Let's solve the given differential equation and find the value of \(y(-3)\).

Solving the Differential Equation

The given differential equation is:

\[ x \, dy = y \, dx + y^2 \, dy \]

We can rearrange the terms to group \(dy\) and \(dx\).

\[ x \, dy - y^2 \, dy = y \, dx \]

Factor out \(dy\) on the left side:

\[ (x - y^2) \, dy = y \, dx \]

To make it easier to solve, let's rewrite this in the form of \(dx/dy\):

\[ \frac{dx}{dy} = \frac{x - y^2}{y} \]

Separate the terms involving \(x\) and \(y\):

\[ \frac{dx}{dy} = \frac{x}{y} - \frac{y^2}{y} \]

\[ \frac{dx}{dy} = \frac{x}{y} - y \]

Rearrange it into the standard form of a first-order linear differential equation with respect to \(y\), which is \( \frac{dx}{dy} + P(y)x = Q(y) \):

\[ \frac{dx}{dy} - \frac{1}{y} x = -y \]

Here, \(P(y) = -\frac{1}{y}\) and \(Q(y) = -y\).

Finding the Integrating Factor

The integrating factor (IF) for this type of equation is given by \( e^{\int P(y) \, dy} \).

\[ IF = e^{\int -\frac{1}{y} \, dy} \]

The integral of \(-\frac{1}{y}\) with respect to \(y\) is \(-\ln|y|\). Since the problem states \(y > 0\), we can use \(-\ln(y)\).

\[ IF = e^{-\ln(y)} = e^{\ln(y^{-1})} = y^{-1} = \frac{1}{y} \]

Solving the Linear Differential Equation

Multiply the differential equation \( \frac{dx}{dy} - \frac{1}{y} x = -y \) by the integrating factor \( \frac{1}{y} \):

\[ \frac{1}{y} \left( \frac{dx}{dy} - \frac{1}{y} x \right) = \frac{1}{y} (-y) \]

\[ \frac{1}{y} \frac{dx}{dy} - \frac{1}{y^2} x = -1 \]

The left side of the equation is the derivative of \((x \times IF)\) with respect to \(y\). So, it is \( \frac{d}{dy} \left( x \cdot \frac{1}{y} \right) \).

\[ \frac{d}{dy} \left( \frac{x}{y} \right) = -1 \]

Now, integrate both sides with respect to \(y\):

\[ \int \frac{d}{dy} \left( \frac{x}{y} \right) \, dy = \int -1 \, dy \]

\[ \frac{x}{y} = -y + C \]

where \(C\) is the constant of integration.

Solve for \(x\):

\[ x = y(-y + C) \]

\[ x = Cy - y^2 \]

This is the general solution to the differential equation.

Using the Initial Condition

We are given the initial condition \(y(1) = 1\). This means that when \(x = 1\), \(y = 1\).

Substitute these values into the general solution \(x = Cy - y^2\):

\[ 1 = C(1) - (1)^2 \]

\[ 1 = C - 1 \]

\[ C = 1 + 1 \]

\[ C = 2 \]

Finding the Particular Solution

Substitute the value of \(C=2\) back into the general solution:

\[ x = 2y - y^2 \]

This is the particular solution that satisfies the initial condition.

Evaluating y(-3)

We need to find the value(s) of \(y\) when \(x = -3\). Substitute \(x = -3\) into the particular solution:

\[ -3 = 2y - y^2 \]

Rearrange this into a standard quadratic equation form \(ay^2 + by + c = 0\):

\[ y^2 - 2y - 3 = 0 \]

Now, solve this quadratic equation for \(y\). We can factor the quadratic expression:

Find two numbers that multiply to -3 and add up to -2. These numbers are -3 and 1.

\[ (y - 3)(y + 1) = 0 \]

This gives two possible solutions for \(y\):

• \(y - 3 = 0 \implies y = 3\)
• \(y + 1 = 0 \implies y = -1\)

Considering the Constraint y > 0

The problem statement includes the condition \(y > 0\).

Let's check our possible values for \(y\):

• \(y = 3\) satisfies the condition \(y > 0\).
• \(y = -1\) does not satisfy the condition \(y > 0\).

Therefore, under the given condition \(y > 0\), the only valid value for \(y\) when \(x = -3\) is 3.

Summary of Steps

Based on our analysis, when \(x = -3\), the value of \(y\) is 3, satisfying the condition \(y > 0\).

Revision Table: Key Concepts for Differential Equations

Additional Information: Constraints in Differential Equation Problems

It is common for differential equation problems to include constraints on the variables, such as \(y > 0\) or \(x \neq 0\). These constraints are important and must be considered when determining the final answer.

• Constraints can affect the domain of the solution or the validity of intermediate steps (like taking logarithms, which requires positive arguments).
• In this problem, the constraint \(y > 0\) was crucial in selecting the correct value of \(y\) from the possible solutions obtained by solving the quadratic equation. Without this constraint, both \(y=3\) and \(y=-1\) would be mathematically valid solutions to \(y^2 - 2y - 3 = 0\) when \(x=-3\). However, the DE is given for \(y > 0\), so we must adhere to that condition.