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Differential Equations — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations5 Marks
Question
Find the particular solution of the differential equation $\frac{\text{dy}}{\text{dx}}=\frac{\text{xy}}{\text{x}^2+\text{y}^2}$ given that y = 1 when x = 0.
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Answer

$\frac{\text{dy}}{\text{dx}}=\frac{\text{xy}}{\text{x}^2+\text{y}^2}\ \dots(1)$
Let y = xv
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$
Substituting the value of y = xv and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$ in (1), we get
$\therefore\ \text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{x}^2\text{v}}{\text{x}^2+\text{x}^2\text{v}^2}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}}{1+\text{v}^2}-\text{v}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{-\text{v}^3}{1+\text{v}^2}$
$\Rightarrow\ \frac{1+\text{v}^2}{-\text{v}^3}\text{dv}=\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \int\frac{1+\text{v}^2}{-\text{v}^3}\text{dv}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \frac{1}{2\text{v}^2}-\log\text{v}=\log\text{x +C}$
$\Rightarrow\ \frac{1}{2\big(\frac{\text{y}}{\text{x}}\big)^2}-\log\frac{\text{y}}{\text{x}}=\log\text{x +C}$
$\Rightarrow\ \frac{\text{x}^2}{2\text{y}^2}-\log\frac{\text{y}}{\text{x}}=\log\text{x + C}\ \dots(2)$
$\Rightarrow\ \frac{0}2-\log\frac{1}0=\log0+\text{C}$
$\Rightarrow\ \text{C}=0$
Substituting the value of C in (2), we get
$\frac{\text{x}^2}{2\text{y}^2}-\log\frac{\text{y}}{\text{x}}=\log{\text{x}}$
$\Rightarrow\ \frac{\text{x}^2}{2\text{y}^2}=\log{\text{x}}+\log\frac{\text{y}}{\text{x}}$
$\Rightarrow\ \frac{\text{x}^2}{2\text{y}^2}=\log{\text{y}}$

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