Differential Equations Question 10
Question 10 - 2024 (29 Jan Shift 1)
If the solution curve $y=y(x)$ of the differential equation $\left(1+y^{2}\right)\left(1+\log _{e} x\right) d x+x d y=0, x>0$ 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
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Answer (3)
Solution
$\int\left(\frac{1}{x}+\frac{\ln x}{x}\right) d x+\int \frac{d y}{1+y^{2}}=0$
$\ln x+\frac{(\ln x)^{2}}{2}+\tan ^{-1} y=C$
Put $x=y=1$
$\therefore \mathrm{C}=\frac{\pi}{4}$
$\Rightarrow \ln x+\frac{(\ln x)^{2}}{2}+\tan ^{-1} y=\frac{\pi}{4}$
Put $\mathrm{x}=\mathrm{e}$
$\Rightarrow \mathrm{y}=\tan \left(\frac{\pi}{4}-\frac{3}{2}\right)=\frac{1-\tan \frac{3}{2}}{1+\tan \frac{3}{2}}$
$\therefore \alpha=1, \beta=1$
$\Rightarrow \alpha+2 \beta=3$
