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Differential Equations — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations5 Marks
Question
For the following differntial equations verify that the accompanying function is a solution:
Differential equation Function
$\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}=\text{y}^2$ $\text{y}=\frac{\text{a}}{\text{x}+\text{a}}$
✓

Answer

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

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