Problem 31 - Advanced Engineering Mathematics

Question

We note that \(\left(M_{y}-N_{x}\right) / N=1 / x,\) so an integrating factor is \(e^{\int d x / x}=x .\) Let \(M=2 x y^{2}+3 x^{2}\) and \(N=2 x^{2} y\) so that \(M_{y}=4 x y=N_{x} .\) From \(f_{x}=2 x y^{2}+3 x^{2}\) we obtain \(f=x^{2} y^{2}+x^{3}+h(y), h^{\prime}(y)=0,\) and \(h(y)=0 . \mathrm{A}\) solution of the differential equation is \(x^{2} y^{2}+x^{3}=c\)

Step-by-Step Solution

Answer

The solution of the differential equation is \(x^2y^2 + x^3 = c\).

Step 1: Verify the Integrating Factor Condition

Given the condition \(\frac{M_y - N_x}{N} = \frac{1}{x}\), we first need to verify that this condition is met. Compute the partial derivatives: \(M_y = \frac{\partial}{\partial y}(2xy^2 + 3x^2) = 4xy\) and \(N_x = \frac{\partial}{\partial x}(2x^2y) = 4xy\). Since \(M_y = N_x\), the condition \(\frac{M_y-N_x}{N} = \frac{1}{x}\) holds as \(\frac{0}{N} = 0 = 0\).

Step 2: Determine the Integrating Factor

We are given that an integrating factor for the differential equation is \(\mu = e^{\int\frac{dx}{x}} = x\). This needs to be applied to both functions \(M\) and \(N\) as the given equation is not exact.

Step 3: Find the Exact Differential Equation

After multiplying by the integrating factor \(x\), redefine \(M\) and \(N\): \(M = x(2xy^2 + 3x^2) = 2x^2y^2 + 3x^3\) and \(N = x(2x^2y) = 2x^3y\). Ensure that the new derivatives are equal to form an exact differential.

Step 4: Construct the Potential Function \(f\)

To find \(f\), solve \(\frac{\partial f}{\partial x} = 2x^2y^2 + 3x^3\) by integrating w.r.t. \(x\): \(f = \int (2x^2y^2 + 3x^3) \, dx = x^2y^2 + x^3 + h(y)\).

Step 5: Solve for \(h(y)\) using \(N\)

Use \(\frac{\partial f}{\partial y} = 2x^2y = N\). From the calculated \(f\), \(\frac{\partial f}{\partial y} = 2x^2y + h'(y)\). Equate \(2x^2y + h'(y) = 2x^3y\) to find \(h'(y) = 0\), thus \(h(y) = 0\).

Step 6: Identify the General Solution

Combine the results from previous steps to determine the general solution of the differential equation: \(f(x, y) = x^2y^2 + x^3\). Set \(f(x, y) = c\) where \(c\) is an arbitrary constant.

Key Concepts

Exact Differential Equations

In the realm of differential equations, an equation is labeled as exact if it can be expressed in the form \( M(x, y)dx + N(x, y)dy = 0 \), where \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \). This equality of mixed partial derivatives implies that there exists a function \( f(x, y) \), known as the potential function, such that \( df = Mdx + Ndy \). Think of it as trying to retrace your path along a surface, where the path aligns perfectly with the gradient of some elevation map

When an equation is not exact, finding an integrating factor can often transform it. This is a function, typically denoted by \( \mu(x, y) \), which when multiplied by the entire equation, causes it to become exact.

However, determining this integrating factor can be challenging and usually involves trial and error or exploiting specific methods, such as recognizing patterns or satisfying specific conditions like \( \frac{M_y - N_x}{N} = \frac{1}{x} \), which we used in this instance. After applying the factor, the new \( M \) and \( N \) satisfy the exactness condition.

Potential Function

The concept of a potential function in differential equations is similar to an energy landscape where each point \( (x, y) \) has a certain potential value \( f(x, y) \). If a differential equation \( Mdx + Ndy = 0 \) is exact, it can be linked to a scalar potential function \( f \) where \( M \) and \( N \) are its partial derivatives with respect to \( x \) and \( y \), respectively.

To find this potential function, one can integrate \( M \) with respect to \( x \) and \( N \) with respect to \( y \), making sure they are consistent through a function \( h(y) \) or \( g(x) \) that only depends on one variable. For example, in our exercise, integrating \( M = 2x^2y^2 + 3x^3 \) with respect to \( x \) while considering \( N \), led us to the potential function \( f(x, y) = x^2y^2 + x^3 + h(y) \).

This function \( f \) encapsulates all solutions of the differential equation, with solutions expressed in terms of a constant \( c \). This makes solving the equation conceptually simpler, as it turns a dynamic problem into a static one.

Partial Derivatives

Partial derivatives represent the rate of change of a multivariable function with respect to one variable, holding the others constant. When dealing with functions like \( f(x, y) \), partial derivatives provide useful insights into how the function behaves as each variable changes.

In the study of exact differential equations, partial derivatives are crucial. To check if \( Mdx + Ndy = 0 \) is exact, we compare the partial derivatives \( \frac{\partial M}{\partial y} \) and \( \frac{\partial N}{\partial x} \). If equal, we can confirm the equation is exact and relate it directly to a potential function \( f \).

Calculating \( M_y \) and \( N_x \) checks the symmetry required for a potential function to exist. As seen with \( M = 2xy^2 + 3x^2 \) and \( N = 2x^2y \), where \( M_y = 4xy \) and \( N_x = 4xy \), their equality ensured that the original equation became exact once modified by an integrating factor. Understanding and manipulating these derivatives allow us to simplify and solve complex equations effectively.