We are given the differential equation: \[ \frac{dy}{dx} = \frac{y}{x}. \]
This is a separable differential equation, so we can rewrite it as: \[ \frac{dy}{y} = \frac{dx}{x}. \]
Now, integrating both sides: \[ \int \frac{1}{y} dy = \int \frac{1}{x} dx, \] \[ \ln |y| = \ln |x| + C. \]
Exponentiating both sides: \[ |y| = e^{\ln |x| + C} = |x| e^C. \] Thus, \( y = Cx \).
Using the initial condition \( y(1) = 2 \), we get: \[ 2 = C(1) \quad \Rightarrow \quad C = 2. \]
Therefore, the solution is \( y = 2x \).
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)