The particular solution of differential equation $\left(1+y^2\right)(1+\log x) \mathrm{d} x+x \mathrm{~d}…

The particular solution of differential equation $\left(1+y^2\right)(1+\log x) \mathrm{d} x+x \mathrm{~d} y=0$ at $x=1, y=1$ is
  1. $\log x-\frac{1}{2}(\log x)^2-\tan ^{-1} y=-\frac{\pi}{4}$
  2. $\quad \log x+\frac{1}{2}(\log x)^2+\tan ^{-1} y=\frac{\pi}{4}$
  3. $\log x-\frac{1}{2}(\log x)^2+\tan ^{-1} y=\frac{\pi}{4}$
  4. $\log x+\frac{1}{2}(\log x)^2-\tan ^{-1} y=\frac{\pi}{4}$

Solution

$\begin{aligned} & \left(1+y^2\right)(1+\log x) \mathrm{d} x+x \mathrm{~d} y=0 \\ & \Rightarrow\left(1+y^2\right)(1+\log x) \mathrm{d} x=-x \mathrm{~d} y \\ & \Rightarrow\left(\frac{1+\log x}{x}\right) \mathrm{d} x=\left(\frac{-1}{1+y^2}\right) \mathrm{d} y \end{aligned}$
Integrating on both sides, we get $\Rightarrow \int \frac{(1+\log x)}{x} \mathrm{~d} x=-1 \int \frac{1}{1+y^2} \mathrm{~d} y$ $\Rightarrow \int \mathrm{tdt}=-\tan ^{-1} y+\mathrm{c} \quad \cdots\left[\begin{array}{l}\text {Let } 1+\log x=\mathrm{t} \\ \frac{1}{x} \mathrm{~d} x=\mathrm{dt}\end{array}\right]$ $\begin{aligned} & \Rightarrow \frac{\mathrm{t}^2}{2}=-\tan ^{-1} y+\mathrm{c}...(i) \\ & \Rightarrow \frac{(1+\log x)^2}{2}=-\tan ^{-1} y+\mathrm{c} \end{aligned}$ $\begin{aligned} & \text { At } x=1, y=1 \\ & \Rightarrow \frac{(1+\log 1)^2}{2}=-\tan ^{-1}(1)+c \\ & \Rightarrow c=\frac{1}{2}+\frac{\pi}{4} \end{aligned}$
Substituting above value in (i), we get $\begin{aligned} & \frac{(1+\log x)^2}{2}=-\tan ^{-1} y+\frac{1}{2}+\frac{\pi}{4} \\ & \frac{1}{2}+\log x+\frac{(\log x)^2}{2}=-\tan ^{-1} y+\frac{1}{2}+\frac{\pi}{4} \\ & \Rightarrow \log x+\frac{(\log x)^2}{2}+\tan ^{-1} y=\frac{\pi}{4} \end{aligned}$

Asked in: MHT CET 2024 (04 May Shift 2)

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