Question

Solve the differential equation \( 2(y + 3) - xy \frac{dy}{dx} = 0 \); given \( y(1) = -2 \).

Hint:

For separable differential equations, rearrange to isolate \( x \) and \( y \) terms before integrating.

Answer 1

To solve the differential equation:

\[ 2(y + 3) - xy \frac{dy}{dx} = 0 \] with initial condition \( y(1) = -2 \)

1. Rearrange the Equation:
Bring all terms to one side: \[ 2(y + 3) = xy \frac{dy}{dx} \Rightarrow \frac{dy}{dx} = \frac{2(y + 3)}{xy} \]

2. Separate the Variables:
\[ \frac{y}{y + 3} dy = \frac{2}{x} dx \]

3. Integrate Both Sides:
We simplify the left-hand side: \[ \frac{y}{y + 3} = \frac{(y + 3) - 3}{y + 3} = 1 - \frac{3}{y + 3} \] So the equation becomes: \[ \int \left(1 - \frac{3}{y + 3}\right) dy = \int \frac{2}{x} dx \]

Now integrate: \[ \int 1\,dy - \int \frac{3}{y + 3}\,dy = \int \frac{2}{x}\,dx \Rightarrow y - 3 \ln |y + 3| = 2 \ln |x| + C \]

4. Apply the Initial Condition \( y(1) = -2 \):
Substitute into the equation: \[ -2 - 3 \ln |-2 + 3| = 2 \ln |1| + C \Rightarrow -2 - 3 \ln 1 = 0 + C \Rightarrow C = -2 \]

5. Final Solution:
Substitute \( C = -2 \) back into the expression: \[ y - 3 \ln |y + 3| = 2 \ln |x| - 2 \]

Final Answer:
\[ \boxed{y - 3 \ln |y + 3| = 2 \ln |x| - 2} \]