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Differential Equations — Maths STD 12 Science — Question

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

Answer

Correct option: D.
$\text{y}=\frac{1-\text{x}}{1+\text{x}}$
We have,
$(\text{x}^{2}+1)\frac{\text{dy}}{\text{dx}}=-(\text{y}^{2}+1)=0$
$\Rightarrow (\text{x}^{2}+1)\frac{\text{dy}}{\text{dx}}=-(\text{y}^{2}+1)$
$\Rightarrow \frac{1}{(\text{y}^{2}+1)}\text{dy}=-\frac{1}{(\text{x}^{2}+1)}\text{dx}$
Intergrating both sides, we get
$\Rightarrow \int\frac{1}{(\text{y}^{2}+1)}\text{dy}=-\int\frac{1}{(\text{x}^{2}+1)}\text{dx}$
$\Rightarrow \tan^{-1}\text{y}=-\tan^{-1}\text{x}+\tan^{-1}\text{C}$
$\Rightarrow \tan^{-1}\text{y}+\tan^{-1}\text{x}=\tan^{-1}\text{C}$
$\Rightarrow \tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big) =\tan^{-1}\text{C}$
$\Rightarrow \frac{\text{x}+\text{y}}{1+\text{xy}}=\text{C}$
Disclaimer : The initial value given,
So the find will be $C = 1,$ So
$\Rightarrow \text{x}+\text{y}=1-\text{xy}$
$\Rightarrow \text{y}+\text{xy}=1-\text{x}$
$\Rightarrow \text{y}(1+\text{x})=1-\text{x}$
$\Rightarrow \text{y}=\frac{1-\text{x}}{1+\text{x}}$

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