1. Tardigrade
  2. Question
  3. Mathematics
  4. Let 0<x<π and y(x) be given by (1+ sin x)y3-( cos x)y2+2(1+ sin x)y-2cos x=0 The derivative of y with respect to (x/2) at x=(π /2) is

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Q. Let $ 0
J & K CETJ & K CET 2004

Solution:

Given equation is $ (1+\sin \,x){{y}^{3}}-(\cos x){{y}^{2}}+2(1+\sin x)\,y-2\cos \,x=0 $ Put $ x=\frac{\pi }{2}, $ we get $ \left( 1+\sin \frac{\pi }{2} \right){{y}^{2}}-\left( \cos \frac{\pi }{2} \right){{y}^{2}}+2\left( 1+\sin \frac{\pi }{2} \right)y-2\cos \frac{\pi }{2}=0 $
$ \Rightarrow $ $ 2{{y}^{3}}+4y=0\,\,\Rightarrow \,\,y=0,\,\pm \sqrt{2}i. $ Let $ \frac{x}{2}=t $
$ \therefore $ $ (1+\sin \,2t){{y}^{3}}-(\cos \,2t){{y}^{2}} $ $ +2(1+\sin \,2t)y-2\,\cos \,2t=0 $
On differentiating w.r.t.t, we get $ 3{{y}^{2}}\frac{dy}{dt}(1+\sin \,2t)+2{{y}^{3}}(\cos \,2t) $ $ -2y\frac{dy}{dt}(\cos \,2t)+2{{y}^{2}}\sin 2t+2\frac{dy}{dt}(1+\sin \,2t) $ $ +2y(2\cos 2t)+4\,\sin \,2t=0 $ when $ 2t=\frac{\pi }{2},\,\,y=0, $ $ 2\frac{dy}{dt}(1+1)+4=0 $
$ \Rightarrow $ $ \frac{dy}{dt}=-1 $
$ \Rightarrow $ $ \frac{dy}{\frac{dx}{2}}=-1 $
$ \Rightarrow $ $ \frac{dy}{dx}=-2 $