Question

Show that the equations are not exact, but become exact when multiplied by the given integrating factor. Then solve the equations. $$ x^{2} y^{3}+x\left(1+y^{2}\right) y^{\prime}=0, \quad \mu(x, y)=1 / x y^{3} $$

Short answer

Based on the solution, explain why multiplying the differential equation by the integrating factor was necessary. Multiplying the differential equation by the integrating factor was necessary because the original equation was not exact. This means that we could not directly solve the equation using the standard techniques for exact differential equations. By multiplying the equation by the integrating factor, we transformed it into an exact differential equation, which allowed us to solve it using the appropriate methods for exact equations.

Step-by-step solution

Check if the original equation is exact

To check if the given differential equation, $$ x^{2} y^{3} + x\left(1+y^{2}\right) y^{\prime} = 0, $$ is exact, we need to find the partial derivative of the first function M with respect to y (i.e., \(\frac{\partial M}{\partial y}\)) and the partial derivative of the second function N with respect to x (i.e., \(\frac{\partial N}{\partial x}\)) and check if they are equal. In this case, M = \(x^2y^3\) and N = \(x(1+y^2)\). $$ \frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(x^2 y^3) = 3x^2y^2, \quad \frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(x(1+y^2)) = 1+y^2 $$ As \(\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}\), the equation is not exact.

Multiply the equation by the integrating factor

Now we have to multiply the entire equation by the given integrating factor, \(\mu(x, y) = \frac{1}{xy^3}\): $$ \frac{1}{xy^3}(x^{2} y^{3} + x\left(1+y^{2}\right) y^{\prime}) = 0 $$ This simplifies to: $$ x + \frac{1+y^{2}}{y^3}y^{\prime} = 0 $$

Verify that the new equation is exact

Now we'll check if the new equation is exact. Let M' = x and N' = \(\frac{1+y^{2}}{y^3}\), then we find \(\frac{\partial M'}{\partial y}\) and \(\frac{\partial N'}{\partial x}\): $$ \frac{\partial M'}{\partial y} = \frac{\partial}{\partial y}(x) = 0, \quad \frac{\partial N'}{\partial x} = \frac{\partial}{\partial x}\left(\frac{1+y^{2}}{y^3}\right) = 0 $$ As \(\frac{\partial M'}{\partial y} = \frac{\partial N'}{\partial x}\), the new equation is exact.

Solve the exact differential equation

Now that we have an exact equation, we can solve it using the following formula: $$ \int M' dx + \int (N' - \frac{\partial}{\partial y} \int M' dx) dy = C $$ First, let's integrate M' with respect to x: $$ \int M' dx = \int x dx = \frac{1}{2}x^2 + g(y) $$ Next, we'll find the partial derivative of this result with respect to y: $$ \frac{\partial}{\partial y}\left(\frac{1}{2}x^2 + g(y)\right) = g'(y) $$ Now, let's subtract this from N': $$ N' - \frac{\partial}{\partial y} \int M' dx = \frac{1+y^{2}}{y^3} - g'(y) $$ Now, we'll integrate this expression with respect to y: $$ \int\left(\frac{1+y^{2}}{y^3} - g'(y)\right) dy = \int\left(\frac{1+y^{2}}{y^3}\right) dy - \int g'(y) dy = \int\left(\frac{1+y^{2}}{y^3}\right) dy - g(y) $$ So, integrating \(\frac{1+y^{2}}{y^3}\) with respect to y, we get: $$ \int\left(\frac{1+y^{2}}{y^3}\right) dy = -\frac{1}{2y^2} + \frac{1}{y} + C $$ Finally, putting everything together, we get the solution: $$ \frac{1}{2}x^2 + g(y) = \frac{1}{2}x^2 -\frac{1}{2y^2} + \frac{1}{y} + C $$ Or simply, $$ g(y) = -\frac{1}{2y^2} + \frac{1}{y} + C $$

Key concepts

Integrating Factor

The concept of an integrating factor is a powerful tool in solving differential equations, particularly when they are not exact. An integrating factor is a function, typically denoted by \(\mu(x, y)\), that, when multiplied by a non-exact differential equation, makes it exact. This transformation allows the equation to be solved using methods for exact equations.

For the example equation \(x^2 y^3+x(1+y^2)y'=0\), it is not initially exact. However, when multiplied by the integrating factor \(\mu(x, y)=1/(xy^3)\), it becomes exact. The updated equation \(x + \frac{1+y^{2}}{y^3}y' = 0\) can then be solved using standard techniques for exact differential equations. Understanding how to find and apply an integrating factor is key to unlocking the solutions to many types of differential equations.

Exactness Condition

For a differential equation of the form \(M(x, y) + N(x, y)y' = 0\), the exactness condition requires that the partial derivative of \(M\) with respect to \(y\) equals the partial derivative of \(N\) with respect to \(x\). Mathematically, this condition is stated as \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\).

With the given exercise, we tested this exactness condition by calculating \(\frac{\partial M}{\partial y} = 3x^2y^2\) and \(\frac{\partial N}{\partial x} = 1+y^2\), which are not equal, indicating the original equation is not exact. This condition is fundamental in determining whether a differential equation can be solved directly or whether applying an integrating factor is necessary to proceed.

Partial Derivatives

Partial derivatives play a significant role in the examination of differential equations, especially for establishing exactness. A partial derivative represents the rate at which a function changes as one of its variables changes, while the others are held constant.

In the provided exercise, the partial derivatives \(\frac{\partial M}{\partial y}\) and \(\frac{\partial N}{\partial x}\) are used to determine if the given differential equation is exact or not. These computations are crucial steps because they help identify whether the integrand is a total differential, enabling the possibility to integrate the function with respect to one variable while treating the other variable as a constant.

Differential Equations Solution

Finding the solution to a differential equation is often the final goal. Exact differential equations are solved by integrating both sides with respect to one variable while treating the other as a constant. This produces a function of the two variables involved, which represents the solution to the differential equation.

In the case of our example after applying the integrating factor, we sought the function \(g(y)\) by integrating the updated \(M'\) and \(N'\), and applying the relationship \(\int M' dx + \int (N' - \frac{\partial}{\partial y} \int M' dx) dy = C\) to find the general solution. This is an illustrative example of solving an exact differential equation and reflects a standard process that can be applied across various exact differential equations.