The solution of the differential equation \(\displaystyle\frac{dy}{dx} = \frac{\lambda^2}{(x+y)^2}\) (λ is constant) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\lambda\tan^{-1}\left(\displaystyle\frac{x+y}{\lambda}\right) = y + C\)
Since x + y appears together, substitute v = x + y. Then
\( \displaystyle \frac{dv}{dx} = 1 + \frac{dy}{dx} = 1 + \frac{\lambda^2}{v^2} = \frac{v^2+\lambda^2}{v^2} \)
Separating the variables:
\( \displaystyle \int \frac{v^2}{v^2+\lambda^2}\,dv = \int dx \)
\( \displaystyle \int \left(1 - \frac{\lambda^2}{v^2+\lambda^2}\right)dv = \int dx \)
\( \displaystyle v - \lambda^2 \cdot \frac{1}{\lambda}\tan^{-1}\frac{v}{\lambda} = x + c \)
Putting back v = x + y:
\( \displaystyle x + y - \lambda\tan^{-1}\left(\frac{x+y}{\lambda}\right) = x + c \)
\( \displaystyle \lambda\tan^{-1}\left(\frac{x+y}{\lambda}\right) = y - c \)
Writing C for the constant −c:
\( \displaystyle \lambda\tan^{-1}\left(\frac{x+y}{\lambda}\right) = y + C \)