Problem 16 - Advanced Engineering Mathematics

Question

Let \(M=-2 y\) and \(N=5 y-2 x\) so that \(M_{y}=-2=N_{x} .\) From \(f_{x}=-2 y\) we obtain \(f=-2 x y+h(y), h^{\prime}(y)=5 y\) and \(h(y)=\frac{5}{2} y^{2} .\) A solution is \(-2 x y+\frac{5}{2} y^{2}=c\)

Step-by-Step Solution

Answer

The solution is \(-2xy + \frac{5}{2}y^2 = c\).

Step 1: Identify Given Information

The problem gives us two expressions, \(M = -2y\) and \(N = 5y - 2x\), along with their derivatives: \(M_y = -2\) and \(N_x = -2\). We also have \(f_x = -2y\) and we need to find a function \(f(x, y)\).

Step 2: Integrate f_x to Find f(x, y)

We start with \(f_x = -2y\). Integrating \(-2y\) with respect to \(x\), we get \(f(x, y) = -2xy + h(y)\), where \(h(y)\) is a function of \(y\) alone.

Step 3: Differentiate with Respect to y

Differentiate the expression \(f(x, y) = -2xy + h(y)\) with respect to \(y\) to check consistency with \(N\). This gives \(f_y = -2x + h'(y)\).

Step 4: Set Up the Equation f_y = N

To ensure that the derivative matches \(N\), set \(-2x + h'(y) = 5y - 2x\). Simplifying, we find \(h'(y) = 5y\).

Step 5: Integrate to Find h(y)

Integrate \(h'(y) = 5y\) with respect to \(y\) to find \(h(y) = \frac{5}{2}y^2\).

Step 6: Write the General Solution

Substituting \(h(y)\) back into \(f(x, y)\), we get \(f(x, y) = -2xy + \frac{5}{2}y^2\). The solution is \(-2xy + \frac{5}{2}y^2 = c\), where \(c\) is a constant.

Step 7: Verify Consistency

Verify that the solution satisfies both the original system \(M_y = N_x\) which is consistent throughout the derivation.

Key Concepts

Exact Equations

When dealing with differential equations, exact equations are a specific type where you can find a common potential function whose mixed partial derivatives match the given expressions. Consider it a method that often simplifies solving differential equations by transforming them into straightforward integration problems. An exact equation is shaped as: \[ M(x, y) + N(x, y) \frac{dy}{dx} = 0 \] The heart of exactness lies in ensuring the condition: \[ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \] This condition confirms that there exists a function \( f(x, y) \) whose partial derivatives with respect to \( x \) and \( y \) correspond to the functions\( M \) and \( N \) respectively.

• For our example, checking the exactness was the first step and was confirmed as \( M_y = N_x = -2 \).
• Recognizing an equation as exact helps in proceeding with the integration process directly.

Integration

Integration is a key mathematical process used to determine the original function from its derivative. In the context of exact equations, integration serves as a bridge to combine the partial derivatives into a cohesive function \( f(x, y) \).

• For instance, given \( f_x = -2y \), integrating with respect to \( x \) involves treating \( y \) as a constant.
• This yields \( f(x, y) = -2xy + h(y) \), where \( h(y) \) is an arbitrary function of \( y \) to satisfy the integration.

Using integration here helps to recover the term that only depends on \( y \) and is necessary for the function to be consistent in terms of both variables. Integration remains an invaluable tool in solving differential equations as it enables us to handle the missing parts of these functions.

Partial Derivatives

Partial derivatives signify how a function changes as its variables shift, independently. In differential equations, these derivatives reveal how sensitive each dependent component is concerning its corresponding variable.

• In our case, from \( f(x, y) = -2xy + h(y) \), the derivative \( f_y \) becomes \( -2x + h'(y) \).
• Adjusting this to match the given \( N \), or \( 5y - 2x \), sets the necessity for \( h'(y) = 5y \).

With this setup, partial derivatives offer a method to confirm that all components of the integrated function align with the original differential equation. Their role is crucial, as they help verify and adjust functions during integration, ensuring solutions correctly model the given relationships.