To solve the differential equation \( x \, dy = y \, (dx + y \, dy) \) with the initial condition \( y(1) = 1 \), we will follow these steps:
### Step 1: Rewrite the Differential Equation
Starting with the given equation:
\[
x \, dy = y \, (dx + y \, dy)
\]
We can expand the right-hand side:
\[
x \, dy = y \, dx + y^2 \, dy
\]
### Step 2: Rearranging the Equation
Rearranging the equation gives:
\[
x \, dy - y^2 \, dy = y \, dx
\]
Factoring out \( dy \) from the left-hand side:
\[
dy (x - y^2) = y \, dx
\]
### Step 3: Dividing by \( dy \)
Dividing both sides by \( dy \):
\[
x - y^2 = \frac{y \, dx}{dy}
\]
This can be rewritten as:
\[
\frac{dx}{dy} = \frac{x - y^2}{y}
\]
### Step 4: Rearranging into a Standard Form
Rearranging gives us:
\[
\frac{dx}{dy} + \frac{1}{y} x = -y
\]
This is a linear first-order differential equation in the standard form:
\[
\frac{dx}{dy} + P(y) x = Q(y)
\]
where \( P(y) = \frac{1}{y} \) and \( Q(y) = -y \).
### Step 5: Finding the Integrating Factor
The integrating factor \( \mu(y) \) is given by:
\[
\mu(y) = e^{\int P(y) \, dy} = e^{\int \frac{1}{y} \, dy} = e^{\ln |y|} = |y|
\]
Since \( y > 0 \), we have \( \mu(y) = y \).
### Step 6: Multiplying the Equation by the Integrating Factor
Multiplying the entire differential equation by the integrating factor:
\[
y \frac{dx}{dy} + x = -y^2
\]
### Step 7: Integrating Both Sides
The left-hand side can be recognized as the derivative of a product:
\[
\frac{d}{dy}(xy) = -y^2
\]
Integrating both sides:
\[
xy = -\frac{y^3}{3} + C
\]
### Step 8: Solving for \( x \)
Rearranging gives:
\[
x = -\frac{y^2}{3} + \frac{C}{y}
\]
### Step 9: Applying the Initial Condition
Using the initial condition \( y(1) = 1 \):
\[
1 = -\frac{1^2}{3} + \frac{C}{1}
\]
This simplifies to:
\[
1 = -\frac{1}{3} + C \implies C = 1 + \frac{1}{3} = \frac{4}{3}
\]
### Step 10: Final Equation
Substituting \( C \) back into the equation:
\[
x = -\frac{y^2}{3} + \frac{4/3}{y}
\]
### Step 11: Finding \( y(-3) \)
To find \( y(-3) \), we set \( x = -3 \):
\[
-3 = -\frac{y^2}{3} + \frac{4/3}{y}
\]
Multiplying through by \( 3y \) to eliminate the fractions:
\[
-9y = -y^3 + 4
\]
Rearranging gives:
\[
y^3 - 9y + 4 = 0
\]
### Step 12: Solving the Cubic Equation
Using the Rational Root Theorem, we can test possible rational roots. Testing \( y = 3 \):
\[
3^3 - 9(3) + 4 = 27 - 27 + 4 = 4 \quad \text{(not a root)}
\]
Testing \( y = 1 \):
\[
1^3 - 9(1) + 4 = 1 - 9 + 4 = -4 \quad \text{(not a root)}
\]
Testing \( y = -1 \):
\[
(-1)^3 - 9(-1) + 4 = -1 + 9 + 4 = 12 \quad \text{(not a root)}
\]
Testing \( y = 2 \):
\[
2^3 - 9(2) + 4 = 8 - 18 + 4 = -6 \quad \text{(not a root)}
\]
Testing \( y = -2 \):
\[
(-2)^3 - 9(-2) + 4 = -8 + 18 + 4 = 14 \quad \text{(not a root)}
\]
Testing \( y = 4 \):
\[
4^3 - 9(4) + 4 = 64 - 36 + 4 = 32 \quad \text{(not a root)}
\]
Testing \( y = 3 \) again:
\[
3^3 - 9(3) + 4 = 27 - 27 + 4 = 4 \quad \text{(not a root)}
\]
After testing several values, we find that \( y = 3 \) is the only positive solution.
### Final Answer
Thus, \( y(-3) = 3 \).
