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DIFFERENTIAL EQUATIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDIFFERENTIAL EQUATIONS1 Mark
Question
Choose the correct answer from the given four option.
The solution of the differential equation $\frac{\text{d}\text{y}}{\text{d}\text{x}}=\frac{1+\text{y}^2}{1+\text{x}^2}$ is:
  1. $\text{y}=\tan^{-1}\text{x}$
  2. $\text{y}-\text{x}=\text{k}(1+\text{xy})$
  3. $\text{x}=\tan^{-1}\text{y}$
  4. $\tan(\text{xy})=\text{k}$
✓

Answer

  1. $\text{y}-\text{x}=\text{k}(1+\text{xy})$

Solution:

Given that, $\frac{\text{d}\text{y}}{\text{d}\text{x}}=\frac{1+\text{y}^2}{1+\text{x}^2}$

$\Rightarrow\frac{\text{dy}}{1+\text{y}^2}=\frac{\text{dx}}{1+\text{x}^2}$

On integrating both sides, we get

$\tan^{-1}\text{y}=\tan^{-1}\text{x}+\text{C}$

$\Rightarrow\tan^{-1}\text{y}-\tan^{-1}\text{x}=\text{C}$

$\Rightarrow\tan^{-1}\Big(\frac{\text{y}-\text{x}}{1+\text{xy}}\Big)=\text{C}$

$\Rightarrow\frac{\text{y}-\text{x}}{1+\text{xy}}=\tan\text{C}$

$\Rightarrow\text{y}-\text{x}=\tan\text{C}(1+\text{xy})$

$\Rightarrow\text{y}-\text{x}=\text{K}(1+\text{xy})$

Where, $\text{k}=\tan\text{C}$

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