Choose the correct answer from the given four option. The solutio…

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:

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

✓

Answer

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

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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