Problem 28 - Calculus

Question

In Exercises \(2-28,\) use separation of variables to find the solutions to the differential equations subject to the given initial conditions. $$x(x+1) \frac{d u}{d x}=u^{2}, \quad u(1)=1$$

Step-by-Step Solution

Answer

The solution is \( u(x) = -\frac{1}{\ln\left(\frac{x}{x+1}\right) - 1 + \ln(2)}.\)

Step 1: Rewrite the Differential Equation

First, we rewrite the given differential equation by separating the variables. The differential equation provided is \(x(x+1) \frac{d u}{d x} = u^2\). Divide both sides by \(u^2\) and multiply both sides by \(dx\) to isolate the terms depending on \(u\) and \(x\): \[ \frac{1}{u^2} du = \frac{1}{x(x+1)} dx. \]

Step 2: Integrate Both Sides

Next, we integrate both sides of the equation. The left-hand side becomes \(\int \frac{1}{u^2} du\), and the right-hand side becomes \(\int \frac{1}{x(x+1)} dx\). This gives:\[ -\frac{1}{u} = \int \frac{1}{x(x+1)} dx. \] Let's find \(\int \frac{1}{x(x+1)} dx\) by partial fraction decomposition:Assume \(\frac{1}{x(x+1)} = \frac{A}{x} + \frac{B}{x+1}\), which simplifies to \(A(x+1) + Bx = 1\). Solving for \(A\) and \(B\), we find \(A = 1\) and \(B = -1\). Thus,\[ \int \frac{1}{x(x+1)} dx = \int \left(\frac{1}{x} - \frac{1}{x+1}\right) dx = \ln|x| - \ln|x+1| + C. \]

Step 3: Solve the Resulting Equation

From the integration, we have:\[ -\frac{1}{u} = \ln|x| - \ln|x+1| + C. \]Combine the logarithm terms on the right: \[ -\frac{1}{u} = \ln\left(\frac{|x|}{|x+1|}\right) + C. \]Exponentiate both sides to solve for \(u\):\[ u = -\frac{1}{\ln\left(\frac{|x|}{|x+1|}\right) + C}. \]

Step 4: Apply the Initial Condition

Now apply the initial condition \(u(1) = 1\) to find the constant \(C\). Replace \(x\) with 1 and \(u\) with 1 in the equation:\[ 1 = -\frac{1}{\ln\left(\frac{1}{2}\right) + C}. \]Solving for \(C\), we find\[ \ln\left(\frac{1}{2}\right) + C = -1 \Rightarrow C = -1 - \ln\left(\frac{1}{2}\right). \]

Step 5: Write the Solution

Substitute the constant \(C\) back into the equation for \(u\):\[ u = -\frac{1}{\ln\left(\frac{x}{x+1}\right) - 1 + \ln(2)}. \]Thus, the solution to the differential equation subject to the given initial condition is:\[ u(x) = -\frac{1}{\ln\left(\frac{x}{x+1}\right) - 1 + \ln(2)}. \]

Key Concepts

Differential equations

Differential equations are mathematical equations that relate a function with its derivatives. In other words, they describe how a particular quantity changes over time or space. These equations are pivotal for modeling a wide variety of phenomena in physics, engineering, biology, and economics.

In the original exercise, the differential equation given was \[ x(x+1) \frac{d u}{d x} = u^{2}, \text{ where } u(1)=1. \]

Our goal is to solve this using the method of separation of variables. This technique involves rearranging the differential equation such that each side contains only one variable. This approach makes it easier to solve because we can integrate each side with respect to its respective variable, leading us towards the solution.

Integration

Integration is the reverse process of differentiation. It is used to find functions when their derivatives are known. In the context of differential equations, we use integration to solve for the unknown function once the variables have been separated.

When the variables in the differential equation are separated, as in our exercise, we end up with:

• Left side: \( \int \frac{1}{u^2} \ du \)
• Right side: \( \int \frac{1}{x(x+1)} \ dx \)

By integrating the left side, we get \( -\frac{1}{u} \). For the right side, the process of partial fraction decomposition is typically used to make integration easier. This eventually yields an expression that links \(u\) to \(x\), crucial for finding the exact solution.

Initial conditions

Initial conditions are specific values that a function must satisfy at a given point. They are essential in finding unique solutions to differential equations. Essentially, they "anchor" the solution so that we can determine the unknown constants that are introduced during the process of integration.

In our scenario, the initial condition is given as \(u(1)=1\). After solving the separated equation, we obtain an expression involving a constant \(C\). Substituting the initial condition \(u(1)=1\) into this expression allows us to solve for \(C\):

\[ 1 = -\frac{1}{\ln\left(\frac{1}{2}\right) + C} \]

Solving for \(C\) ensures that the final solution \(u(x)\) fits the initial condition requirement.

Partial fraction decomposition

Partial fraction decomposition is a method used to express a complex rational function as a sum of simpler fractions. This technique is particularly useful when integrating rational functions, as it breaks them down into easier-to-integrate components.

In our exercise, we need to integrate \( \frac{1}{x(x+1)} \). We start by expressing this as:

\[ \frac{1}{x(x+1)} = \frac{A}{x} + \frac{B}{x+1} \]

Solving for the constants \(A\) and \(B\) involves setting up an equation from which we find \(A=1\) and \(B=-1\). Therefore, the integral can be written as:

\[ \int \left(\frac{1}{x} - \frac{1}{x+1}\right) \ dx = \ln|x| - \ln|x+1| + C \]

This decomposition ultimately facilitates the integration process, making it possible to find the solution to the differential equation.