Question

If the solution curve of the differential equation $\frac{dy}{dx} = \frac{x + y - 2}{x - y}$ passing through the point $(2, 1)$ is $\tan^{-1}\left(\frac{y - 1}{x - 1}\right) - \frac{1}{\beta} \log_e\left(\alpha + \left(\frac{y - 1}{x - 1}\right)^2\right) = \log_e |x - 1|,$ then $5\beta + \alpha$ is equal to

Answer 1

Given:
$\frac{dy}{dx} = \frac{x + y - 2}{x - y}.$

Substitute:
$x = X + h, \quad y = Y + k.$

Let:
$h + k = 2, \quad h - k = 0 \implies h = k = 1.$

So:
$Y = vX, \quad \frac{dv}{dX} = \frac{1 + v^2}{1 - v}.$

Integrating and applying the condition (2, 1):
$\tan^{-1} \left(\frac{y - 1}{x - 1}\right) = \frac{1}{2} \log_e \left(1 + \left(\frac{y - 1}{x - 1}\right)^2\right) - \log_e |x - 1|.$

From the equation:
$\alpha = 1, \quad \beta = 2.$

Calculating $5\beta + \alpha$:
$5\beta + \alpha = 5 \times 2 + 1 = 11.$

Thus, the Correct Answer is 11.

Let me know if further clarification is needed!
The Correct answer is: 11