Let α ∣ x ∣ = ∣ y ∣ e x y − β , α , β ∈ N be the solution of the differential equation x d y − y d x + x y ( x d y + y d x ) = 0 , y ( 1 ) = 2 . Then α + β is equal to ________

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​ a ∣ x ∣ = ∣ y ∣ e y x − β , a , b ∈ N x d y − y d x + x y ( x d y + y d x ) = 0 y d y ​ − x d x ​ + ( x d y + y d x ) = 0 ℓ n ∣ y ∣ − ℓ n ∣ x ∣ + x y = c y ( 1 ) = 2 ℓ n ∣2∣ − 0 + 2 = c c = 2 + ℓ n 2 ℓ n ∣ y ∣ − ℓ n ∣ x ∣ + x y = 2 + ℓ n 2 ℓ n ∣ x ∣ = ℓ n ​ 2 y ​ ​ − 2 + x y ∣ x ∣ = ​ 2 y ​ ​ e x y − 2 2∣ x ∣ = ∣ y ∣ e x y − 2 α = 2 β = 2 α + β = 4 ​

Mathematics - Differential Equation Question with Solution | TestHub

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