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QJEEM23Maf375: JEE Main 2023 Mathematics - Class 12 JEEM23Maf375 Solution: If is the solution of the differential equation such that :

If y=y(x)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

y(2)=29log⁡e(2+3) and y(2)=αlog⁡e(α+β)+β−γ,α,β,γ∈N, then αβγ is equal to y(2)=\frac{2}{9} \log _{e}(2+\sqrt{3}) \text { and } y(\sqrt{2})=\alpha \log _{e}(\sqrt{\alpha}+\beta)+\beta-\sqrt{\gamma}, \alpha, \beta, \gamma \in \mathbb{N} \text {, then } \alpha \beta \gamma \text { is equal to } :

Step-by-Step Verified Solution

We can solve the given differential equation using an integrating factor.

The integrating factor is given by :

μ(x)=e∫4xx2−1dx=e2ln⁡(x2−1)=(x2−1)2\mu(x) = e^{\int \frac{4x}{x^2 - 1} dx} = e^{2\ln(x^2 - 1)} = (x^2 - 1)^2

Multiplying both sides of the differential equation by μ(x)\mu(x) , we get :

(x2−1)2dydx+4xy=x+2(x2−1)12(x^2-1)^2 \frac{d y}{d x} + 4x y = \frac{x+2}{\left(x^{2}-1\right)^{\frac{1}{2}}}

We can rewrite the left-hand side using the product rule:

ddx((x2−1)2y)=x+2(x2−1)12\frac{d}{dx} \left((x^2-1)^2 y\right) = \frac{x+2}{\left(x^{2}-1\right)^{\frac{1}{2}}}

Integrating both sides with respect to xx , we get:

(x2−1)2y=∫x+2(x2−1)12dx=x2−1+2ln⁡∣x+x2−1∣+C(x^2-1)^2 y = \int \frac{x+2}{\left(x^{2}-1\right)^{\frac{1}{2}}} dx = \sqrt{x^2-1}+2 \ln\left|x+\sqrt{x^2-1}\right|+C

where CC is the constant of integration. Using the initial condition y(2)=29log⁡e(2+3)y(2) = \frac{2}{9} \log_e(2+\sqrt{3}) , we can solve for CC :

At x=2x=2 ,

9⋅29ln⁡(2+3)=2ln⁡(2+3)+3+C9 \cdot \frac{2}{9} \ln (2+\sqrt{3})=2 \ln (2+\sqrt{3})+\sqrt{3}+C

C=−3C=-\sqrt{3}

At x=2x=\sqrt{2}

y(2)=2ln⁡(1+2)+1−3y(\sqrt{2})=2 \ln (1+\sqrt{2})+1-\sqrt{3}

β=1,α=2,γ=3\beta=1, \alpha=2, \gamma=3

⇒αβγ=6\Rightarrow \alpha \beta \gamma=6

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