Question

The solution of the differential equation $ \frac{dy}{dx}+\frac{1}{x}\tan y=\frac{1}{{{x}^{2}}}\tan y\sin y $ is

• 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

Answer 1

Answer: (B)

$ \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}}) $