Understanding the Differential Equation

The given equation is a first-order differential equation: \( y \, dx - (x + 2y^2) \, dy = 0 \). This equation involves differentials \(dx\) and \(dy\). To solve it, we need to recognize its type and apply the appropriate method.

Rearranging the Differential Equation

Let's rearrange the equation to see if it fits a standard form. We can rewrite it as:

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

To potentially transform it into a linear differential equation, let's divide by \(dy\) (assuming \(dy \neq 0\)):

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

Now, let's rearrange to group the \(x\) term:

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

Divide by \(y\) (assuming \(y \neq 0\)) to get the coefficient of \(\frac{dx}{dy}\) as 1:

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

This equation is in the form of a linear differential equation in \(x\) with respect to \(y\), which is \(\frac{dx}{dy} + P(y)x = Q(y)\).

Identifying P(y) and Q(y)

Comparing the rearranged equation \(\frac{dx}{dy} - \frac{1}{y} x = 2y\) with the standard form \(\frac{dx}{dy} + P(y)x = Q(y)\), we can identify:

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

Calculating the Integrating Factor (IF)

For a linear differential equation of the form \(\frac{dx}{dy} + P(y)x = Q(y)\), the integrating factor is given by \( IF = e^{\int P(y) dy} \). Let's calculate the integral of \(P(y)\):

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

Now, calculate the integrating factor:

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

We can use \( IF = \frac{1}{y} \) (assuming \(y \neq 0\)).

Finding the General Solution

The general solution for a 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 constant of integration.

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

\[ x \cdot \frac{1}{y} = \int \left( 2y \cdot \frac{1}{y} \right) dy + C \]

Simplify the integrand:

\[ \frac{x}{y} = \int 2 \, dy + C \]

Perform the integration:

\[ \frac{x}{y} = 2y + C \]

Finally, multiply by \(y\) to solve for \(x\):

\[ x = y(2y + C) \]

\[ x = 2y^2 + Cy \]

This is the general solution to the given differential equation \( y \, dx - (x + 2y^2) \, dy = 0 \).

Checking the Options

Let's compare our derived general solution \( x = 2y^2 + Cy \) with the provided options:

• Option 1: \( x = y^2 + cy \) - Does not match.
• Option 2: \( x = 2cy^2 \) - Does not match.
• Option 3: \( x = 2y^2 + cy \) - Matches our derived solution (using \(c\) as the constant of integration).
• Option 4: None of the above - Incorrect, as Option 3 matches.

General Solution Finding Summary

To find the general solution of \( y \, dx - (x + 2y^2) \, dy = 0 \), we transformed it into a first-order linear differential equation in \(x\) with respect to \(y\). We identified \(P(y)\) and \(Q(y)\), calculated the integrating factor, and used the standard formula for the general solution of linear differential equations.

Revision Table: Key Differential Equation Concepts

Additional Information: Solving Differential Equations

Differential equations are mathematical equations that relate a function with its derivatives. They are used to model many real-world processes in physics, engineering, economics, biology, and more. Solving a differential equation means finding the function that satisfies the equation. There are various types of differential equations and corresponding methods to solve them.

• Separable Equations: Can be written in the form \( f(x) dx + g(y) dy = 0 \).
• Exact Equations: \( M(x,y)dx + N(x,y)dy = 0 \) is exact if \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \).
• Homogeneous Equations: \( \frac{dy}{dx} = f(\frac{y}{x}) \).
• Linear Equations: Discussed above (\( \frac{dy}{dx} + P(x)y = Q(x) \) or \( \frac{dx}{dy} + P(y)x = Q(y) \)).
• Bernoulli's Equation: \( \frac{dy}{dx} + P(x)y = Q(x)y^n \). Can be transformed into a linear equation.

Understanding how to identify the type of differential equation is the first crucial step towards finding its general solution.