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 \( y = 2x \).
Let a line passing through the point $ (4,1,0) $ intersect the line $ L_1: \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} $ at the point $ A(\alpha, \beta, \gamma) $ and the line $ L_2: x - 6 = y = -z + 4 $ at the point $ B(a, b, c) $. Then $ \begin{vmatrix} 1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c \end{vmatrix} \text{ is equal to} $
20 mL of sodium iodide solution gave 4.74 g silver iodide when treated with excess of silver nitrate solution. The molarity of the sodium iodide solution is _____ M. (Nearest Integer value) (Given : Na = 23, I = 127, Ag = 108, N = 14, O = 16 g mol$^{-1}$)