Question

Show that the equations are not exact, but become exact when multiplied by the given integrating factor. Then solve the equations. $$ \left(\frac{\sin y}{y}-2 e^{-x} \sin x\right) d x+\left(\frac{\cos y+2 e^{-x} \cos x}{y}\right) d y=0, \quad \mu(x, y)=y e^{x} $$

Short answer

Question: Verify the integrating factor and solve the given differential equation with the integrating factor \(ye^x\). Answer: The integrating factor \(ye^x\) makes the given differential equation exact, and therefore, the general solution of the given differential equation is \(e^x\sin y - 2y\cos x = C\).

Step-by-step solution

Verify integrating factor

To verify that the given integrating factor \(\mu(x, y) = ye^x\) indeed makes the equation exact, we multiply both sides of the equation by \(\mu(x,y)\): $$ (ye^x)\left(\frac{\sin y}{y}-2e^{-x}\sin x\right)dx+(ye^x)\left(\frac{\cos y+2e^{-x}\cos x}{y}\right)dy=0 $$ Now simplify the equation: $$ (e^x\sin y - 2y\sin x) dx + (e^x\cos y + 2\cos x) dy = 0 $$ Next, check if the new equation is exact.

Check if the new equation is exact

To check if the equation is exact, we find the partial derivatives of the terms with respect to \(x\) and \(y\): $$ \frac{\partial}{\partial y}(e^x\sin y - 2y\sin x) = e^x\cos y - 2\sin x $$ $$ \frac{\partial}{\partial x}(e^x\cos y + 2\cos x) = e^x\cos y - 2\sin x $$ Since the partial derivatives match, the equation is exact.

Find the potential function

To find the potential function, integrate the expression for \(M\) with respect to \(x\), and the expression for \(N\) with respect to \(y\): $$ \int (e^x\sin y - 2y\sin x) dx = e^x\sin y - 2y\cos x + g(y) $$ $$ \int (e^x\cos y + 2\cos x) dy = e^x\sin y + 2y\cos x + f(x) $$ Now, compare the two potential functions and find a single function that satisfies both: $$ e^x\sin y - 2y\cos x + g(y) = e^x\sin y + 2y\cos x + f(x) $$ We find \(f(x) = 0\) and \(g(y) = 0\). Thus, the potential function is \(\phi(x, y) = e^x\sin y - 2y\cos x\).

Solve the exact differential equation

The exact differential equation can now be represented as: $$ \phi(x, y) = C $$ Substitute the potential function: $$ e^x\sin y - 2y\cos x = C $$ This is the general solution of the given differential equation, where C is an arbitrary constant.

Key concepts

Integrating Factor

In the realm of differential equations, an integrating factor is a function that, when multiplied by a non-exact differential equation, transforms it into an exact equation. This tool is immensely useful for solving first-order linear differential equations that do not initially satisfy the condition for exactness.

An integrating factor, denoted \(\mu(x, y)\), depends on the independent variables of the equation and is designed to balance the partial derivatives of the terms, allowing us to find a potential function—the ingredient we need for the equation to be exact. In the exercise provided, the integrating factor \(\mu(x, y) = y e^x\) is given. To validate its effect, we multiply the original differential equation by \(\mu(x, y)\), simplifying it thereafter. The new equation's terms are then examined for exactness, successfully turning an initially troublesome equation into one that is tractable.

Partial Derivatives

When dealing with multivariable functions, partial derivatives are a cornerstone concept. They represent the rates at which the function changes with respect to one variable, holding others constant. For an equation to be exact, the condition that the partial derivative of \(M\) with respect to \(y\) must equal the partial derivative of \(N\) with respect to \(x\).

In our exercise, we assess exactness by differentiating the \(M\) term, which is associated with \(dx\), with respect to \(y\), and the \(N\) term, which is associated with \(dy\), with respect to \(x\). Upon examination, we find that these derivatives are identical, which confirms that the integrating factor has done its job, and the equation has now become exact. This matching of partial derivatives is a green light for us to pursue a potential function that will lead us to the general solution.

Potential Function

A potential function, often represented by \(\phi(x, y)\), is a pivotal concept in solving exact differential equations. It's a scalar function whose partial derivatives with respect to \(x\) and \(y\) give us back the terms of the equation. To find this potential function, we integrate \(M\) with respect to \(x\) and \(N\) with respect to \(y\), keeping an eye out for arbitrary functions that may arise from each integration, \(f(x)\) and \(g(y)\) respectively.

When we complete these integrations in our exercise's context, we compare the resulting expressions to tease out the form of these unknown functions, often finding that they vanish, as in our case. Once the comparison is made, we distill a single potential function that encapsulates the nature of the exact differential equation.

General Solution

The general solution to an exact differential equation described by a potential function is derived by setting the potential function equal to an arbitrary constant \(C\). This equation encapsulates all possible solutions to the differential equation. The constant represents the family of curves, or solution set, that fulfills the equation for all possible initial conditions.

In the example provided, we establish the general solution by substituting our potential function, \(\phi(x, y)\), into the equation \(\phi(x, y) = C\). The result, \(e^x\sin y - 2y\cos x = C\), embodies the entire spectrum of solutions for the differential equation. Any particular solution can be isolated by determining the constant \(C\) with given initial conditions, effectively decoding the specific behavior of the system described by the equation.