A) \[-\frac{1}{2}\]
B) \[\frac{1}{2}\]
C) \[-2\]
D) \[2\]
Correct Answer: C
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 \[2t=\frac{\pi }{2},\,\,y=0,\] \[2\frac{dy}{dt}(1+1)+4=0\] \[\Rightarrow \] \[\frac{dy}{dt}=-1\] \[\Rightarrow \] \[\frac{dy}{\frac{dx}{2}}=-1\] \[\Rightarrow \] \[\frac{dy}{dx}=-2\]
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