Problem 18 - Advanced Engineering Mathematics

Question

Determine whether the given differential equation is exact. If it is exact, solve it. $$ \left(2 y \sin x \cos x-y+2 y^{2} e^{x^{2}}\right) d x=\left(x-\sin ^{2} x-4 x y e^{x^{2}}\right) d y $$

Step-by-Step Solution

Answer

The differential equation is not exact.

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

Start by identifying the terms associated with \( dx \) and \( dy \) from the differential equation. Our given equation is:\[(2y \sin x \cos x - y + 2y^2 e^{x^2}) \, dx = (x - \sin^2 x - 4xy e^{x^2}) \, dy\]This gives us:\[M(x, y) = 2y \sin x \cos x - y + 2y^2 e^{x^2}\]\[N(x, y) = x - \sin^2 x - 4xy e^{x^2}\]

Step 2: Calculate the partial derivatives

To check if the equation is exact, calculate the partial derivatives of \( M \) and \( N \). Specifically, compute \( \frac{\partial M}{\partial y} \) and \( \frac{\partial N}{\partial x} \):\[\frac{\partial M}{\partial y} = 2 \sin x \cos x - 1 + 4y e^{x^2}\]\[\frac{\partial N}{\partial x} = 1 - 2\sin x \cos x - 4y^2 e^{x^2}\]

Step 3: Check for Exactness

The equation is exact if \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \). Compare the results from Step 2:\[\frac{\partial M}{\partial y} = 2 \sin x \cos x - 1 + 4y e^{x^2}\]\[\frac{\partial N}{\partial x} = 1 - 2\sin x \cos x - 4y^2 e^{x^2}\]Since these are not equal, the equation is not exact.

Step 4: Conclusion

The differential equation is not exact as the condition \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \) does not hold. Exactness is necessary for solving an exact differential equation.

Key Concepts

Partial Derivatives

Partial derivatives are a core concept in calculus, especially when dealing with functions of multiple variables. They help us understand how a function changes as one variable changes, while the others are held constant.

For a function of two variables, such as \( f(x, y) \), we often calculate \( \frac{\partial f}{\partial x} \) or \( \frac{\partial f}{\partial y} \). These notations represent the partial derivatives of \( f \) with respect to \( x \) and \( y \), respectively.

In the context of differential equations, partial derivatives are used to determine the exactness condition. This involves checking if the partial derivatives of certain expressions, associated with differentials, are equal. Although this might sound complicated, it helps in verifying whether a differential equation can be expressed as a simple derivative.

Exactness Condition

The exactness condition is a crucial criterion for solving differential equations. It tells us whether a differential equation is 'exact', meaning it can be derived from a potential function.

To verify the exactness of an equation of the form \( M(x, y) \, dx + N(x, y) \, dy = 0 \), we need to compare the partial derivatives: \( \frac{\partial M}{\partial y} \) and \( \frac{\partial N}{\partial x} \). If these derivatives are equal, the equation is exact.

In our exercise, the two derivatives were found to be unequal: \( 2 \sin x \cos x - 1 + 4y e^{x^2} \) and \( 1 - 2\sin x \cos x - 4y^2 e^{x^2} \). Since they do not match, the equation fails the exactness condition and is not exact.

Differential Equation Solution

The solution to a differential equation involves finding a function, or set of functions, that satisfy the original equation. For exact differential equations, this usually means integrating to find a potential function.

When an equation is determined to be exact, integration of the components \( M(x, y) \) and \( N(x, y) \) should lead directly to a function \( \psi(x, y) \). This function represents the solution to the differential equation.

Unfortunately, when a differential equation is not exact, like in our case, the typical straightforward method of integrating cannot be directly applied. Other techniques or transformations may be needed to solve the equation, such as finding an integrating factor.

Mathematical Analysis

Mathematical analysis is the study of limits, continuity, and the infinite, providing the theoretical underpinnings of calculus and differential equations. It is crucial for understanding the behavior of functions and ensuring solutions to differential equations are properly derived.

In differential equation analysis, we often deal with conditions like exactness, rely on calculus concepts, and utilize partial derivatives. Analysis allows us to rigorously prove that the solutions found actually satisfy the equations they were derived from.

The verification process we applied involves mathematical analysis by testing the exactness condition and computing partial derivatives. Even if a differential equation isn't exact, understanding its mathematical properties is important for exploring alternative solution strategies.