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 \( y = 2x \).
Given below are two statements:
Statement (I):
are isomeric compounds.
Statement (II):
are functional group isomers.
In the light of the above statements, choose the correct answer from the options given below:
The effect of temperature on the spontaneity of reactions are represented as: Which of the following is correct?
