The solution of the differential equation \( \displaystyle\frac{dy}{dx}=\displaystyle\frac{x^2+xy+y^2}{x^2} \) for \( x\gt 0,y\gt 0 \) at \( x=1,y=2 \) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \tan^{-1}\displaystyle\frac{y}{x}=\log_e x+\tan^{-1}2 \)
The equation is homogeneous:
\( \displaystyle \frac{dy}{dx}=1+\frac{y}{x}+\frac{y^2}{x^2} \)
Let y = vx, so \( \frac{dy}{dx}=v+x\frac{dv}{dx} \). Then
\( \displaystyle v+x\frac{dv}{dx}=1+v+v^2 \)
\( \displaystyle \frac{dv}{1+v^2}=\frac{dx}{x} \)
\( \displaystyle \tan^{-1}v=\log_e x+C \)
\( \displaystyle \tan^{-1}\frac{y}{x}=\log_e x+C \)
At x = 1, y = 2: \( \tan^{-1}2=0+C \), so \( C=\tan^{-1}2 \).
\( \displaystyle \tan^{-1}\frac{y}{x}=\log_e x+\tan^{-1}2 \)
Why the other options are incorrect
Option (1) \( \tan\left(\displaystyle\frac{y}{x}\right)=\log_e x+\tan 2 \) — Incorrect. Integrating \( \frac{dv}{1+v^2} \) gives \( \tan^{-1}v \), not tan v; this option only happens to hold at x = 1, y = 2.
Option (2) \( \tan^{-1}\left(\displaystyle\frac{x}{y}\right)=\log_e x+\tan^{-1}2 \) — Incorrect. It has \( \frac{x}{y} \) in place of \( \frac{y}{x} \); at x = 1, y = 2 it gives \( \tan^{-1}\frac{1}{2}=\tan^{-1}2 \), which is false.
Option (4) \( \tan^{-1}\displaystyle\frac{y}{x}=\log_e x-\tan^{-1}2 \) — Incorrect. The constant has the wrong sign; at x = 1, y = 2 it gives \( \tan^{-1}2=-\tan^{-1}2 \), which is false.