Problem 31 - Differential Equations and Linear Algebra

Question

Determine the values of the constants \(r\) and \(s\) such that \(I(x, y)=x^{r} y^{s}\) is an integrating factor for the given differential equation. $$2 y\left(y+2 x^{2}\right) d x+x\left(4 y+3 x^{2}\right) d y=0$$

Step-by-Step Solution

Answer

The values of the constants are \(r = -2\) and \(s = 2\), which gives the integrating factor \(I(x, y) = x^{-2} y^2\).

Step 1: Rewrite the differential equation in the form Mdx + Ndy = 0

The given differential equation is: $$2 y(y+2 x^{2}) dx+x(4 y+3 x^{2}) dy=0$$ We can rewrite this as: $$Mdx + Ndy = 0$$ Where \( M = 2y(y+2x^2) \) and \( N = x(4y+3x^2) \)

Step 2: Calculate partial derivatives of M and N

We need to find the partial derivatives of M and N to determine the conditions for an integrating factor I(x, y). The partial derivatives we need are: $$\frac{\partial M}{\partial y}$$ and $$\frac{\partial N}{\partial x}$$ Calculating these derivatives, we get: \(\) \( \frac{\partial M}{\partial y} = 4y + 4x^2 \\\ \frac{\partial N}{\partial x} = 4y + 6x^2 \)

Step 3: Set up the equation with Ix and Iy

The condition for an integrating factor I(x, y) can be written as: \[ \frac{1}{I}\frac{\partial I}{\partial y} = \frac{\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}}{M - N} \] Substituting the values for the partial derivatives we calculated in Step 2, as well as knowing the integrating factor I(x, y) = \( x^r y^s \), we get: \[ \frac{1}{x^r y^s}\frac{\partial (x^r y^s)}{\partial y} = \frac{(4y + 6x^2) - (4y + 4x^2)}{2y(y+2x^2) - x(4y+3x^2)} \]

Step 4: Calculate Iy and simplify the equation

To find the partial derivative of I(x, y) with respect to y, recall that I(x, y) = \( x^r y^s \): \( \frac{\partial (x^r y^s)}{\partial y} = s x^r y^{s-1} \) Now, we substitute this expression for I_y into the equation obtained in step 3: \[ \frac{1}{x^r y^s}(s x^r y^{s-1}) = \frac{2x^2}{2y(y+2x^2) - x(4y+3x^2)} \] We can now simplify this equation by canceling some terms: \[ s y^{s-1} = \frac{2x^2}{2y(y+2x^2) - x(4y+3x^2)} \]

Step 5: Equate coefficients

Now, note that the powers of x and y must be equal on both sides of the equation. We have: \( s y^{s-1} = \frac{2x^2}{2y(y+2x^2) - x(4y+3x^2)} \) Comparing the coefficients, we can notice that s = 2.

Step 6: Calculate r and the integrating factor I(x, y)

With the value of s found above, we can now find the value for r. We can rewrite I(x, y) in the light of this information: \( I(x, y) = x^r y^2 \) At this point, we notice that the equation is homogeneous, meaning both M and N are functions of y/x, so we can rewrite the integrating factor as: \( I \propto (y/x)^r \) Taking the constant of proportionality to be 1 (which is a valid choice for an integrating factor), we can write the integrating factor as: \( I(x, y) = (y/x)^2 \) In conclusion, the integrating factor we found is: \( I(x, y) = x^{-2} y^2 \)

Key Concepts

Partial Derivatives

Partial derivatives are instrumental when we deal with functions of two or more variables, such as in our exercise dealing with differential equations. They represent the rate of change of the function with respect to one of its variables while holding the other variables constant. In the context of our problem, we need to find the partial derivatives of functions M and N.
The partial derivative of M with respect to y, denoted as \( \frac{\partial M}{\partial y} \), and the partial derivative of N with respect to x, denoted as \( \frac{\partial N}{\partial x} \), are key to finding an integrating factor that simplifies the given differential equation. This process involves differentiating the components of the equation and looking for a pattern or condition that allows the differential equation to become exact. Recognizing how these derivatives relate to each other leads us to the desired integrating factor that simplifies the equation.

Homogeneous Equations

Homogeneous equations are a class of differential equations that exhibit a specific symmetry: the functions involved are homogeneous of the same degree. This means that each term is a product of variables raised to the same power. For example, our equation from the exercise has terms like \( y^2 \) and \( x^2y \), indicating it can be manipulated into a homogeneous form.
When we identify a differential equation as homogeneous, we can employ a change of variables to simplify the problem, making it far more tractable. In our exercise, the insight that the equation is homogeneous allows us to assert the integrating factor \( I(x, y) \) possesses a certain structure, easing the process of identifying the constants r and s. This characteristic is essential for solving a broad class of differential equations encountered not only in mathematics but also in physical sciences and engineering.

Exact Differential Equations

Exact differential equations are those where an integrating factor can be found, enabling the equation to be expressed in a precise differential form \( dU = Mdx + Ndy = 0 \) for some potential function U(x, y). Therefore, exactness is a crucial property which signifies that the path integral of a conservative vector field is path-independent.
The road to finding an integrating factor involves assessing whether the cross derivatives of M and N, or their difference, fulfills certain conditions. If these conditions are met, the differential equation can be simplified and solved. In the textbook problem, our quest is to determine the constants r and s such that \( x^r y^s \) becomes an integrating factor, transforming the given equation into an exact differential equation. This alteration not only makes the equation easier to solve but also, conceptually, signifies we've found a potential function whose differential reflects the original equation.