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

The Correct Option is A

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^–1t + e^t = lnx + C
\(⇒sin^{-1} ( \frac{y}{x} ) + e^\frac{y}{x} = In x + C\)
at x = 1, y = 0
So, 0 + e⁰ = 0 + C⇒C = 1
at (2α, α)
\(sin^{-1} ( \frac{y}{x} ) + e^\frac{y}{x} = In x + 1\)
\(⇒ \frac{π}{6} + e^\frac{1}{2} - 1 = In (2α)\)
\(⇒ α = \frac{1}{2} e^{( \frac{π}{6} + e^\frac{1}{2} - 1 )}\)