Question

Determine whether or not each of the equations is exact. If it is exact, find the solution. $$ \left(e^{x} \sin y-2 y \sin x\right) d x+\left(e^{x} \cos y+2 \cos x\right) d y=0 $$

Short answer

Question: Determine whether the given equation is exact, and if so, find its solution: $$e^x\sin y dx - 2y\sin x dx + e^x\cos y dy + 2\cos x dy = 0$$ Answer: The given equation is exact, and its solution is $$e^x\sin y - 2y\cos x = C$$ where C is the constant of integration.

Step-by-step solution

Identify M and N

The given equation is in the form of \(M(x, y) dx + N(x, y) dy = 0\), where $$M(x, y) = e^x\sin y - 2y\sin x$$ $$N(x, y) = e^x\cos y + 2\cos x$$

Calculate the partial derivatives

Now we will calculate the partial derivatives \(\frac{\partial M}{\partial y}\) and \(\frac{\partial N}{\partial x}\): $$\frac{\partial M}{\partial y} = e^x\cos y - 2\sin x$$ $$\frac{\partial N}{\partial x} = e^x\cos y - 2\sin x$$

Determine if the equation is exact

Since the mixed partial derivatives are equal: $$\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$$ This confirms that the given equation is exact.

Integration

Since the equation is exact, we now integrate M with respect to x and N with respect to y: $$\int M dx = \int (e^x\sin y - 2y\sin x) dx$$ $$\int N dy = \int (e^x\cos y + 2\cos x) dy$$ $$\int M dx = e^x\sin y - 2y\cos x + g(y)$$ $$\int N dy = e^x\cos y + 2y\cos x + h(x)$$

Combine the integration results

Now, we combine the results of the integration and compare the constants of integration \(g(y)\) and \(h(x)\). The constants should be equal, but possibly with a scalar constant involved. In this case, they match exactly: $$g(y) = h(x) = 0$$ So, the solution is: $$e^x\sin y - 2y\cos x = C$$ where C is the constant of integration.

Key concepts

Partial Derivatives

Partial derivatives are a fundamental concept when dealing with functions of multiple variables. They help determine how a function changes as we vary one of the variables while keeping the rest constant. In the context of exact differential equations, this is crucial to understand.

When we have a function like \( M(x, y) = e^x \sin y - 2y \sin x \), the partial derivative \( \frac{\partial M}{\partial y}\) tells us how \( M \) changes with respect to \( y \), treating \( x \) as constant. Similarly, \( \frac{\partial N}{\partial x}\) for \( N(x, y) = e^x \cos y + 2\cos x \) indicates the change in \( N \) with respect to \( x \) while keeping \( y \) constant.

Partial derivatives are calculated by differentiating the given function with respect to the variable of interest, just as we do in single-variable calculus, but treating all other variables as constants. This concept helps in checking whether a differential equation is exact, as it involves comparing the partial derivatives of different terms.

Integration

Integration is the process of finding a function whose derivative is the given function. In exact differential equations, integration comes into play once we establish that the equation is indeed exact.

For example, to integrate the function \( M(x, y) = e^x \sin y - 2y \sin x \) with respect to \( x \), we treat \( y \) as a constant. We perform the integral \( \int M dx \) by integrating each term separately:

• The integral of \( e^x \sin y \) with respect to \( x \) is \( e^x \sin y \).
• The integral of \( -2y \sin x \) with respect to \( x \) is \(-2y \cos x\).

Similarly, for \( N(x, y) \), we integrate with respect to \( y \), holding \( x \) constant:

• The integral of \( e^x \cos y \) with respect to \( y \) is \(-e^x \sin y\).
• The integral of \( 2\cos x \) with respect to \( y \) is \( 2y \cos x \).

These integrals help us form the potential function whose level curve represents the solution to the differential equation.

Exactness Condition

The exactness condition in a differential equation is crucial for determining whether we can proceed with methods applicable to exact equations. A differential equation of the form \( M(x, y) dx + N(x, y) dy = 0 \) is exact if the mixed partial derivatives of its terms satisfy \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \).

This condition ensures that both \( M \) and \( N \) come from the gradient of a single potential function. If the equation is exact, a potential function exists whose differential equals \( M dx + N dy \). Recognizing the exactness condition allows us to pair functions to find a single, coherent solution as we did with the integrals.

When the mixed partial derivatives are equal, it implies symmetry in the ways the function grows in response to changes in \( x \) and \( y \). Thus, it confirms that integrating either \( M \) with \( x \) or \( N \) with \( y \) will eventually lead to finding a common potential function – the key to solving the problem.

Mixed Partial Derivatives

Mixed partial derivatives involve taking the derivative of one partial derivative with respect to another variable. In multi-variable functions, they provide information on how the rate of change of one partial derivative is affected by another variable.

For an exact differential equation, checking if \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \) is necessary. It tells us whether the function \( M \) and \( N \) derive from a common potential function. The equality of these mixed partial derivatives suggests that the function is "well-behaved" across its variables and confirms that the derivatives are intertwined.

This is why in our earlier example, finding \( \frac{\partial M}{\partial y} = e^x \cos y - 2\sin x \) and \( \frac{\partial N}{\partial x} = e^x \cos y - 2\sin x \) and having them equal, ensures the equation is exact. It symbolizes a mutual dependency between the variables, facilitating the integration process and pointing towards a robust solution. Such properties are pivotal in determining the solution path for these equations efficiently.