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DIFFERENTIAL EQUATIONS — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDIFFERENTIAL EQUATIONS3 Marks
Question
For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.
$\text{x}^{2}=2\text{y}^2\log\text{y}$ : $(\text{x}^2+\text{y}^2)\frac{\text{dy}}{\text{dx}}-\text{xy}=0$
✓

Answer

The given differential equation is
$\text{x}^2=2\text{y}^2\log\text{y}\ \ ...(1)$
$\therefore\ \ 2\text{x}=2\Big\{\text{y}^2\times\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}+\log\text{y}\Big[2\text{y}\frac{\text{dy}}{\text{dx}}\Big]\Big\}$
$\text{or}\ \ 2\text{x}=2(\text{y}+2\text{y}\log\text{y})\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\ \ \text{x}=(\text{y}+2\text{y}\log\text{y})\frac{\text{dy}}{\text{dx}}$
Multiplying both sides by y, we get
$\text{xy}=(\text{y}^2+2\text{y}^2\log\text{y})\frac{\text{dy}}{\text{dx}}$
$\text{or}\ \ \text{xy}=(\text{y}^2+\text{x}^2)\frac{\text{dy}}{\text{dx}}\ \ [\because \text{of}\ (1)]$
$\text{or}\ \ (\text{x}^2+\text{y}^2)\frac{\text{dy}}{\text{dx}}-\text{xy}=0$
Hence the result.

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