If the solution curve $y=y(x)$ of the differential equation $\left(1+y^2\right)\left(1+\log _{\mathrm{e}} x\right) d x+x d y=0, x > 0$ passes through the point $(1,1)$ and $$y(e)=\frac{\alpha-\tan \left(\frac{3}{2}\right)}{\beta+\tan \left(\frac{3}{2}\right)}$$, then $\alpha+2 \beta$ is _________.
Answer (integer)
3
Solution
<p>$$\begin{aligned}
& \int\left(\frac{1}{x}+\frac{\ln x}{x}\right) d x+\int \frac{d y}{1+y^2}=0 \\
& \ln x+\frac{(\ln x)^2}{2}+\tan ^{-1} y=C
\end{aligned}$$</p>
<p>Put $x=y=1$</p>
<p>$$\begin{aligned}
& \therefore C=\frac{\pi}{4} \\
& \Rightarrow \ln x+\frac{(\ln x)^2}{2}+\tan ^{-1} y=\frac{\pi}{4}
\end{aligned}$$</p>
<p>Put $x=e$</p>
<p>$$\begin{aligned}
& \Rightarrow \mathrm{y}=\tan \left(\frac{\pi}{4}-\frac{3}{2}\right)=\frac{1-\tan \frac{3}{2}}{1+\tan \frac{3}{2}} \\
& \therefore \alpha=1, \beta=1 \\
& \Rightarrow \alpha+2 \beta=3
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Differential Equations · Topic: Order and Degree
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