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Differential Equations — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations5 Marks
Question
Show that $\text{y}=\frac{\text{c}-\text{x}}{1+\text{cx}}$ is a solution of the differential equation $(1+\text{x}^2)\frac{\text{dy}}{\text{dx}}+(1+\text{y}^2)=0.$
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Answer

We have,
$\text{y}=\frac{\text{c}-\text{x}}{1+\text{cx}}\ ...(1)$
Differential both sides of (1) with respect to x, we get
$\frac{\text{dy}}{\text{dx}}=\frac{(1+\text{cx})(-1)-(\text{c}-\text{x})(\text{c})}{(1+\text{cx})^2}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-1-\text{cx}-\text{c}^2+\text{cx}}{(1+\text{cx})^2}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1+\text{c}^2}{(1+\text{cx})^2}\ ...(2)$
Now,
$(1+\text{x}^2)\frac{\text{dy}}{\text{dx}}+(1+\text{y}^2)$
$=-(1+\text{x}^2)\frac{(1+\text{c}^2)}{(1+\text{cx})^2}+\Big\{1+\frac{(1+\text{c}^2)}{(1+\text{cx})^2}\Big\}$
$=-\frac{(1+\text{x}^2)(1+\text{c}^2)}{(1+\text{cx})^2}+\frac{(1+\text{cx})^2+(\text{c}-\text{x})^2}{(1+\text{cx})^2}$
$=-\frac{(1+\text{x}^2)(1+\text{c}^2)}{(1+\text{cx})^2}+\frac{1+2\text{cx}+\text{c}^2\text{x}^2+\text{c}^2-2\text{cx}+\text{x}^2}{(1+\text{cx})^2}$
$=-\frac{(1+\text{x}^2)(1+\text{c}^2)}{(1+\text{cx})^2}+\frac{(1+\text{x})^2+\text{c}^2(1+\text{x})^2}{(1+\text{cx})^2}$
$=-\frac{(1+\text{x}^2)(1+\text{c}^2)}{(1+\text{cx})^2}+\frac{(1+\text{x}^2)(1+\text{c}^2)}{(1+\text{cx})^2}=0$
Hence, the given function is the solution to the given differential equation.

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