+xdy=0,x>0$ passes through the point $(1,1)$ and $y(e)=\frac{\alpha -	an (\frac{3}{2})}{\beta +	an (\frac{3}{2})}$, then $\alpha +2\beta $ is \_\_\_\_\_\_\_\_\_.

Answer: 3

Accepted Answer

$\begin{gathered} \int (\frac{1}{x}+\frac{\ln x}{x})dx+\int \frac{dy}{1+{y}^{2}}=0 \ \ln x+\frac{(\ln x{)}^{2}}{2}+{	an }^{-1}y=C \end{gathered}$

Put $x=y=1$

$\begin{gathered} 	herefore C=\frac{\pi }{4} \ \Rightarrow \ln x+\frac{(\ln x{)}^{2}}{2}+{	an }^{-1}y=\frac{\pi }{4} \end{gathered}$

Put $x=e$

$\begin{gathered} \Rightarrow y=	an (\frac{\pi }{4}-\frac{3}{2})=\frac{1-	an \frac{3}{2}}{1+	an \frac{3}{2}} \ 	herefore \alpha =1,\beta =1 \ \Rightarrow \alpha +2\beta =3 \end{gathered}$

Solutions · 1