A) \[y\sin \,x+\frac{1}{2}\,\text{cosec x}\]
B) \[y=\tan \,(x/2)+\cot (x/2)\]
C) \[y=(1/\sqrt{2})\,\sec \,(x/2)+\sqrt{2}\,\cos \,\,(x/2)\]
D) None of the above
Correct Answer: A
Solution :
Given, \[dy=\cos \,x(2-y\,\text{cosec x)dx}\] \[\Rightarrow \] \[\frac{dy}{dx}=2\,\cos \,x-y\,cos\,x\,cosec\,x\,\] \[\Rightarrow \] \[\frac{dy}{dx}+y\,\cot \,x=2\,\cos x\] Here, \[\frac{dy}{dx}+y\,\cot \,x=2\,\cos x\] \[\therefore \] IF \[{{e}^{\int{P\,\,dx}}}={{e}^{\int{\cot \,x\,dx}}}={{e}^{\log (\sin x)}}=\sin x\] Now, solution is y. IF \[=\int{Q.\,IF\,\,dx+c}\] \[\Rightarrow \] \[y\,\sin \,x=\int{2\,\,\cos x.\sin x\,dx+c}\] \[\Rightarrow \] \[y\,\sin \,x=\int{\sin \,2x\,\,dx+c}\] \[\Rightarrow \] \[y\,\,\sin x=\frac{-\cos \,2x}{2}+c\] At \[x=\frac{\pi }{4},y=\sqrt{2}\] \[\therefore \] \[\sqrt{2}\sin \frac{\pi }{4}=\frac{-\cos \,2(\pi /4)}{2}+c\] \[\Rightarrow \] \[c=\frac{\sqrt{2}}{\sqrt{2}}=1\] \[\therefore \] \[y\,\sin x=-\frac{1}{2}\cos \,2x+1\] \[\Rightarrow \] \[y=-\frac{1}{2}\frac{\cos \,\,2x}{\sin \,x}+\text{cosec x}\] \[=-\frac{1}{2\,\sin x}(1-2\,{{\sin }^{2}}x)+\operatorname{cosec}\,x\] \[=-\frac{1}{2}\text{cosec x+sin x+cosec x}\] \[y=\frac{1}{2}\,\text{cosec x + sin x}\]
You need to login to perform this action.
You will be redirected in
3 sec
