Question

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

Short answer

If not, prove that multiplying it by the provided integrating factor makes it exact, and find the general solution. Answer: The given equation is not exact. When multiplied by the integrating factor, it becomes exact. The general solution of the resulting exact differential equation is \(y^2x - ye^y + e^y = C.\)

Step-by-step solution

Check the exactness of the given equation

To check the exactness of a differential equation in the form $$ M(x, y) dx + N(x, y) dy = 0 $$ we need to compute the partial derivatives: $$ \frac{\partial M}{\partial y} \quad \text{and} \quad \frac{\partial N}{\partial x}. $$ For our given equation, $$ M(x, y) = y, \quad N(x, y) = 2x - ye^y. \\ $$ Let's compute the partial derivatives, $$ \frac{\partial M}{\partial y} = \frac{\partial y}{\partial y} = 1, \quad \frac{\partial N}{\partial x} = \frac{\partial (2x - ye^y)}{\partial x} = 2. $$ Since \(\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}\), the given equation is not exact.

Multiply the given equation by the integrating factor

Now, we will multiply the given equation by the provided integrating factor \(\mu(x, y) = y\). This will yield: $$ \mu(x, y)M(x, y) dx + \mu(x, y)N(x, y) dy = y^2 dx +(2xy - y^2e^y) dy = 0. $$

Verify the exactness of the modified equation

Now, we need to check if the modified equation is exact. To do this, we'll compute the new partial derivatives. Let $$ M'(x, y) = y^2, \quad N'(x, y) = 2xy - y^2e^y. \\ $$ Computing the partial derivatives gives us: $$ \frac{\partial M'}{\partial y} = \frac{\partial (y^2)}{\partial y} = 2y, \quad \frac{\partial N'}{\partial x} = \frac{\partial (2xy - y^2e^y)}{\partial x} = 2y. $$ Since \(\frac{\partial M'}{\partial y} = \frac{\partial N'}{\partial x}\), the modified equation is exact.

Solve the modified exact differential equation

To solve an exact differential equation, we need to find the potential function \(\phi(x, y)\) such that: $$ \frac{\partial \phi}{\partial x} = M'(x, y) \quad \text{and} \quad \frac{\partial \phi}{\partial y} = N'(x, y). \\ $$ Let's find the potential function by integrating both \(M'(x, y)\) and \(N'(x, y)\) in respective variables \(x\) and \(y\): Integrating \(M'(x, y)\) w.r.t \(x\) gives us: $$ \phi(x, y) = \int M'(x, y) dx = \int y^2 dx = y^2x + f(y), $$ where \(f(y)\) is an arbitrary function of \(y\). Now, we'll differentiate \(\phi(x, y)\) w.r.t \(y\): $$ \frac{\partial \phi}{\partial y} = 2yx + f'(y). \\ $$ Comparing this with \(N'(x, y) = 2xy - y^2e^y\), we need to have: $$ f'(y) = - y^2e^y. $$ Integrating w.r.t \(y\) gives us: $$ f(y) = \int (-y^2e^y) dy = -ye^y + e^y + C. $$ The potential function is then $$ \phi(x, y) = y^2x - ye^y + e^y + C. $$ Thus, the general solution of the given differential equation is: $$ y^2x - ye^y + e^y = C. \\ $$

Key concepts

Integrating Factor

An integrating factor is a powerful tool used to transform a non-exact differential equation into an exact one, making it easier to solve. In the given exercise, the equation \( y \, dx + (2x - ye^y) \, dy = 0 \) is initially not exact, as demonstrated by the different partial derivatives of the functions involved. By identifying a suitable integrating factor, we can achieve exactness.
Typically, the integrating factor depends on the variables \( x \) and \( y \) involved in the equation. Here, the integrating factor proposed is \( \mu(x, y) = y \).
Multiplying the entire equation by this factor changes it to \( y^2 \, dx + (2xy - y^2e^y) \, dy = 0 \). By this multiplication, the partial derivatives become equal, verifying the exactness condition. An integrating factor can sometimes be identified by analyzing the equation's structure or using the knowledge of standard forms. After the transformation, we proceed to solve the equation using methods applicable to exact differential equations.

Partial Derivatives

Partial derivatives are essential in determining whether a differential equation is exact. They measure how a function changes as each variable is varied separately, holding the other variables constant. In the context of differential equations, if a given equation in two variables \( M(x, y) \, dx + N(x, y) \, dy = 0 \) is exact, it must satisfy the condition \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \).

• For the initial equation \( M(x, y) = y \), its partial derivative \( \frac{\partial M}{\partial y} = 1 \).
• For \( N(x, y) = 2x - ye^y \), its derivative with respect to \( x \) is \( \frac{\partial N}{\partial x} = 2 \).

These derivatives do not match, indicating the equation is not exact. Upon multiplying by the integrating factor, we reassess: for \( M'(x, y) = y^2 \), \( \frac{\partial M'}{\partial y} = 2y \), and for \( N'(x, y) = 2xy - y^2e^y \), \( \frac{\partial N'}{\partial x} = 2y \), confirming the equation's exactness after adjustment.

Potential Function

In solving an exact differential equation, a potential function \( \phi(x, y) \) is sought, which helps encapsulate the whole equation into a single expression. This potential function is equivalent to the total differential of a function, representing the idea that the differential equation stems from a gradient field.
For the modified exact equation, we determine \( \phi(x, y) \) such that:

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

In this example, the integration of \( M'(x, y) = y^2 \) with respect to \( x \) results in \( \phi(x, y) = y^2x + f(y) \), where \( f(y) \) is an arbitrary function of \( y \).
Continuing, \( f(y) \) is determined by matching the derivative with \( N'(x, y) \) as \( f'(y) = - y^2e^y \). By integrating \( f'(y) \), we find \( f(y) = -ye^y + e^y + C \). Thus, the potential function becomes \( \phi(x, y) = y^2x - ye^y + e^y + C \), and hence the solution is derived, encapsulated as \( y^2x - ye^y + e^y = C \).

Exactness Condition

The exactness condition is a critical requirement for a differential equation to be solved as an exact equation. It states that for a differential equation \( M(x, y) \, dx + N(x, y) \, dy = 0 \) to be exact, the partial derivatives \( \frac{\partial M}{\partial y} \) and \( \frac{\partial N}{\partial x} \) must be equal.
This condition ensures that there exists a potential function \( \phi(x, y) \) such that its total derivative accounts for both \( M \) and \( N \).
In practice, checking this condition involves calculating these partial derivatives and comparing them. If they are not equal, as seen initially in the exercise, the equation isn't exact. The integrating factor method is then employed to adjust the equation to meet this condition.
Once exactness is established, the solution process simplifies as the existence of a potential function streamlines finding the general solution, making this condition a pivotal gateway in solving such equations efficiently.