Differential Equations — Class 12 Maths Solution

: The given equation is $\left[ {x{{\sin }^2}\left( {\cfrac{y}{x}} \right) - y} \right]dx + xdy = 0$

$\Rightarrow \cfrac{{dy}}{{dx}} = - \cfrac{{x{{\sin }^2}\cfrac{y}{x} - y}}{x} = \cfrac{{y - x{{\sin }^2}\cfrac{y}{x}}}{x} = \cfrac{y}{x} - {\sin ^2}\cfrac{y}{x}$

…(1)
Put $y = vx$, so that $\cfrac{{dy}}{{dx}} = v + x\cfrac{{dv}}{{dx}}$

$\therefore$ (1) becomes $v + x\cfrac{{dv}}{{dx}} = v - {\sin ^2}v$

$\Rightarrow x\cfrac{{dv}}{{dx}} = - {\sin ^2}v \Rightarrow {\rm{cose}}{{\rm{c}}^2}vdv = - \cfrac{{dx}}{x}$

Integrating both sides,

we get $\int {{\rm{cose}}{{\rm{c}}^2}vdv} = - \int {\cfrac{{dx}}{x}}$

$\Rightarrow - \cot v = - \log |x| + C \Rightarrow \log |x| - \cot v = C$

$\Rightarrow \log |x| - \cot \cfrac{y}{x} = C$

…(2)
When $x = 1,y = \cfrac{\pi }{4}\therefore \log |1| - \cot \cfrac{\pi }{4} = C \Rightarrow 0 - 1 = C \Rightarrow C = - 1$

Putting in (2), $\log |x| - \cot \cfrac{y}{x} = - 1 \Rightarrow \cot \left( {\cfrac{y}{x}} \right) = \log |x| + \log |e|$

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