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  • The solution of the differential equation \frac{d y}{d x}=\frac{1+y^{2}}{1+x^{2}} is (a) y=\tan ^{-1} x

The solution of the differential equation \frac{d y}{d x}=\frac{1+y^{2}}{1+x^{2}} is
(a) y=\tan ^{-1} x
(b) y-x=k(1+x y)
(c) x=\tan ^{-1} y
(d) \tan (x y)=k

Answers (1)

The answer is the option (b) y – x = k(1 + xy)

Explanation: -

\begin{aligned} &\Rightarrow \quad \frac{d y}{1+y^{2}}=\frac{d x}{1+x^{2}}\\ &\text { On integrating both sides, we get }\\ &\tan ^{-1} y=\tan ^{-1} x+C\\ &\Rightarrow \quad \tan ^{-1} y-\tan ^{-1} x=C\\ &\Rightarrow \quad \tan ^{-1}\left(\frac{y-x}{1+x y}\right)=C\\ &\Rightarrow \quad \frac{y-x}{1+x y}=\tan C\\ &\Rightarrow \quad y-x=\tan C(1+x y)\\ &\Rightarrow \quad y-x=k(1+x y), \text { where, } k=\tan C \end{aligned}

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