Problem 12 - Advanced Engineering Mathematics

Question

Let \(M=3 x^{2} y+e^{y}\) and \(N=x^{3}+x e^{y}-2 y\) so that \(M_{y}=3 x^{2}+e^{y}=N_{x} .\) From \(f_{x}=3 x^{2} y+e^{y}\) we obtain \(f=x^{3} y+x e^{y}+h(y), h^{\prime}(y)=-2 y,\) and \(h(y)=-y^{2} .\) A solution is \(x^{3} y+x e^{y}-y^{2}=c\)

Step-by-Step Solution

Answer

The solution is \( x^3 y + xe^y - y^2 = c \).

Step 1: Differentiate M with respect to y

Start by differentiating the function \( M = 3x^2 y + e^y \) with respect to \( y \). The partial derivative \( M_y = \frac{\partial}{\partial y}(3x^2 y + e^y) = 3x^2 + e^y \).

Step 2: Differentiate N with respect to x

Next, differentiate the function \( N = x^3 + xe^y - 2y \) with respect to \( x \). The partial derivative \( N_x = \frac{\partial}{\partial x}(x^3 + xe^y - 2y) = 3x^2 + e^y \). Notice that \( N_x = M_y \), confirming the condition for being an exact differential equation.

Step 3: Integrate M with respect to x

Integrate \( M = 3x^2 y + e^y \) with respect to \( x \) to find the potential function \( f \).\[ f(x, y) = \int (3x^2 y + e^y) \, dx = x^3 y + xe^y + h(y) \] The function \( h(y) \) is an arbitrary function to include any terms solely dependent on \( y \).

Step 4: Find h(y) using derivative constraint

Since \( f(x, y) = x^3 y + xe^y + h(y) \), differentiate \( f \) with respect to \( y \):\( \frac{\partial f}{\partial y} = x^3 + xe^y + h'(y) \).Set this equal to \( M = 3x^2 y + e^y \) to solve for \( h'(y) = -2y \).

Step 5: Integrate to find h(y)

Integrate \( h'(y) = -2y \) with respect to \( y \) to find \( h(y) \):\[ h(y) = \int -2y \, dy = -y^2 + C \]. Since \( C \) is a constant, it can be absorbed into other terms later.

Step 6: Write the full solution

Substitute \( h(y) = -y^2 \) back into the potential function \( f(x, y) \): \[ f(x, y) = x^3 y + xe^y - y^2 \]The equation of the solution is \( x^3 y + xe^y - y^2 = c \), where \( c \) is a constant.

Key Concepts

Partial Derivatives

Partial derivatives are a crucial concept in calculus, especially when dealing with functions of multiple variables. When you have a function like \( M(x, y) = 3x^2 y + e^y \), the partial derivative with respect to one variable considers how the function changes while keeping the other variable constant.

For instance, to find \( M_y \), the partial derivative of \( M \) with respect to \( y \), you treat \( x \) as a constant. This means:

• In the term \( 3x^2 y \), \( 3x^2 \) is constant, so the derivative is simply \( 3x^2 \).
• The derivative of \( e^y \) with respect to \( y \) is \( e^y \).

Combining them gives us \( M_y = 3x^2 + e^y \).
This same method applies when differentiating \( N \) with respect to \( x \), where you consider \( y \) constant to find that \( N_x = 3x^2 + e^y \). Recognizing these derivatives as equal is key to determining the exactness of the differential equation.

Integration

Integration is the process of finding the integral of a function. This is a fundamental technique in solving exact differential equations.

In our exercise, we integrate the function \( M(x, y) = 3x^2 y + e^y \) with respect to \( x \) to find the potential function \( f(x, y) \). Here's a simple breakdown:

• The integral of \( 3x^2 y \) with respect to \( x \) is \( x^3 y \).
• The integral of \( e^y \) with respect to \( x \) is \( xe^y \), even though \( e^y \) seems dependent on \( y \), it behaves as a constant in this context.

The resulting integral, \( f(x, y) = x^3 y + xe^y + h(y) \), includes an arbitrary function \( h(y) \), which accounts for any additional dependency on \( y \) that could not be identified in this process.

Potential Functions

Potential functions are solutions to exact differential equations and represent the "integral" of the differential system. In this context, once we have an exact equation, we look for a potential function that satisfies it.

For the equation in this problem, the potential function \( f \) is derived from integrating the component \( M \). The purpose of \( f(x, y) = x^3 y + xe^y + h(y) \) is to encapsulate the dynamics of the differential equation in a single equation.

This function must be verified by differentiating it with respect to \( y \) to confirm the consistency with the original function component \( N_y \). Any leftover terms in \( y \) are accounted for by determining the derivative of \( h(y) \). Hence, locating \( h(y) = -y^2 \) provides a complete potential function that aligns entirely with the original differential equations.

Arbitrary Functions

Arbitrary functions, such as \( h(y) \) in our exercise, play a crucial role in the integration process of differential equations. They appear when integrating functions with respect to one variable, allowing flexibility to account for unspecified dependencies on other variables.

In our problem, when integrating \( M(x, y) \) with respect to \( x \), \( h(y) \) is introduced because it represents any function purely dependent on \( y \) that doesn't affect the differentiation with respect to \( x \).

To determine \( h(y) \), you match the derivative of the potential function with the other function component, \( N \). Solving \( h'(y) = -2y \) and integrating gives \( h(y) = -y^2 + C \). Since \( C \) is a constant, it can be omitted in the context of our solution or absorbed elsewhere. This method ensures the solution remains valid for the entire set of conditions imposed by the original differential equation.