JEE Main · 2024 · 08 Apr 2024 · Shift 2 · Differential Equations

Let

α ∣ x ∣ = ∣ y ∣ e^{x y - β}

α, β \in 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 ______.

Answer: 4

Solution

x d y - y d x + x y (x d y + y d x) = 0

. Divide by

x y

\frac{d y}{y} - \frac{d x}{x} + x d y + y d x = 0

, i.e.

\frac{d y}{y} - \frac{d x}{x} + d (x y) = 0

.
Integrating:

ln ∣ y ∣ - ln ∣ x ∣ + x y = c ⟹ ln \frac{y}{x} = c - x y

y (1) = 2 ⟹ ln 2 = c - 2 ⟹ c = 2 + ln 2

.
So

ln \frac{y}{x} = 2 + ln 2 - x y ⟹ \frac{y}{x} = 2 e^{2 - x y} ⟹ 2∣ x ∣ = ∣ y ∣ e^{x y - 2}

.
Comparing with

α ∣ x ∣ = ∣ y ∣ e^{x y - β}

α = 2, β = 2 ⟹ α + β = 4

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