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Differential Equations — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations1 Mark
MCQ
The solution of the differention equation $(1+\text{x}^{2})\frac{\text{dy}}{\text{dx}}+1+\text{y}^{2}=0$ is:
  • A
    $\tan^{-1}\text{x}-\tan^{-1}\text{y}=\tan^{-1}\text{C}$
  • B
    $\tan^{-1}\text{y}-\tan^{-1}\text{x}=\tan^{-1}\text{C}$
  • C
    $\tan^{-1}\text{y}\pm\tan^{-1}\text{x}=\tan^{-1}\text{C}$
  • ✓
    $\tan^{-1}\text{y}+\tan^{-1}\text{x}=\tan^{-1}\text{C}$
✓

Answer

Correct option: D.
$\tan^{-1}\text{y}+\tan^{-1}\text{x}=\tan^{-1}\text{C}$
We have,
$(1+\text{x}^{2})\frac{\text{dy}}{\text{dx}}+1+\text{y}^{2}=0$
$\Rightarrow (1+\text{x}^{2})\frac{\text{dy}}{\text{dx}}=-(1+\text{y}^{2})$
$\Rightarrow \frac{1}{(1+\text{y}^{2})}\text{dy}=-\frac{1}{(1+\text{x}^{2})}\text{dx}$
Integrating both sides, we get
$\int\frac{1}{(1+\text{y}^{2})}\text{dy}=-\int\frac{1}{(1+\text{x}^{2})}\text{dx}$
$\Rightarrow \tan^{-1}\text{y}=-\tan^{-1}\text{x}+\tan^{-1}\text{C}$
$\Rightarrow \tan^{-1}\text{y}+\tan^{-1}\text{x}=\tan^{-1}\text{C}$

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