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Differential Equations — Maths STD 12 Science — Question

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

We have, $\text{y dx}+\Big\{\text{x}\log\Big(\frac{\text{y}}{\text{x}}\Big)\Big\}\text{dy}-2\text{x dy}=0$ $\Rightarrow\ \Big\{2\text{x}-\text{x}\log\Big(\frac{\text{y}}{\text{x}}\Big)\Big\}\text{dy}=\text{y dx}$ $\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{2\text{x}-\text{x}\log\big(\frac{\text{y}}{\text{x}}\big)}$ This is a homogeneous differential equation. Put y = vx and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}},$ we get$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{vx}}{2\text{x}-\text{x}\log\text{v}}$
$\Rightarrow\ \text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}}{2-\log\text{v}}$ $\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}}{2-\log\text{v}}-\text{v}$ $\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}-2\text{v + v}\log\text{v}}{2-\log\text{v}}$ $\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}\log\text{v}-\text{v}}{2-\log\text{v}}$ $\Rightarrow\ \frac{2-\log\text{v}}{\text{v}\log\text{v}-\text{v}}\text{dv}=\frac{1}{\text{x}}\text{dx}$ integrating both sides, we get $\int\frac{2-\log\text{v}}{\text{v}\log\text{v}-\text{v}}\text{dv}=\int\frac{1}{\text{x}}\text{dx}$ $\Rightarrow\ \int\frac{1-(\log\text{v}-1)}{\text{v}(\log\text{v}-1)}\text{dv}=\int\frac{1}{\text{x}}\text{dx}$ Putting $\log\text{v}-1=\text{t}$ $\Rightarrow\ \frac{1}{\text{v}}\text{dv}=\text{dt}$ $\therefore\ \int\frac{1-\text{t}}{\text{t}}\text{dt}=\int\frac{1}{\text{x}}\text{dx}$ $\Rightarrow\ \int\Big(\frac{1}{\text{t}}-1\Big)\text{dt}=\int\frac{1}{\text{x}}\text{dx}$ $\Rightarrow\ \log|\text{t}|-\text{t}=\log|\text{x}|+\log\text{C}$ $\Rightarrow\ \log|\log\text{v}-1|-(\log\text{v}-1)=\log|\text{x}|+\log\text{C}$ $\Rightarrow\ \log|\log\text{v}-1|-\log\text{v}=\log|\text{x}|+\log\text{C}$ $\big($where, $\log\text{C}_1=\log\text{C}-1\big)$ $\Rightarrow\ \log\Big|\frac{\log\text{v}-1}{\text{v}}\Big|=\log|\text{C}_1\text{x}|$ $\Rightarrow\ \frac{\log\text{v}-1}{\text{v}}=\text{C}_1\text{x}$ $\Rightarrow\ \log\text{v}-1=\text{C}_1\text{xv}$ Putting $\text{v}=\frac{\text{y}}{\text{x}},$ we get $\log\frac{\text{y}}{\text{x}}-1=\text{C}_1\text{x}\times\frac{\text{y}}{\text{x}}$ $\Rightarrow\ \log\frac{\text{y}}{\text{x}}-1=\text{C}_1\text{y}$ Hence, $\log\frac{\text{y}}{\text{x}}-1=\text{C}_1\text{y}$ is the required solution.

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