Question

Let the solution curve y = y(x) of the differential equation
$[ \frac{x}{\sqrt{x² -y²}} + e^\frac{y}{x} ] x \frac{dy}{dx} = x + [ \frac{x}{\sqrt{x² -y²}} + e^\frac{y}{x} ]y$
pass through the points (1, 0) and (2α, α), α> 0. Then α is equal to

• A. $\frac{1}{2} exp ( \frac{π}{6} + \sqrt{e} - 1 )$
• B. $\frac{1}{2} exp ( \frac{π}{3} + e - 1 )$
• C. $exp ( \frac{π}{6} + \sqrt{e} + 1 )$
• D. $2 exp ( \frac{π}{3} + \sqrt{e} - 1 )$

Answer 1

Answer: (A)

$\frac{1}{2} exp ( \frac{π}{6} + \sqrt{e} - 1 )$

Answer 2

The correct answer is (A) : $\frac{1}{2} exp ( \frac{π}{6} + \sqrt{e} - 1 )$
$\left( \frac{1}{ \sqrt{1 - \frac{y²}{x²}}} + e^\frac{y}{x} \right) \frac{dy}{dx} = 1 + \left( \frac{1}{\sqrt{1 - \frac{y²}{x²}}} + e^\frac{y}{x} \right) \frac{y}{x}$
Putting y = tx
$\left( \frac{1}{\sqrt{1 - t²}} + e^t \right) \left( t + x \frac{dt}{dx} \right) = 1 + \left( \frac{1}{\sqrt{1 - t²}} + e^t \right)t$
$⇒ x ( \frac{1}{\sqrt{1 - t²}} + e^t ) \frac{dt}{dx} = 1$
⇒ sin–1 t + e t = ln x + C
$⇒sin^{-1} ( \frac{y}{x} ) + e^\frac{y}{x} = In x + C$
at x = 1, y = 0
So, 0 + e 0 = 0 + C ⇒ C = 1
at (2α, α)
$sin^{-1} ( \frac{y}{x} ) + e^\frac{y}{x} = In x + 1$
$⇒ \frac{π}{6} + e^\frac{1}{2} - 1 = In (2α)$
$⇒ α = \frac{1}{2} e^{( \frac{π}{6} + e^\frac{1}{2} \- 1 )}$