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Differential Equations — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations5 Marks
Question
Solve the following differential equation:
$\text{xy}\log\Big(\frac{\text{y}}{\text{x}}\Big)\text{dx}+\Big\{\text{y}^2-\text{x}^2\log\Big(\frac{\text{y}}{\text{x}}\Big)\Big\}\text{dy}=0$
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Answer

$\text{xy}\log\Big(\frac{\text{y}}{\text{x}}\Big)\text{dx}+\Big\{\text{y}^2-\text{x}^2\log\Big(\frac{\text{y}}{\text{x}}\Big)\Big\}\text{dy}=0$ $\Rightarrow\ \text{xy}\log\Big(\frac{\text{y}}{\text{x}}\Big)\text{dx}=-\Big\{\text{y}^2-\text{x}^2\log\Big(\frac{\text{y}}{\text{x}}\Big)\Big\}\text{dy}$$\Rightarrow\ \frac{\text{dx}}{\text{dy}}=\frac{-\big\{\text{y}^2-\text{x}^2\log\big(\frac{\text{y}}{\text{x}}\big)\big\}}{\text{xy}\log\big(\frac{\text{x}}{\text{y}}\big)}$
$=\frac{\text{x}^2\log\big(\frac{\text{x}}{\text{y}}\big)-\text{y}^2}{\text{xy}\log\big(\frac{\text{x}}{\text{y}}\big)}$
It is a homogeneous equation. We put x = vy $\frac{\text{dx}}{\text{dy}}=\text{v + y}\frac{\text{dv}}{\text{dy}}$ So, $\text{v + y}\frac{\text{dv}}{\text{dy}}=\frac{\text{v}^2\text{y}^2\log(\text{v})-\text{y}^2}{\text{vy}^2\log(\text{v})}$ $\text{v + y}\frac{\text{dv}}{\text{dy}}=\frac{\text{v}^2\log(\text{v})-1}{\text{v}\log(\text{v})}$ $\Rightarrow\ \text{y}\frac{\text{dv}}{\text{dy}}=\frac{\text{v}^2\log(\text{v})-1}{\text{v}\log(\text{v})}-\text{v}$ $\Rightarrow\ \text{y}\frac{\text{dv}}{\text{dy}}=\frac{\text{v}^2\log(\text{v})-1-\text{v}^2\log(\text{v})}{\text{v}\log(\text{v})}$ $\Rightarrow\ \text{y}\frac{\text{dv}}{\text{dy}}=\frac{-1}{\text{v}\log(\text{v})}$ $\Rightarrow\ \text{v}\log(\text{v})\text{dv}=\frac{-1}{\text{y}}\text{dy}$ On integrating both sides we get, $\int\text{v}\log(\text{v})\text{dv}=-\int\frac{1}{\text{y}}\text{dy}$ $\Rightarrow\ \frac{\text{v}^2}2\log(\text{v})-\int\frac{\text{v}}2\text{dv}=-\log\text{y + C}$ $\Rightarrow\ \frac{\text{v}^2}2\log(\text{v})-\frac{\text{v}^2}4=-\log\text{y + C}$ $\Rightarrow\ \frac{\text{v}^2}2\Big[\log(\text{v})-\frac{1}2\Big]=-\log\text{y + C}$ $\Rightarrow\ \text{v}^2\Big[\log(\text{v})-\frac{1}2\Big]=-2\log\text{y + C}$ Now putting back the values of v as $\frac{\text{x}}{\text{y}}$ we get, $\frac{\text{x}^2}{\text{y}^2}\Big[\log(\text{v})-\frac{1}2\Big]+\log\text{y}^2=\text{C}$

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