A) \[2x=\sin y(1+2x)\]
B) \[2x=\sin y(1+c{{x}^{2}})\]
C) \[2x+\sin y(1+c{{x}^{2}})=0\]
D) None of these
Correct Answer: B
Solution :
[b] We have \[\frac{dy}{dx}+\frac{1}{x}\tan y=\frac{1}{{{x}^{2}}}\tan y\sin y\] \[\Rightarrow \,\,\cot y\,\,\text{cosec }y\frac{dy}{dx}+\frac{1}{x}\cos ecy=\frac{1}{{{x}^{2}}}\] Putting cosec \[y=t,\]we get \[\frac{dt}{dx}-\frac{t}{x}=-\frac{1}{{{x}^{2}}},\] which is linear differential dx x equation. \[I.F.={{e}^{\int{\frac{1}{x}dx}}}=\frac{1}{x}\] Therefore, solution is: \[\frac{t}{x}=\int{-\frac{1}{{{x}^{3}}}}\,dx+c'\] \[\Rightarrow \,\,\,\frac{t}{x}=\frac{1}{2{{x}^{2}}}+c'\] \[\Rightarrow \,\,\,2xt=1+2e'{{x}^{2}}\] \[\Rightarrow \,\,\,2x=\sin y(1+c{{x}^{2}})\]
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