Question

Find the particular solution of the differential equation

\(xy\dfrac{dy}{dx} = (x+2)(y+2)\), given that \(y(1) = -1\).

Hint:

Separate as y/(y+2) dy = (x+2)/x dx, integrate, and use y(1) = -1 to get the constant C = -2.

Answer 1

Step 1: Separate the variables:
Divide both sides by \(x(y+2)\) and multiply by \(dx\). The variable \(y\) goes with \(dy\) and \(x\) with \(dx\).
\[ \frac{y}{y+2}\,dy = \frac{x+2}{x}\,dx \]

Step 2: Split both fractions:
Write \(\dfrac{y}{y+2} = 1 - \dfrac{2}{y+2}\) and \(\dfrac{x+2}{x} = 1 + \dfrac{2}{x}\).
\[ \left(1 - \frac{2}{y+2}\right)dy = \left(1 + \frac{2}{x}\right)dx \]

Step 3: Integrate both sides:
Each term is a standard integral.
\[ y - 2\log|y+2| = x + 2\log|x| + C \]

Step 4: Use the condition y(1) = -1:
Put \(x = 1\) and \(y = -1\). Then \(\log|y+2| = \log 1 = 0\) and \(\log|x| = \log 1 = 0\).
\[ -1 - 0 = 1 + 0 + C \quad \Rightarrow \quad C = -2 \]

Step 5: Write the particular solution:
Substitute \(C = -2\) back.
\[ y - 2\log|y+2| = x + 2\log|x| - 2 \]
This can also be written as \(y - x + 2 = 2\log|x(y+2)|\).

Final Answer:
The particular solution is below. \[ \boxed{y - x + 2 = 2\log|x(y+2)|} \]