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DIFFERENTIAL EQUATIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDIFFERENTIAL EQUATIONS4 Marks
Question
For each of the differential equation in find the particular solution satisfying the given condition:.

$\frac{\text{dy}}{\text{dx}}-\frac{\text{y}}{\text{x}}+\text{cosec}\Big(\frac{\text{y}}{\text{x}}\Big)=0;\ \text{y}=0\ \text{when x}=1$

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Answer

Given: Differential equation $\frac{\text{dy}}{\text{dx}}-\frac{\text{y}}{\text{x}}+\cos\text{ec}\ \frac{\text{y}}{\text{x}}=0;\text{y}=0,\text{x}=1$
$\Rightarrow\ \ \frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}-\cos\text{ec}\ \frac{\text{y}}{\text{x}}=f\Big(\frac{\text{y}}{\text{x}}\Big)\ \ ....(\text{i})$
Therefore, the given differential equation is homogeneous.
$\text{Putting}\frac{\text{y}}{\text{x}}=\text{v}\ \ \Rightarrow\ \ \text{y}=\text{vx}\ \ $ $\Rightarrow\ \ \frac{\text{dy}}{\text{dx}}=\text{v}+\text{x}\frac{\text{dv}}{\text{dx}}$
$\text{Putting these values of}\ \frac{\text{y}}{\text{x}}\ \text{and}\ \frac{\text{dy}}{\text{dx}}\ \text{in eq. (ii), we have}$
$\text{v}+\text{x} \frac{\text{dv}}{\text{dx}}=\text{v}-\cos\text{ec v}\ \ \Rightarrow\ \ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{-1}{\sin\text{v}}$
$\Rightarrow\ \ \sin\text{v dv}=-\frac{\text{dx}}{\text{x}}\ \ \big[\text{Separating variables}\big]$
$\text{Integrating both sides,}$ $\int\sin\text{v dv}=-\int\frac{1}{\text{x}}\text{dx}\ \ \Rightarrow\ \ -\cos\text{v}=-\log|\text{x}|+\text{c}$
$\Rightarrow\ \ \cos\text{v}=\log|\text{x}|-\text{c}\ \ $ $\Rightarrow\ \ \cos\frac{\text{y}}{\text{x}}=\log|\text{x}|+\text{c}\ \ \big[\text{Putting}\ \frac{\text{y}}{\text{x}}=\text{v}\big]\ \ ...\text{(ii)}$
Now putting y = 0, x = 1 in eq. (ii), $\cos0=\log1-\text{c}\ \ \Rightarrow\ \ \text{c}=-1$
Putting the value of C in eq. (ii),
$\cos\frac{\text{y}}{\text{x}}=\log|\text{x}|+1\ \ $ $\Rightarrow\ \ \cos\frac{\text{y}}{\text{x}}=\log|\text{x}|+\log\text{e}\ \ \Rightarrow\ \ \cos\frac{\text{y}}{\text{x}}=\log\text{xe}$

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