Problem 2 - Differential Equations and Linear Algebra

Question

Determine whether the given differential equation is exact. $$[\cos (x y)-x y \sin (x y)] d x-x^{2} \sin (x y) d y=0$$

Step-by-Step Solution

Answer

The given differential equation is not exact since \(\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}\), where \(M(x, y) = \cos(xy) - xy\sin(xy)\) and \(N(x, y) = -x^2\sin(xy)\).

Step 1: Identify M(x, y) and N(x, y)

M(x, y) is the coefficient of dx, which is: $$M(x, y) = \cos(xy) - xy\sin(xy)$$ N(x, y) is the coefficient of dy, which is: $$N(x, y) = -x^2\sin(xy)$$

Step 2: Calculate the partial derivatives

We need to find the following partial derivatives: $$\frac{\partial M}{\partial y}$$ and $$\frac{\partial N}{\partial x}$$ For \(\frac{\partial M}{\partial y}\), we have: $$\frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(\cos(xy) - xy\sin(xy))$$ Use the product rule: $$\frac{\partial M}{\partial y} = -\sin(xy)\left(x\right) - x\sin(xy) + x^2\cos(xy)$$ Simplify the result: $$\frac{\partial M}{\partial y} = -x\sin(xy) - x\sin(xy) + x^2\cos(xy)$$ For \(\frac{\partial N}{\partial x}\), we have: $$\frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(-x^2\sin(xy))$$ Use the product rule: $$\frac{\partial N}{\partial x} = -2x\sin(xy) - x^2\cos(xy)y$$ Simplify the result: $$\frac{\partial N}{\partial x} = -2x\sin(xy) - x^2y\cos(xy)$$

Step 3: Compare the partial derivatives

Now, compare \(\frac{\partial M}{\partial y}\) and \(\frac{\partial N}{\partial x}\) to see if the given equation is exact. We have: $$\frac{\partial M}{\partial y} = -x\sin(xy) - x\sin(xy) + x^2\cos(xy)$$ and $$\frac{\partial N}{\partial x} = -2x\sin(xy) - x^2y\cos(xy)$$ Since \(\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}\), the given differential equation is not exact.

Key Concepts

Partial Derivatives

Partial derivatives are used to evaluate the rate at which a function changes as its input variables change. In the context of exact differential equations, these derivatives help determine how each component of the equation changes relative to each other.

When calculating partial derivatives, it's essential to consider the variable relative to which you're differentiating while treating all other variables as constants. For example, given a function \(M(x, y) = \cos(xy) - xy\sin(xy)\), to find the partial derivative with respect to \(y\), we treat \(x\) as a constant. Thus, we evaluate how \(M(x, y)\) changes as \(y\) changes while maintaining \(x\) constant.

The concept of partial derivatives is crucial in exact differential equations as we need to compare partial derivatives of the terms associated with \(dx\) and \(dy\) to determine exactness. Ensuring that the partial derivative of \(M(x, y)\) with respect to \(y\) is equal to the partial derivative of \(N(x, y)\) with respect to \(x\) confirms that the equation is exact or points out that additional steps are needed if the partial derivatives don't match.

Product Rule

The product rule is a fundamental technique in calculus used when differentiating products of two functions. It's expressed as \((uv)' = u'v + uv'\), where \(u\) and \(v\) are functions of a variable.

In our example, the product rule is utilized twice: when calculating \(\frac{\partial M}{\partial y}\) and \(\frac{\partial N}{\partial x}\).

• For \(\frac{\partial M}{\partial y}\), we differentiate \(xy\sin(xy)\), treating \(xy\) as a product of \(x\) and \(y\). This requires differentiating \(\cos(xy)\) and \(-xy\sin(xy)\) separately and applying the product rule.
• For \(\frac{\partial N}{\partial x}\), the product rule is used again on \(-x^2\sin(xy)\), involving derivatives of \(-x^2\) and \(\sin(xy)\).

By slicing the functions into parts and applying the product rule, we accurately assess the contribution of each part to the overall derivative, simplifying complex expressions into more manageable parts.

Differential Equation Analysis

Analyzing whether a differential equation is exact involves comparing specific partial derivatives. In the given problem, we're interested in whether the differential equation can be rewritten as a total derivative of some potential function.

An exact differential equation has the form \(M(x, y)dx + N(x, y)dy = 0\) where the partial derivative \(\frac{\partial M}{\partial y}\) equals \(\frac{\partial N}{\partial x}\). If these conditions are met, it signifies that \(M\) and \(N\) derive from the same function, indicating exactness.

In our exercise, calculating these partial derivatives revealed that \(\frac{\partial M}{\partial y} eq \frac{\partial N}{\partial x}\). This discrepancy indicates non-exactness, suggesting the equation might not directly stem from a potential function, or it may require more advanced solving techniques like integrating factors to potentially construct a solution. Understanding this analysis allows us to interpret how components of the differential equation interplay and why exactness (or lack thereof) dictates the solution method.