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DIFFERENTIAL EQUATIONS — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDIFFERENTIAL EQUATIONS5 Marks
Question
Solve the following initial value problems:
$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}(\text{x}+2\text{y})}{\text{x}(2\text{x}+\text{y})},\text{y}(1)=2$
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Answer

$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}(\text{x}+2\text{y})}{\text{x}(2\text{x}+\text{y})},\text{y}(1)=2$
This is a homogeneous equation, put y = vx
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$
$\Rightarrow\ \text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}(\text{x}+2\text{vx})}{(2\text{x + vx})}$
$\Rightarrow\ \text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}(1+2\text{v})}{(2+\text{v})}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}+2\text{v}^2-2\text{v}-\text{v}^2}{(2+\text{v})}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}^2-\text{v}}{(2+\text{v})}$
$\Rightarrow\ \frac{(2+\text{v})\text{dv}}{(\text{v}^2-\text{v})}=\frac{\text{dx}}{\text{x}}$
On integrating both sides of the equation we get,
$\int\frac{2+\text{v}}{(\text{v}^2-\text{v})}\text{dv}=\int\frac{\text{dx}}{\text{x}}$
$\Rightarrow\ \int\frac{2}{\text{v}(\text{v}-1)}\text{dv}+\int\frac{\text{v}}{\text{v}(\text{v}-1)}\text{dv}=\int\frac{\text{dx}}{\text{x}}$
$\Rightarrow\ 2\Big[\int\frac{1}{(1-\text{v})}\text{dv}-\int\frac{1}{\text{v}}\text{dv}\Big]+\int\frac{1}{\text{v}-1}\text{dv}=\log_{\text{e}}\text{x + C}$
$\Rightarrow\ 2\big[\log_{\text{e}}(\text{v}-1)-\log_{\text{e}}\text{v}\big]+\log_{\text{e}}(\text{v}-1)=\log_{\text{e}}\text{x + C}$
$2\Big[\log_{\text{e}}\Big(\frac{\text{v}-1}{\text{v}}\Big)\Big]+\log_{\text{e}}(\text{v}-1)=\log_{\text{e}}\text{x + C}$
$2\log_{\text{e}}\Big(\frac{\text{y}-\text{x}}{\text{y}}\Big)+\log_{\text{e}}\Big(\frac{\text{y}-\text{x}}{\text{x}}\Big)=\log_{\text{e}}\text{x + C}$
As y(1) = 2
$2\log_{\text{e}}\Big(\frac{2-1}{2}\Big)+\log_{\text{e}}\Big(\frac{2-1}{1}\Big)=\log_{\text{e}}1+\text{C}$
$2\log_{\text{e}}\frac{1}2+\log_{\text{e}}1=\log_{\text{e}}1+\text{C}$
$-2\log_{\text{e}}2+0=0+\text{C}$
$-2\log_{\text{e}}2=\text{C}$
$\therefore\ 2\log_{\text{e}}\Big(\frac{\text{y}-\text{x}}{\text{y}}\Big)+\log_{\text{e}}\Big(\frac{\text{y}-\text{x}}{\text{x}}\Big)=\log_{\text{e}}\text{x}-2\log_{\text{e}}2$

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