Question

Solve the differential equation \( (x - \sin y) \, dy + \tan y \, dx = 0 \).

Hint:

If an equation isn't linear in \( y \), check if it's linear in \( x \) by isolating \( dx/dy \). This is a common technique when \( y \) is inside trigonometric functions.

Answer 1

Concept:
• A first-order linear differential equation in \( x \) is of the form \( \frac{dx}{dy} + P(y)x = Q(y) \).
• Solution is given by \( x \cdot (IF) = \int Q(y) \cdot (IF) \, dy + C \), where \( IF = e^{\int P(y) \, dy} \).

Step 1: Rearrange the equation into standard linear form
The given equation is: \( \tan y \, dx = (\sin y - x) \, dy \).
Divide by \( dy \cdot \tan y \):
\[ \frac{dx}{dy} = \frac{\sin y - x}{\tan y} \]
\[ \frac{dx}{dy} = \frac{\sin y}{\tan y} - \frac{x}{\tan y} \]
\[ \frac{dx}{dy} + (\cot y)x = \cos y \]

Step 2: Find the Integrating Factor (IF)
Here \( P(y) = \cot y \) and \( Q(y) = \cos y \).
\[ IF = e^{\int \cot y \, dy} = e^{\log |\sin y|} = \sin y \]

Step 3: Apply the general solution formula
The solution is \( x \cdot IF = \int Q(y) \cdot IF \, dy + C \):
\[ x \sin y = \int \cos y \sin y \, dy + C \]
Multiply and divide the integral by 2 to use \( 2\sin y \cos y = \sin 2y \):
\[ x \sin y = \frac{1}{2} \int \sin 2y \, dy + C \]
\[ x \sin y = \frac{1}{2} \left( -\frac{\cos 2y}{2} \right) + C \]
\[ x \sin y = -\frac{1}{4} \cos 2y + C \]