Problem 9 - Advanced Engineering Mathematics

Question

Let \(M=y^{3}-y^{2} \sin x-x\) and \(N=3 x y^{2}+2 y \cos x\) so that \(M_{y}=3 y^{2}-2 y \sin x=N_{x} .\) From \(f_{x}=y^{3}-y^{2} \sin x-x\) we obtain \(f=x y^{3}+y^{2} \cos x-\frac{1}{2} x^{2}+h(y), h^{\prime}(y)=0,\) and \(h(y)=0 .\) A solution is \(x y^{3}+y^{2} \cos x-\frac{1}{2} x^{2}=c\)

Step-by-Step Solution

Answer

The solution is given by \( xy^3 + y^2 \cos x - \frac{1}{2}x^2 = c \).

Step 1: Analyze the Problem

We are given two functions, \( M = y^3 - y^2 \sin x - x \) and \( N = 3xy^2 + 2y \cos x \). We need to verify if these satisfy the condition \( M_y = N_x \). We also have a proposed solution form \( f(x,y) = c \) and we need to solve for \( f(x,y) \).

Step 2: Differentiate M with Respect to y

Compute the partial derivative of \( M \) with respect to \( y \): \[ \frac{\partial}{\partial y}(M) = \frac{\partial}{\partial y}(y^3 - y^2 \sin x - x) = 3y^2 - 2y \sin x \]

Step 3: Differentiate N with Respect to x

Compute the partial derivative of \( N \) with respect to \( x \): \[ \frac{\partial}{\partial x}(N) = \frac{\partial}{\partial x}(3xy^2 + 2y \cos x) = 3y^2 - 2y \sin x \]

Step 4: Verify Equation M_y = N_x

We check if \( M_y = N_x \): Both derivatives in Steps 2 and 3 result in \( 3y^2 - 2y \sin x \), confirming that \( M_y = N_x \) holds true.

Step 5: Find the Potential Function f(x,y)

The problem suggests \( f_y = N \) and \( f_x = M \). Starting with \( f_x = y^3 - y^2 \sin x - x \), integrate with respect to \( x \): \[ f(x,y) = xy^3 + y^2 \cos x - \frac{1}{2}x^2 + h(y) \] where \( h(y) \) is some function of \( y \).

Step 6: Determine h(y)

Since \( h'(y) = 0 \) as mentioned in the problem, this implies \( h(y) \) is a constant. But if \( h(y) = 0 \), then there’s no additional term to consider.

Step 7: Construct the Solution Equation

Using the expression from Step 5 and \( h(y) = 0 \), the solution to the problem is \( f(x,y) = xy^3 + y^2 \cos x - \frac{1}{2}x^2 \). Set \( f(x,y) = c \) where \( c \) is a constant for the solution to the differential equation.

Key Concepts

Partial Derivatives

Partial derivatives are a key tool in calculus for examining functions of multiple variables. They represent how a function changes as one of the variables changes, while keeping the other variables constant. For example, given a function \( M(y, x) = y^3 - y^2 \sin x - x \), the partial derivative \( \frac{\partial}{\partial y}(M) \) indicates how \( M \) changes with respect to \( y \), assuming \( x \) is constant.
This is done by differentiating only with respect to \( y \). Performing this on our function yields \( 3y^2 - 2y \sin x \), highlighting how \( M \) varies as \( y \) changes. Similarly, the partial derivative \( \frac{\partial}{\partial x}(N) \) on another function \( N(x, y) = 3xy^2 + 2y \cos x \) provides insights on changes concerning \( x \).

• Partial derivatives can be thought of as slices through the surface of a multi-variable function, giving a slope in one direction.
• In exact differential equations, matching partial derivatives helps verify certain conditions for finding solutions.

Potential Functions

The concept of a potential function \( f(x, y) \) in the context of differential equations is paramount. A potential function can simplify the process of solving differential equations.
It is a function whose partial derivatives match with the terms of the differential equation, thereby allowing us to express the original form in a simpler way.
For example, the suggested potential function starts with \( f_x = y^3 - y^2 \sin x - x \). Integrating this with respect to \( x \) gives our potential function \( f(x, y) = xy^3 + y^2 \cos x - \frac{1}{2}x^2 + h(y) \), where \( h(y) \) is purely a function of \( y \).

• Finding a potential function converts the problem of solving a differential equation into one of finding constants or functions of the remaining variable.
• When partial derivatives in question match, it guarantees some underlying function exists that generates both expressions.

Integration

Integration is a powerful mathematical operation that finds the anti-derivative or area under a curve of a function. In exact differential equations, integration helps in constructing the potential function.
From \( f_x = y^3 - y^2 \sin x - x \), we integrate with respect to \( x \), assuming \( y \) to be constant in this particular process. This results in \( f(x, y) = xy^3 + y^2 \cos x - \frac{1}{2}x^2 + h(y) \).

• Integration resembles the reverse process of differentiation, effectively ‘undoing’ the derivative to retrieve an original function or potential form.
• When integrating, terms that depend solely on the other variable could appear, represented here by \( h(y) \).
• Ensuring all terms are correctly retrieved is crucial as incomplete integration can cause errors in the solution construction process.

Exactness Condition

The exactness condition is a criterion that determines whether a given differential equation is exact, meaning it can be solved using a potential function.
We verify exactness by checking if \( M_y = N_x \), where \( M \) and \( N \) are components from the differential equation.
In our example, both \( \frac{\partial}{\partial y}(M) \) and \( \frac{\partial}{\partial x}(N) \) resulted in \( 3y^2 - 2y \sin x \), confirming the equation's exactness.

• An exact differential equation possesses a potential function, meaning it can be expressed as a total differential \( df \).
• This verification step ensures that the solutions we calculate are valid for given expressions of \( M \) and \( N \).
• The condition simplifies, ensuring consistent methods weigh equally derived solutions applicable to dynamic scenarios.