Let $\alpha x=\exp \left(x^{\beta} y^{\gamma}\right)$ be the solution of the differential equation $2 x^{2} y dy-\left(1-x y^{2}\right) dx=0$, $x>0, y(2)=\sqrt{\log _{e} 2}$. Then $\alpha+\beta-\gamma$ equals :

JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Differential Equations

Let

α x = exp (x^{β} y^{γ})

be the solution of the differential equation

2 x^{2} y d y - (1 - x y^{2}) d x = 0

x > 0, y (2) = lo g_{e} 2

. Then

α + β - γ

equals :

Answer: (A)

With

t = y^{2}

the equation

2 x^{2} y d y - (1 - x y^{2}) d x = 0

becomes

x^{2} \frac{d t}{d x} = 1 - x t

, that is

\frac{d t}{d x} + \frac{t}{x} = \frac{1}{x ^{2}}

, a linear equation with integrating factor

x

. Then

\frac{d}{d x} (x t) = \frac{1}{x}

gives

x t = ln x + C

, i.e.

x y^{2} = ln x + C

. The condition

y (2) = ln 2

gives

2 ln 2 = ln 2 + C

, so

C = ln 2

and

x y^{2} = ln (2 x)

, which rearranges to

2 x = e^{x y^{2}}

. Comparing with

α x = exp (x^{β} y^{γ})

gives

α = 2

β = 1

γ = 2

, so

α + β - γ = 1

JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Single correct option

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