Problem 23 - Calculus for Biology and Medicine

Question

For each of the Problems 13-24 you should determine whether the problem needs to be solved using separation of variables or integrating factors (some of the problems may be solved using \mathrm{\\{} e i t h e r ~ m e t h o d ) . ~ T h e n ~ s o l v e ~ t h e ~ d i f f e r e n t i a l ~ e q u a t i o n . ~. $$ \frac{d y}{d x}=\frac{x y}{x+1} $$

Step-by-Step Solution

Answer

Use separation of variables to solve, resulting in \( y = C_1 \cdot \frac{e^{x+1}}{|x+1|} \).

Step 1: Analyze the Differential Equation

The given differential equation is \( \frac{dy}{dx} = \frac{xy}{x+1} \). We need to determine if it can be solved by separation of variables or integrating factors. Since the equation is not in the form \( \frac{dy}{dx} + p(x)y = q(x) \), it is not immediately suitable for integrating factors. Let's check if it can be separated.

Step 2: Separate Variables

Rearrange the equation to check if the variables can be separated:\[ \frac{dy}{y} = \frac{x}{x+1}dx \]This form shows that the variables can indeed be separated, allowing us to proceed with separation of variables.

Step 3: Integrate Both Sides

Integrate both sides separately:\[ \int \frac{1}{y} \, dy = \int \frac{x}{x+1} \, dx \]This results in:\[ \ln |y| = \int \frac{x}{x+1} \, dx \]

Step 4: Solve the Right-Side Integral

Use substitution for the integral on the right-hand side. Let \( u = x + 1 \), then \( du = dx \) and \( x = u - 1 \). The integral becomes:\[ \int \frac{u-1}{u} \, du = \int \left(1 - \frac{1}{u}\right) \, du \]This splits into:\[ \int 1 \, du - \int \frac{1}{u} \, du = u - \ln|u| + C \]

Step 5: Solve for y

Substitute back \( u = x + 1 \) into the integral result to get:\[ u - \ln |u| = x + 1 - \ln |x+1| + C \]Thus, the equation is:\[ \ln |y| = x + 1 - \ln |x+1| + C \]To solve for \( y \), exponentiate both sides:\[ y = e^{x + 1 - \ln |x+1| + C} \]

Step 6: Simplify the Final Expression

Simplify the expression, noting that \( e^{-\ln |x+1|} = \frac{1}{|x+1|} \):\[ y = e^{C} \cdot e^{x+1} \cdot \frac{1}{|x+1|} \]Let \( e^{C} = C_1 \) (a constant), then:\[ y = C_1 \cdot \frac{e^{x+1}}{|x+1|} \]

Key Concepts

Separation of Variables

When faced with a differential equation like \( \frac{dy}{dx} = \frac{xy}{x+1} \), separation of variables is a powerful technique that can save the day. This method allows us to rearrange terms so that each side of the equation contains only one variable. For our specific case, by dividing both sides by \( y \) and multiplying by \( dx \), we end up with:

• \( \frac{dy}{y} = \frac{x}{x+1} \, dx \)

Once separated, each side can be integrated independently. Why is this helpful? Well, it allows you to simplify and handle complex equations piece by piece, making integration more manageable. By dividing up the equation, you turn one difficult problem into two easier ones.
Try to focus on rearranging terms accurately so that you can apply this method efficiently whenever possible. Look out for this technique particularly when your differential equation can be expressed in a product of distinct functions of \( x \) and \( y \). This method effectively uncouples the variables, simplifying the solving process tremendously.

Integrating Factors

Integrating factors are another arsenal in solving differential equations, particularly effective when equations aren't easily separable. However, for the given equation \( \frac{dy}{dx} = \frac{xy}{x+1} \), integrating factors aren't the best fit since the equation doesn't initially resemble the standard linear form of \( \frac{dy}{dx} + P(x)y = Q(x) \).
But, what does an integrating factor do? In cases where it is applicable, multiplying the entire equation by a certain function, known as the integrating factor, transforms it into an exact derivative form. This simplifies the equation to something more solvable. For instance:

• Consider a linear form differential equation.
• Find the integrating factor \( \mu(x) = e^{\int P(x) \, dx} \).
• Multiply this factor through the equation to obtain the desired form.

Integrating factors excel at dealing with non-separable linear equations, turning a tough differential problem into a simple integration challenge. Although not used in our specific exercise, understanding how and when to apply this tool is invaluable in broader contexts.

Integration by Substitution

Integration by substitution is like changing gears while riding a bicycle; it helps you manage more challenging paths ahead. In our differential equation problem, after separating the variables and setting up the integral \( \int \frac{x}{x+1} \, dx \), substitution becomes highly useful.
Let's see how it works:

• Identify a part of the integrand that can be simplified. Here, choosing \( u = x + 1 \) simplifies the fraction.
• Express \( du = dx \) and \( x = u - 1 \).
• Transform the integral into a friendlier form for integration: \( \int \left(1 - \frac{1}{u}\right) \, du \).

The trick is to simplify the expression as much as possible, turning an otherwise complex integral into a straightforward task with basic rules. Substitution also ensures you don't get tangled in complicated algebra. By rethinking parts of the equation intelligently, you smooth the path to a solution, making integration more intuitive and less error-prone. This technique is crucial for unlocking more complex integrals that appear throughout calculus.