Question

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

Short answer

Based on the given information and the step by step solution above, the final answer for the differential equation after applying the integrating factor is: $$xe^x \sin y -2e^x \cos y = C$$

Step-by-step solution

Check if the given differential equation is exact

To check if the given equation is exact, we will use the test of exactness. The test requires computing the partial derivatives. Let's first assign the functions to M and N: $$M(x, y) = (x + 2) \sin y$$ $$N(x, y) = x \cos y$$ Now, compute the partial derivatives: $$\frac{\partial M}{\partial y} = (x + 2) \cos y$$ $$\frac{\partial N}{\partial x} = \cos y$$ Since \(\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}\), the given differential equation is not exact.

Multiply the given equation by the integrating factor

To make the given differential equation exact, we need to multiply it by the integrating factor, \(\mu(x, y) = x e^x\): $$\Rightarrow \mu(x, y) (x+2) \sin y dx + \mu(x,y) x \cos y dy = 0$$ $$\Rightarrow x e^x (x+2) \sin y dx + x e^x x \cos y dy = 0$$

Verify that the modified equation is exact

We will now check if the modified equation is exact by calculating the partial derivatives. Let's redefine the functions M and N with the integrating factor: $$M'(x, y) = x e^x (x+2) \sin y$$ $$N'(x, y) = x^2 e^x \cos y$$ Now, compute the partial derivatives: $$\frac{\partial M'}{\partial y} = x e^x (x+2) \cos y$$ $$\frac{\partial N'}{\partial x} = x e^x [(x+2) \cos y + x \cos y] = x e^x (x+2) \cos y$$ Since \(\frac{\partial M'}{\partial y} = \frac{\partial N'}{\partial x}\), the modified differential equation is exact.

Solve the exact differential equation

To solve the exact differential equation, we need to find the function f(x,y) such that: $$\frac{\partial f}{\partial x} = x e^x (x+2) \sin y$$ $$\frac{\partial f}{\partial y} = x^2 e^x \cos y$$ Integrate the first equation with respect to x: $$f(x, y) = \int x e^x (x+2) \sin y dx = xe^x \sin y-2e^x \cos y + g(y)$$ Now differentiate f(x, y) with respect to y: $$\frac{\partial f}{\partial y} = xe^x \cos y - 2e^x(-\sin y) + g'(y) = x^2 e^x \cos y$$ Comparing this with our second equation, we can see that: $$g'(y) = 0 \Rightarrow g(y) = C$$ Thus, the solution to the exact differential equation is: $$xe^x \sin y -2e^x \cos y = C$$

Key concepts

Integrating Factor

When solving a differential equation that is not exact, an integrating factor can sometimes make it exact. An integrating factor is a function, usually denoted by \( \mu(x, y) \), which when multiplied by the differential equation, transforms it into an exact equation. This means the form of the equation can be adjusted such that the partial derivatives are equal, satisfying the condition for exactness.

• To find the appropriate integrating factor for a given equation, it often involves intuition or trial based on the structure of the equation.
• In our exercise, the integrating factor \( \mu(x, y) = x e^x \) makes the original non-exact equation into an exact equation.

The process involves multiplying each term of the differential equation by \( \mu(x, y) \) to ensure the recalculated partial derivatives become equal. Once the equation is exact, it becomes significantly easier to solve systematically.

Partial Derivatives

Partial derivatives play a critical role in checking the exactness of a differential equation. They represent the rate of change of a function with respect to one variable while keeping all other variables constant. For a differential equation, we define functions \( M(x, y) \) and \( N(x, y) \) from the terms involving \( dx \) and \( dy \) respectively.

• To determine if an equation is exact, we compute \( \frac{\partial M}{\partial y} \) and \( \frac{\partial N}{\partial x} \).
• If \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \), the equation is exact.

In the provided exercise, the initial equations did not satisfy this equality until they were multiplied by the integrating factor \( \mu(x, y) \). The recalculated partial derivatives proved the transformed equation is exact.

Exactness Criterion

The exactness criterion is crucial in identifying whether a differential equation is exact. It revolves around the equality of specific partial derivatives as mentioned in the previous section. In mathematical terms, for a differential equation of the form \( M(x, y)dx + N(x, y)dy = 0 \), the criterion checks whether:

• \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \)

If this condition is met, it means that there exists a scalar function \( f(x, y) \) such that \( \frac{\partial f}{\partial x} = M \) and \( \frac{\partial f}{\partial y} = N \).
In our case, multiplying by the integrating factor \( x e^x \) aligns the partial derivatives, changing the equation’s form to meet the exactness criterion.

Solving Differential Equations

Once a differential equation has been confirmed to be exact, or made exact by an integrating factor, solving it involves finding the potential function \( f(x, y) \). This function is the solution to the equation, usually expressed implicitly. The steps include:

• Integrate \( M(x, y) \) with respect to \( x \) to find \( f(x, y) \) plus an arbitrary function of \( y \).
• Differentiate the obtained \( f(x, y) \) concerning \( y \) and equate it to \( N(x, y) \) to find and adjust any remaining functions or constants.

In the textbook solution, after obtaining \( f(x, y) \), we determined that the arbitrary function \( g(y) = C \) where \( C \) is a constant. Therefore, the solution was expressed as \( xe^x \sin y - 2e^x \cos y = C \), which completes the process.