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Differential Equations — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations1 Mark
MCQ
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 :
  • A
    $\frac{\text{d}\text{y}}{\text{d}\text{x}}=\frac{1+\text{y}^2}{1+\text{x}^2}$
  • B
    $\frac{\text{d}\text{y}}{\text{d}\text{x}}=\frac{1+\text{x}^2}{1+\text{y}^2}$
  • ✓
    $(1+\text{x}^2)\text{dy}+(1+\text{y}^2)\text{dx}=0$
  • D
    $(1+\text{x}^2)\text{dx}+(1+\text{y}^2)\text{dy}=0$
✓

Answer

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