The solution of the differential equation d y d x + 1 x = e y x 2 , ∀ x > 0 is λ x e - y = 1 - 2 c x 2 (where c is an arbitrary constant). Then, the value of λ is equal to

Step-by-step solution

The given equation is e - y d y d x + e - y ⋅ 1 x = 1 x 2 Let, - e - y = t ⇒ e - y d y d x = d t d x Thus we have, d t d x - t 1 x = 1 x 2 Integrating factor = e - ∫ 1 x d x = e - ln ⁡ x = 1 x Thus, the solution is t ⋅ 1 x = ∫ 1 x 2 ⋅ 1 x d x ⇒ - e - y x = x - 2 - 2 + c ⇒ - e - y x = - 1 2 x 2 + c ⇒ - 2 x e - y = - 1 + 2 c x 2 or 2 x e - y = 1 - 2 c x 2 Hence, λ = 2