class 12 maths differential equations

$\cfrac{{dy}}{{dx}} - \cfrac{y}{x} + {\rm{cosec}}\left( {\cfrac{y}{x}} \right) = 0;y = 0$ when $x = 1$

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📘 Differential Equations NCERT Ex.9.5,Q.14,Page 406 SA

$\cfrac{{dy}}{{dx}} - \cfrac{y}{x} + {\rm{cosec}}\left( {\cfrac{y}{x}} \right) = 0;y = 0$ when $x = 1$

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: The given equation is $\cfrac{{dy}}{{dx}} - \cfrac{y}{x} + {\rm{cosec}}\left( {\cfrac{y}{x}} \right) = 0$

…(1)
Put $y = vx$,

so that $\cfrac{{dy}}{{dx}} = v + x\cfrac{{dv}}{{dx}}$

$\therefore$ (1) becomes $v + x\cfrac{{dv}}{{dx}} - v + {\rm{cosec}}v = 0$

$\Rightarrow x\cfrac{{dv}}{{dx}} = - {\rm{cosec}}v \Rightarrow - \sin vdv = \cfrac{{dx}}{x}$

Integrating both sides,

we get
$\cos v = \log |x| + C \Rightarrow \cos \cfrac{y}{x} = \log |x| + C$

…(2)
When $x = 1,y = 0\therefore \cos 0 = \log |1| + C \Rightarrow C = 1$

Putting $C = 1$ in (2) ,

we get $\cos \cfrac{y}{x} = \log |x| + 1$
$\Rightarrow \cos \cfrac{y}{x} = \log |x| + \log |e|$

$\Rightarrow \cos \left( {\cfrac{y}{x}} \right) = \log |ex|$, which is the required solution.

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