Step 1: Separating the Variables Rewriting the equation: \[ \frac{1}{y} \, dy = \cot 2x \, dx \]
Step 2: Integrating Both Sides Integrate both sides: \[ \int \frac{1}{y} \, dy = \int \cot 2x \, dx \] The left-hand side becomes: \[ \log |y| \] The right-hand side uses the integral of \( \cot 2x \): \[ \int \cot 2x \, dx = \frac{1}{2} \log |\sin 2x| \] So the equation becomes: \[ \log |y| = \frac{1}{2} \log |\sin 2x| + \log c \] Here, \( \log c \) is the constant of integration.
Step 3: Simplify the Expression} Combine the logarithms: \[ \log |y| = \log \left( c \sqrt{\sin 2x} \right) \] Exponentiate both sides to remove the logarithm: \[ y = c \sqrt{\sin 2x} \]
Step 4: Finding the Particular Solution} We are given the condition \( y\left( \frac{\pi}{4} \right) = 2 \). Substitute \( x = \frac{\pi}{4} \) and \( y = 2 \) into the solution: \[ 2 = c \sqrt{\sin\left( 2 \cdot \frac{\pi}{4} \right)} \] Simplify: \[ 2 = c \sqrt{\sin\left( \frac{\pi}{2} \right)} \] Since \( \sin\left( \frac{\pi}{2} \right) = 1 \), we have: \[ 2 = c \cdot 1 \quad \Rightarrow \quad c = 2 \]
Step 5: Final Solution} Substitute \( c = 2 \) back into the solution: \[ y = 2 \sqrt{\sin 2x} \]
Final Answer: \[ \boxed{y = 2 \sqrt{\sin 2x}} \] This is the required particular solution to the given differential equation.
Let $ y(x) $ be the solution of the differential equation $$ x^2 \frac{dy}{dx} + xy = x^2 + y^2, \quad x > \frac{1}{e}, $$ satisfying $ y(1) = 0 $. Then the value of $ 2 \cdot \frac{(y(e))^2}{y(e^2)} $ is ________.
According to the generally accepted definition of the ideal solution there are equal interaction forces acting between molecules belonging to the same or different species. (This is equivalent to the statement that the activity of the components equals the concentration.) Strictly speaking, this concept is valid in ecological systems (isotopic mixtures of an element, hydrocarbons mixtures, etc.). It is still usual to talk about ideal solutions as limiting cases in reality since very dilute solutions behave ideally with respect to the solvent. This law is further supported by the fact that Raoult’s law empirically found for describing the behaviour of the solvent in dilute solutions can be deduced thermodynamically via the assumption of ideal behaviour of the solvent.
Answer the following questions:
(a) Give one example of miscible liquid pair which shows negative deviation from Raoult’s law. What is the reason for such deviation?
(b) (i) State Raoult’s law for a solution containing volatile components.
OR
(ii) Raoult’s law is a special case of Henry’s law. Comment.
(c) Write two characteristics of an ideal solution.