The general solution of the differential equation $\frac{1}{x} \frac{\mathrm{~d} y}{\mathrm{~d} x}=\tan…

The general solution of the differential equation $\frac{1}{x} \frac{\mathrm{~d} y}{\mathrm{~d} x}=\tan ^{-1} x$ is
  1. $y+\frac{x^2 \tan ^{-1} x}{2}+\mathrm{c}=0$, where c is a constant of integration.
  2. $y+x \tan ^{-1} x+\mathrm{c}=0$, where c is a constant integration.
  3. $y-x-\tan ^{-1} x+\mathrm{c}=0$, where is a constant of integration.
  4. $y=\frac{x^2 \tan ^{-1} x}{2}-\frac{1}{2}\left(x-\tan ^{-1} x\right)+\mathrm{c}$, where c is constant of integration.

Solution

$\begin{aligned} & \frac{1}{x} \frac{\mathrm{~d} y}{\mathrm{~d} x}=\tan ^{-1} x \\ & \mathrm{~d} y=x \cdot \tan ^{-1} x \mathrm{~d} x \end{aligned}$
Integrating on both sides we get, $\begin{aligned} & y=\int x \cdot \tan ^{-1} x \mathrm{~d} x \\ & y=\tan ^{-1} x \int x \mathrm{~d} x-\int\left(\frac{\mathrm{d}}{\mathrm{~d} x} \tan ^{-1} x \cdot \int x \mathrm{~d} x\right) \mathrm{d} x \\ & y=\tan ^{-1} x \cdot \frac{x^2}{2}-\int\left(\frac{1}{1+x^2} \times \frac{x^2}{2}\right) \mathrm{d} x \\ & y=\tan ^{-1} x \cdot \frac{x^2}{2}-\frac{1}{2} \int\left(\frac{x^2}{1+x^2}\right) \mathrm{d} x \\ & =\frac{x^2 \cdot \tan ^{-1} x}{2}-\frac{1}{2}\left(\int \frac{x^2+1}{x^2+1}-\int \frac{1}{1+x^2} \mathrm{~d} x\right) \\ & \Rightarrow y=\frac{x^2 \tan ^{-1} x}{2}-\frac{1}{2}\left(x-\tan ^{-1} x\right)+\mathrm{c} \end{aligned}$

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

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