A) \[2x=\frac{(1+2c{{x}^{2}})}{\sin y}\]
B) \[2x=\sin y(1+2c{{x}^{2}})\]
C) \[2x+\sin y(1+c{{x}^{2}})=0\]
D) None of the above
Correct Answer: B
Solution :
\[\frac{dy}{dx}+\frac{1}{x}\tan y=\frac{1}{{{x}^{2}}}\tan y\sin y\] \[\Rightarrow \] \[\cot y\cos ec\,y\frac{dy}{dx}+\frac{1}{x}\cos ecy=\frac{1}{{{x}^{2}}}\] ...(i) Let cosec \[y=-v\] \[\Rightarrow \] \[-\cos ecy.\cot y\frac{dy}{dx}=\frac{-dv}{dx}\] Then \[\frac{dv}{dx}-\frac{1}{x}v=\frac{1}{{{x}^{2}}}\] Here, \[P=-\frac{1}{x},Q=\frac{1}{{{x}^{2}}}\] \[IF={{e}^{\int{P}\,dx}}\] \[={{e}^{\int{\left( -\frac{1}{x} \right)}\,dx}}={{e}^{-\log x}}\] \[={{x}^{-1}}=1/x\] Hence, solution of the given differential equation is \[v.\left( \frac{1}{x} \right)=\int{\frac{1}{{{x}^{2}}}.\left( \frac{1}{x} \right)}dx-c\] \[=\int{\frac{1}{{{x}^{3}}}}dx-c\] \[\Rightarrow \] \[\frac{v}{x}=\left( -\frac{1}{2{{x}^{2}}} \right)-c\] \[\Rightarrow \] \[\frac{v}{x}=-\frac{1}{2{{x}^{2}}}-c\] \[\Rightarrow \] \[2xv=-(1+2c{{x}^{2}})\] \[\Rightarrow \] \[-2x\cos ecy=-(1+2c{{x}^{2}})\] \[\Rightarrow \] \[2x=\sin y(1+2c{{x}^{2}})\]
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