If \(y=y(x)\) is the solution of the differential equation \(\frac{d y}{d x}+\frac{4 x}{\left(x^{2}-1\right)} y=\frac{x+2}{\left(x^{2}-1\right)^{\frac{5}{2}}}, x>1\) such that \(\mathrm{y}(2)=\frac{2}{9} \log_{\mathrm{e}}(2+\sqrt{3})\) and \(\mathrm{y}(\sqrt{2})= \alpha \log_{\mathrm{e}}(\sqrt{\alpha}+\beta)+\beta-\sqrt{\gamma}, \alpha, \beta, \gamma \in \mathrm{N}\), then \(\alpha \beta \gamma\) is equal to ______.

JEE Main · 2023 · 13 Apr 2023 · Shift 2 · Differential Equations

y = y (x)

is the solution of the differential equation

\frac{d y}{d x} + \frac{4 x}{( x ^{2} - 1 )} y = \frac{x + 2}{( x ^{2} - 1 ) ^{\frac{5}{2}}}, x > 1

such that

y (2) = \frac{2}{9} lo g_{e} (2 + 3)

and

y (2) = α lo g_{e} (α + β) + β - γ, α, β, γ \in N

, then

α β γ

is equal to ______.

Answer: 6

This is a first-order linear ODE. The integrating factor is:

I . F . = e^{\int \frac{4 x}{x ^{2} - 1} d x} = e^{2 l n ∣ x^{2} - 1∣} = (x^{2} - 1)^{2}

Multiplying through:

\frac{d}{d x} [y (x^{2} - 1)^{2}] = \frac{x + 2}{( x ^{2} - 1 ) ^{1/2}}

Integrating:

y (x^{2} - 1)^{2} = \int \frac{x + 2}{x ^{2} - 1} d x = x^{2} - 1 + 2 ln (x + x^{2} - 1) + C

Using

y (2) = \frac{2}{9} ln (2 + 3)

\frac{2}{9} ln (2 + 3) \cdot 9 = 3 + 2 ln (2 + 3) + C

C = - 3

x = 2

y (2) \cdot 1 = 1 + 2 ln (2 + 1) - 3

= 2 ln (2 + 1) + 1 - 3

Comparing with

α ln (α + β) + β - γ

α = 2, β = 1, γ = 3

.

Therefore:

α β γ = 2 \cdot 1 \cdot 3 = 6

JEE Main · 2023 · 13 Apr 2023 · Shift 2 · Integer answer