Solving the Differential Equation \(\frac{{dx}}{{dy}} + \frac{x}{y} - {y^2} = 0\)

The given differential equation is:

\(\frac{{dx}}{{dy}} + \frac{x}{y} - {y^2} = 0\)

We can rewrite this equation in the standard form of a first-order linear differential equation with respect to \(y\), which is \(\frac{dx}{dy} + P(y)x = Q(y)\).

Rearranging the given equation, we get:

\(\frac{{dx}}{{dy}} + \frac{1}{y}x = {y^2}\)

Comparing this to the standard form, we can identify \(P(y)\) and \(Q(y)\):

• \(P(y) = \frac{1}{y}\)
• \(Q(y) = {y^2}\)

To solve a first-order linear differential equation, we use the integrating factor method. The integrating factor (IF) is given by the formula \(IF = e^{\int P(y) dy}\).

Let's calculate the integrating factor:

\(IF = e^{\int \frac{1}{y} dy}\)

\(IF = e^{\ln|y|}\)

Assuming \(y > 0\), we have:

\(IF = y\)

The general solution for a first-order linear differential equation \(\frac{dx}{dy} + P(y)x = Q(y)\) is given by:

\(x \cdot (IF) = \int Q(y) \cdot (IF) \, dy + C\)

Where \(C\) is the arbitrary constant of integration.

Now, we substitute the values of \(x\), \(IF\), and \(Q(y)\) into the general solution formula:

\(x \cdot y = \int {y^2} \cdot y \, dy + C\)

\(xy = \int {y^3} \, dy + C\)

Next, we integrate \(y^3\) with respect to \(y\):

\(xy = \frac{y^{3+1}}{3+1} + C\)

\(xy = \frac{y^4}{4} + C\)

We need to match this solution to the given options. Let's rearrange our solution by multiplying the entire equation by 4:

\(4 \cdot (xy) = 4 \cdot \left(\frac{y^4}{4}\right) + 4 \cdot C\)

\(4xy = y^4 + 4C\)

Since \(C\) is an arbitrary constant, \(4C\) is also an arbitrary constant. We can replace \(4C\) with a new arbitrary constant, let's call it \(c\), as used in the options.

\(4xy = y^4 + c\)

This matches one of the provided options.

Comparing with Options

• Option 1: \(xy = x^4+ c\) - Does not match our solution.
• Option 2: \(xy = y^4+ c\) - Does not match our solution (differs by a factor of 4 on the \(y^4\) term).
• Option 3: \(4xy = y^4+ c\) - Matches our derived solution.
• Option 4: \(3xy = y^3+ c\) - Does not match our solution.

Therefore, the solution to the differential equation \(\frac{{dx}}{{dy}} + \frac{x}{y} - {y^2} = 0\) is \(4xy = y^4 + c\).

Key Steps in Solving the Differential Equation

Revision Table: First-Order Linear Differential Equations

Additional Information: The Integrating Factor Method

The integrating factor method is a powerful technique used to solve first-order linear differential equations. A differential equation is considered linear if it is linear in the dependent variable and its derivatives. For a first-order equation, this means the dependent variable and its first derivative appear only to the power of one, and are not multiplied together.

The goal of the integrating factor method is to transform the left-hand side of the differential equation into the derivative of a product of the dependent variable and the integrating factor. This makes the equation easy to integrate directly.

For the standard form \(\frac{dx}{dy} + P(y)x = Q(y)\), multiplying by the integrating factor \(IF = e^{\int P(y) dy}\) gives:

\(e^{\int P(y) dy} \frac{dx}{dy} + P(y) e^{\int P(y) dy} x = Q(y) e^{\int P(y) dy}\)

The left side is exactly the derivative of the product \(x \cdot e^{\int P(y) dy}\) with respect to \(y\), based on the product rule. That is:

\(\frac{d}{dy} \left( x \cdot e^{\int P(y) dy} \right) = e^{\int P(y) dy} \frac{dx}{dy} + x \cdot \frac{d}{dy} \left( e^{\int P(y) dy} \right)\)

Using the chain rule, \(\frac{d}{dy} \left( e^{\int P(y) dy} \right) = e^{\int P(y) dy} \cdot \frac{d}{dy} \left( \int P(y) dy \right) = e^{\int P(y) dy} \cdot P(y)\).

So, the equation becomes:

\(\frac{d}{dy} (x \cdot IF) = Q(y) \cdot IF\)

Integrating both sides with respect to \(y\) gives:

\(\int \frac{d}{dy} (x \cdot IF) \, dy = \int Q(y) \cdot IF \, dy\)

\(x \cdot IF = \int Q(y) \cdot IF \, dy + C\)

This is the general solution formula we used.