Question

If y=y(x) is the solution of the differential equation \(\frac{dy}{dx}+\frac{4x}{(x^2-1)}y=\frac{x+2}{(x^2-1)^{\frac{5}{2}}},x>1\) such that \(y(2)=\frac{2}{9}\log_e(2+\sqrt3)\ \text{and}\ y(\sqrt2)=\alpha\log_e(\sqrt{\alpha}+\beta)+\beta-\sqrt{\gamma}∈\N,\) then αβγ is equal to

Answer 1

Correct Answer: 6

The given differential equation is: \[ \frac{dy}{dx} + \frac{4x}{x^2 - 1} y = \frac{x + 2}{(x^2 - 1)^{5/2}} \] We first calculate the integrating factor (I.F.): \[ \text{I.F.} = e^{\int \frac{4x}{x^2 - 1} dx} \] Using substitution, we get: \[ \int \frac{4x}{x^2 - 1} dx = 2 \ln(x^2 - 1) \] Thus, the integrating factor is: \[ e^{\int \frac{4x}{x^2 - 1} dx} = e^{2 \ln(x^2 - 1)} = (x^2 - 1)^2 \] Step 1: Solve the equation using the integrating factor. Now, the equation becomes: \[ (x^2 - 1)^2 \frac{dy}{dx} + (x^2 - 1)^2 \frac{4x}{x^2 - 1} y = (x^2 - 1)^2 \frac{x + 2}{(x^2 - 1)^{5/2}} \] This simplifies to: \[ \frac{d}{dx} \left[ y (x^2 - 1)^2 \right] = (x + 2)(x^2 - 1)^{1/2} \] Now integrate both sides: \[ y(x^2 - 1)^2 = \int (x + 2)(x^2 - 1)^{1/2} dx \] Performing the integration, we get: \[ y(x^2 - 1)^2 = \int x (x^2 - 1)^{1/2} dx + 2 \int (x^2 - 1)^{1/2} dx \] The results are: \[ \int x (x^2 - 1)^{1/2} dx = \frac{1}{2} (x^2 - 1)^{3/2}, \quad \int (x^2 - 1)^{1/2} dx = \frac{1}{2} x (x^2 - 1)^{1/2} \] Thus, the solution becomes: \[ y(x^2 - 1)^2 = \frac{1}{2} (x^2 - 1)^{3/2} + x (x^2 - 1)^{1/2} + C \] Step 2: Apply the initial condition. Substitute \( x = 2 \) and \( y(2) = \frac{2}{9} \log_e \left( 2 + \sqrt{3} \right) \): \[ \frac{2}{9} \log_e \left( 2 + \sqrt{3} \right) (2^2 - 1)^2 = \frac{1}{2} (2^2 - 1)^{3/2} + 2 (2^2 - 1)^{1/2} + C \] Simplifying: \[ \frac{2}{9} \log_e \left( 2 + \sqrt{3} \right) \cdot 9 = \sqrt{3} + 2\sqrt{3} + C \] Thus: \[ C = -\sqrt{3} \] Step 3: Find the value of \( \alpha \beta \gamma \). Substitute \( x = \sqrt{2} \) into the equation for \( y \): \[ y(\sqrt{2}) = 1 + 2 \ln \left( \sqrt{2 + 1} \right) - \sqrt{3} \] We find: \[ \alpha = 2, \, \beta = 1, \, \gamma = 3 \] Thus, \( \alpha \beta \gamma = 2 \times 1 \times 3 = 6 \).