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DIFFERENTIAL EQUATIONS — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDIFFERENTIAL EQUATIONS1 Mark
Question
Choose the correct answer from the given four option.
$\tan^{-1}+\tan^{-1}\text{y}=\text{C}$ is the general solution of the differential equation:
  1. $\frac{\text{d}\text{y}}{\text{d}\text{x}}=\frac{1+\text{y}^2}{1+\text{x}^2}$
  2. $\frac{\text{d}\text{y}}{\text{d}\text{x}}=\frac{1+\text{x}^2}{1+\text{y}^2}$
  3. $(1+\text{x}^2)\text{dy}+(1+\text{y}^2)\text{dx}=0$
  4. $(1+\text{x}^2)\text{dx}+(1+\text{y}^2)\text{dy}=0$
✓

Answer

  1. $(1+\text{x}^2)\text{dy}+(1+\text{y}^2)\text{dx}=0$
Solution:
Given is, $\tan^{-1}+\tan^{-1}\text{y}=\text{C}$
On differentiating above eqaution w.r. t. x, we get
$\frac{1}{1+\text{x}^2}+\frac{1}{1+\text{y}^2}.\frac{\text{d}\text{y}}{\text{d}\text{x}}=0$
$\Rightarrow\frac{1}{1+\text{y}^2}.\frac{\text{d}\text{y}}{\text{d}\text{x}}=-\frac{1}{1+\text{x}^2}$
$\Rightarrow(1+\text{x}^2)\text{dy}+(1+\text{y}^2)\text{dx}=0$

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