The general solution of the differential equation \( \tan\left(\displaystyle\frac{(1+x^{2})y'+2xy}{2x}\right)=1+x^{2} \), is
(Consider \( \log_{e}x=\log x \))
(where C is an arbitrary constant)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( y=\tan^{-1}(1+x^{2})-\displaystyle\frac{1}{2(1+x^{2})}\log\left|x^{4}+2x^{2}+2\right|+\displaystyle\frac{C}{1+x^{2}} \)
Taking the inverse tangent of both sides and clearing the denominator,
\( \displaystyle \frac{(1+x^{2})y'+2xy}{2x}=\tan^{-1}(1+x^{2}) \)
\( \displaystyle (1+x^{2})y'+2xy=2x\tan^{-1}(1+x^{2}) \)
The left side is exactly the derivative of a product, so no integrating factor is needed.
\( \displaystyle \frac{d}{dx}\left[(1+x^{2})y\right]=2x\tan^{-1}(1+x^{2}) \)
Integrating, with \( t=1+x^{2} \), \( dt=2x\,dx \):
\( \displaystyle (1+x^{2})y=\int\tan^{-1}t\,dt \)
By parts,
\( \displaystyle \int\tan^{-1}t\,dt=t\tan^{-1}t-\int\frac{t}{1+t^{2}}\,dt \)
\( \displaystyle =t\tan^{-1}t-\frac{1}{2}\log(1+t^{2}) \)
Put back \( t=1+x^{2} \), where \( 1+t^{2}=x^{4}+2x^{2}+2 \).
\( \displaystyle (1+x^{2})y=(1+x^{2})\tan^{-1}(1+x^{2})-\frac{1}{2}\log\left|x^{4}+2x^{2}+2\right|+C \)
Dividing by \( 1+x^{2} \),
\( \displaystyle y=\tan^{-1}(1+x^{2})-\frac{1}{2(1+x^{2})}\log\left|x^{4}+2x^{2}+2\right|+\frac{C}{1+x^{2}} \)