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The solution of the differential equation dydx=λ2(x+y)2\frac{dy}{dx} = \frac{\lambda^2}{(x+y)^2} (λ\lambda is constant) is:

Solution

✅ Correct Option: 3

Let v=x+yv = x + y, so dvdx=1+dydx=1+λ2v2=v2+λ2v2\frac{dv}{dx} = 1 + \frac{dy}{dx} = 1 + \frac{\lambda^2}{v^2} = \frac{v^2 + \lambda^2}{v^2}.

Separate: v2v2+λ2dv=dx\frac{v^2}{v^2 + \lambda^2}dv = dx.

Write as (1−λ2v2+λ2)dv=dx\left(1 - \frac{\lambda^2}{v^2 + \lambda^2}\right)dv = dx.

Integrate: v−λtan⁡−1(v/λ)=x+C1v - \lambda\tan^{-1}(v/\lambda) = x + C_1.

Substitute back: λtan⁡−1(x+yλ)=y+C\lambda\tan^{-1}\left(\frac{x+y}{\lambda}\right) = y + C.

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