Problem 20 - Differential Equations and Linear Algebra

Question

Determine whether the given function is an integrating factor for the given differential equation. $$I(x, y)=\cos (x y),[\tan (x y)+x y] d x+x^{2} d y=0$$

Step-by-Step Solution

Answer

The given function \(I(x, y) = \cos(xy)\) is NOT an integrating factor for the differential equation \([\tan(xy) + xy]dx + x^2dy = 0\) as \[\frac{(\sec^2(xy) + x) \cdot x - 2x}{x^2 - [\tan(xy) + xy]} \neq \cos(xy)\].

Step 1: Rewrite the differential equation

Identify and write the differential equation in the form \(M(x, y)dx + N(x, y)dy = 0\) \[M(x, y) = [\tan(xy) + xy]\] \[N(x, y) = x^2\] So the given differential equation is: \[[\tan(xy) + xy]dx + x^2dy = 0\]

Step 2: Find the partial derivatives of M and N

Find \[\frac{\partial M}{\partial y} = \frac{\partial [\tan(xy) + xy]}{\partial y}\] and \[\frac{\partial N}{\partial x} = \frac{\partial x^2}{\partial x}\]. First, let's find the partial derivative of \(M\) with respect to \(y\). Using the product rule and chain rule, \[\frac{\partial M}{\partial y} = (\sec^2(xy) + x) \cdot x\] Now, let's find the partial derivative of \(N\) with respect to \(x\). This is a simple power rule application: \[\frac{\partial N}{\partial x} = 2x\]

Step 3: Check if the function is an integrating factor

We'll now determine if the given function \(I(x, y) = \cos(xy)\) is an integrating factor by checking if \[\frac{\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}}{N - M} = I(x, y).\] \[\frac{(\sec^2(xy) + x) \cdot x - 2x}{x^2 - [\tan(xy) + xy]} = \cos(xy)\] Since the left-hand side of the equation does not equal the integrating factor, \(\cos (xy)\), as given: \[\frac{(\sec^2(xy) + x) \cdot x - 2x}{x^2 - [\tan(xy) + xy]} \neq \cos(xy)\] Therefore, the given function \(I(x, y) = \cos(xy)\) is NOT an integrating factor for the given differential equation.

Key Concepts

Differential Equations

Differential equations are mathematical equations that involve functions and their derivatives. They are fundamental in describing various physical phenomena like motion, heat, and fluid flow. In general, a differential equation expresses a relationship between some function \( y(x) \) and its derivatives, such as \( y'(x) \) or \( y''(x) \).

- Ordinary Differential Equations (ODEs): These involve functions of a single variable and their derivatives. For example, \( \frac{dy}{dx} = y \) is a simple ordinary differential equation. - Partial Differential Equations (PDEs): These involve functions of multiple variables and their partial derivatives, such as \( \frac{\partial^2 u}{\partial x^2} = \frac{\partial u}{\partial t} \).

In solving these equations, we often seek an integrating factor, especially for linear first-order differential equations, which can simplify the equation to its solvable form.

Partial Derivatives

Partial derivatives are used when we have functions of multiple variables. They represent the rate of change of the function with respect to one of these variables while keeping the others constant. Consider a function \( f(x, y) \); the partial derivative with respect to \( x \) is represented as \( \frac{\partial f}{\partial x} \), and with respect to \( y \), it is \( \frac{\partial f}{\partial y} \).

In the given problem, the partial derivatives of \( M(x, y) \) and \( N(x, y) \) are computed to assess if the function is an integrating factor. - For \( M(x, y) = [\tan(xy) + xy] \), \( \frac{\partial M}{\partial y} \) is calculated considering the change in \( M \) due to changes in \( y \), applying both the product and chain rules.- \( N(x, y) = x^2 \) is simpler and involves only the power rule for its partial derivative with respect to \( x \).

Product Rule

The product rule is a fundamental principle in calculus used when differentiating the product of two functions. If you have two functions, \( u(x) \) and \( v(x) \), their product \( u(x) \cdot v(x) \) is differentiated as follows:

\[\frac{d}{dx}[u(x) \cdot v(x)] = u'(x) \cdot v(x) + u(x) \cdot v'(x)\]

In the context of our differential equation, we applied the product rule when calculating the partial derivative of \( M(x, y) = \tan(xy) + xy \) with respect to \( y \). It helps us deal with the composite nature of its terms, such as \( xy \) being a product of \( x \) and \( y \).

Chain Rule

The chain rule is another essential rule in calculus for finding the derivative of composite functions. It states that if you have a composite function \( f(g(x)) \), its derivative is given by:

\[\frac{df}{dx} = \frac{df}{dg} \cdot \frac{dg}{dx}\]

In the study of differential equations, the chain rule works in tandem with the product rule when functions are nested or when calculating derivatives of composite terms.

For instance, in calculating \( \frac{\partial M}{\partial y} \) where \( M(x, y) = \tan(xy) + xy \), the chain rule is used on terms like \( \tan(xy) \) since it involves a composite function \( \tan \) of the product \( xy \). It ensures we correctly derive how changes in \( y \) affect our function as a whole.