Question

Find the particular solution of the differential equation \( \frac{dy}{dx} - 3y \cot x = \sin 2x \), given that \( y = 2 \) when \( x = \frac{\pi}{2} \).

Hint:

Remember the log property rule: \( e^{\log(f(x))} = f(x) \). Bring numerical coefficients inside the log as exponents first, otherwise they prevent correct cancellation of the base \( e \).

Answer 1

Concept: The given differential equation is of the standard form \( \frac{dy}{dx} + P(x)y = Q(x) \), which describes a first-order linear differential equation. We can solve this system using the Integrating Factor method: \[ \text{I.F.} = e^{\int P(x) \, dx} \] The general solution is then given by: \[ y \cdot (\text{I.F.}) = \int Q(x) \cdot (\text{I.F.}) \, dx + C \]

Step 1: Identify components and compute the Integrating Factor (I.F.).
Comparing with the standard linear form: \[ P(x) = -3\cot x \quad \text{and} \quad Q(x) = \sin 2x \] Calculate the integral of \( P(x) \): \[ \int P(x) \, dx = \int -3\cot x \, dx = -3\log|\sin x| = \log|(\sin x)^{-3}| = \log\left(\frac{1}{\sin^3 x}\right) \] Now calculate the Integrating Factor: \[ \text{I.F.} = e^{\log\left(\frac{1}{\sin^3 x}\right)} = \frac{1}{\sin^3 x} = \csc^3 x \]

Step 2: Write out the solution format.
\[ y \cdot \frac{1}{\sin^3 x} = \int (\sin 2x) \cdot \frac{1}{\sin^3 x} \, dx \] Expand using the double-angle identity \( \sin 2x = 2\sin x \cos x \): \[ y \cdot \frac{1}{\sin^3 x} = \int \frac{2\sin x \cos x}{\sin^3 x} \, dx = 2 \int \frac{\cos x}{\sin^2 x} \, dx \]

Step 3: Integrate the right-hand side using substitution.
For the integral \( \int \frac{\cos x}{\sin^2 x} \, dx \), substitute \( u = \sin x \), which gives \( du = \cos x \, dx \): \[ 2 \int \frac{du}{u^2} = 2 \left(-\frac{1}{u}\right) = -\frac{2}{\sin x} \] Thus, our general solution equation is: \[ \frac{y}{\sin^3 x} = -\frac{2}{\sin x} + C \] Multiplying through by \( \sin^3 x \): \[ y = -2\sin^2 x + C\sin^3 x \]

Step 4: Use the initial conditions to find the particular solution parameter \( C \).
We are given that \( y = 2 \) when \( x = \frac{\pi}{2} \). Substitute these boundary values: \[ 2 = -2\sin^2\left(\frac{\pi}{2}\right) + C\sin^3\left(\frac{\pi}{2}\right) \] Since \( \sin\left(\frac{\pi}{2}\right) = 1 \): \[ 2 = -2(1)^2 + C(1)^3 \quad \Rightarrow \quad 2 = -2 + C \quad \Rightarrow \quad C = 4 \] Substituting \( C = 4 \) back into the general equation yields our unique particular solution: \[ y = -2\sin^2 x + 4\sin^3 x \]