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Differential Equations — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations4 Marks
Question
Solve the following initial value problems:
$\frac{\text{dy}}{\text{dx}}-\frac{\text{y}}{\text{x}}+\text{cosec}\frac{\text{y}}{\text{x}}=0,\text{y}(1)=0$
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Answer

$\frac{\text{dy}}{\text{dx}}-\frac{\text{y}}{\text{x}}+\text{cosec}\frac{\text{y}}{\text{x}}=0,\text{y}(1)=0$
This is a homogeneous equation, Put y = vx
$\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$
$\text{v + x}\frac{\text{dv}}{\text{dx}}-\text{v + cosec v}=0$
$\text{x}\frac{\text{dv}}{\text{dx}}=\text{cosec v}$
$\frac{\text{dv}}{\text{cosec v}}=\frac{\text{dx}}{\text{x}}$
$\sin\text{v dv}=\frac{\text{dx}}{\text{x}}$
On integrating both sides, we get
$\int\sin\text{v dv}=\int\frac{\text{dx}}{\text{x}}$
$-\cos\text{v}=\log_{\text{e}}\text{x + C}$
$-\cos\text{v}+\log_{\text{e}}\text{x}=\text{C}$
$\cos\text{v}+\log_{\text{e}}\text{x}=-\text{C}$
$\cos\Big(\frac{\text{y}}{\text{x}}\Big)+\log_{\text{e}}\text{x}=-\text{C}$
As y(1) = 0
$\cos\Big(\frac{0}1\Big)=0+\log_{\text{e}}1=-\text{C}$
$1+0=-\text{C}$
$\Rightarrow\ \text{C}=-1$
$\Rightarrow\ \cos\Big(\frac{\text{y}}{\text{x}}\Big)+\log_{\text{e}}\text{x}=1$

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