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 \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32