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Differential Equations — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations5 Marks
Question
Solve the following differential equations $(1+\text{x}^2)\frac{\text{dy}}{\text{dx}}+(1+\text{y}^2)=0,$ given that $\text{y}=1,$ when $\text{x}=0.$
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Answer

We have,
$(1+\text{x}^2)\frac{\text{dy}}{\text{dx}}+(1+\text{y}^2)=0,\text{y}=1$ when $\text{x}=0.$
$\Rightarrow(1+\text{x}^2)\frac{\text{dy}}{\text{dx}}=-(1+\text{y}^2)$
$\Rightarrow\frac{1}{1+\text{y}^2}\text{dy}=-\frac{1}{(1+\text{x}^2)}\text{dx}$
Integrating both sides, we get
$\int \frac{1}{1+\text{y}^2}\text{dy}=-\int\frac{1}{(1+\text{x}^2)}\text{dx}$
$\Rightarrow\tan^{-1}\text{y}=-\tan^{-1}\text{x + C}$
$\Rightarrow\tan^{-1}\text{y}+\tan^{-1}\text{x = C}...(1)$
Given: $\text{x}=0,\text{y}=1.$
Substituting the valuse of x and y in (1), we get
$\frac{\pi}{4}+0=\text{C}$
$\Rightarrow\text{C}=\frac{\pi}{4}$
Substituting the value of C in (1), we get
$\tan^{-1}\text{y}+\tan^{-1}\text{x}=\frac{\pi}{4}$
$\Rightarrow\tan^{-1}\text{x}+\tan^{-1}\text{y}=\frac{\pi}{4}$
$\Rightarrow\tan^{-1}\Big(\frac{\text{x + y}}{1-\text{xy}}\Big)=\frac{\pi}{4}$
$\Rightarrow\frac{\text{x}+\text{y}}{1-\text{xy}}=1$
$\Rightarrow\text{x+y}=1-\text{xy}$
Hence, $\text{x+y}=1-\text{xy}$ is the required solution.

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