Understanding the Differential Equation

The given differential equation is \(x dy = y(dx + ydy)\). We are also provided with an initial condition \(y(1) = 1\) and the constraint \(y(x) > 0\) for all relevant \(x\). Our goal is to find the value of \(y(-3)\).

First, let's rearrange the given equation to identify its type.

\(x dy = y dx + y^2 dy\)

Move the \(y^2 dy\) term to the left side:

\(x dy - y^2 dy = y dx\)

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

\((x - y^2) dy = y dx\)

We can rewrite this in the form \(\frac{dx}{dy}\) to see if it fits a standard type:

\(\frac{dx}{dy} = \frac{x - y^2}{y}\)

Separate the terms on the right side:

\(\frac{dx}{dy} = \frac{x}{y} - \frac{y^2}{y}\)

\(\frac{dx}{dy} = \frac{x}{y} - y\)

Rearrange it into the standard form of a first-order linear differential equation in \(x\) with respect to \(y\):

\(\frac{dx}{dy} - \frac{1}{y} x = -y\)

This equation is of the form \(\frac{dx}{dy} + P(y)x = Q(y)\), where \(P(y) = -\frac{1}{y}\) and \(Q(y) = -y\).

Solving the First-Order Linear Differential Equation

To solve this linear differential equation, we use an integrating factor. The integrating factor (IF) is given by \(e^{\int P(y) dy}\).

Calculate the integral of \(P(y)\):

\(\int P(y) dy = \int -\frac{1}{y} dy = -\ln|y|\)

Since we are given \(y(x) > 0\), we can use \(\ln(y)\).

So, \(\int P(y) dy = -\ln(y) = \ln(y^{-1}) = \ln\left(\frac{1}{y}\right)\).

Now, find the integrating factor:

IF \(= e^{\int P(y) dy} = e^{\ln\left(\frac{1}{y}\right)} = \frac{1}{y}\).

The general solution of a linear differential equation \(\frac{dx}{dy} + P(y)x = Q(y)\) is given by \(x \cdot (\text{IF}) = \int (Q(y) \cdot \text{IF}) dy + C\).

Substitute the values of IF and \(Q(y)\):

\(x \cdot \frac{1}{y} = \int \left(-y \cdot \frac{1}{y}\right) dy + C\)

\(\frac{x}{y} = \int (-1) dy + C\)

\(\frac{x}{y} = -y + C\)

Multiply by \(y\) to solve for \(x\):

\(x = -y^2 + Cy\)

This is the general solution to the differential equation.

Applying the Initial Condition y(1) = 1

We are given that when \(x=1\), \(y=1\). Substitute these values into the general solution \(x = -y^2 + Cy\) to find the value of the constant \(C\).

\(1 = -(1)^2 + C(1)\)

\(1 = -1 + C\)

\(1 + 1 = C\)

\(C = 2\)

Now substitute the value of \(C\) back into the general solution to get the particular solution:

\(x = -y^2 + 2y\)

Finding y(-3)

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

\(-3 = -y^2 + 2y\)

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

\(y^2 - 2y - 3 = 0\)

We can solve this quadratic equation for \(y\) by factoring. We need 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 \(y\):

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

However, the problem states that \(y(x) > 0\). We must choose the solution for \(y\) that is positive.

• \(y = 3\) is positive (\(3 > 0\)).
• \(y = -1\) is not positive (\(-1 \ngtr 0\)).

Therefore, based on the constraint \(y(x) > 0\), the only valid solution for \(y(-3)\) is 3.

Revision Table: Key Steps

Additional Information: Linear Differential Equations

A first-order linear differential equation has the general form \(\frac{dy}{dx} + P(x)y = Q(x)\) or \(\frac{dx}{dy} + P(y)x = Q(y)\).

• If the form is \(\frac{dy}{dx} + P(x)y = Q(x)\), the integrating factor is \(e^{\int P(x) dx}\), and the general solution is \(y \cdot (\text{IF}) = \int (Q(x) \cdot \text{IF}) dx + C\).
• If the form is \(\frac{dx}{dy} + P(y)x = Q(y)\), the integrating factor is \(e^{\int P(y) dy}\), and the general solution is \(x \cdot (\text{IF}) = \int (Q(y) \cdot \text{IF}) dy + C\).

The integrating factor method is a standard technique for solving these types of differential equations. It works by multiplying the entire equation by a function (the integrating factor) that makes the left side the derivative of a product, which can then be easily integrated.