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.
A racing track is built around an elliptical ground whose equation is given by \[ 9x^2 + 16y^2 = 144 \] The width of the track is \(3\) m as shown. Based on the given information answer the following: 
(i) Express \(y\) as a function of \(x\) from the given equation of ellipse.
(ii) Integrate the function obtained in (i) with respect to \(x\).
(iii)(a) Find the area of the region enclosed within the elliptical ground excluding the track using integration.
OR
(iii)(b) Write the coordinates of the points \(P\) and \(Q\) where the outer edge of the track cuts \(x\)-axis and \(y\)-axis in first quadrant and find the area of triangle formed by points \(P,O,Q\).