Question

Solving initial value problems Solve the following initial value problems. $$p^{\prime}(x)=\frac{2}{x^{2}+x}, p(1)=0$$

Short answer

Question: Determine the particular solution to the initial value problem \(p'(x) = \frac{2}{x^2 + x}\) with \(p(1) = 0\). Answer: The particular solution to the given initial value problem is: \(p(x) = 2\ln|x| - 2\ln|x+1| + 2\ln|2|\).

Step-by-step solution

Integrate both sides of the differential equation

Integrate both sides of the given differential equation: $$\int p'(x) \, dx = \int \frac{2}{x^2 + x} \, dx.$$

Separate the right-side fraction into partial fractions

To integrate the right-side fraction, we first need to rewrite it as the sum of partial fractions. Since the denominator is \((x)(x+1)\), we can rewrite the fraction with the form: $$\frac{2}{x^2 + x} = \frac{A}{x} + \frac{B}{x+1}.$$ Now we can set up a system of equations to solve for A and B.

Solve for A and B using cross-multiplication

Multiply both sides of the equation by \(x^2 + x\) to cancel out the denominators: $$2 = A(x+1) + B(x).$$ Now set up the system of equations to find A and B by letting \(x=0\) and \(x=-1\). When \(x=0\), we get the equation \(2 = A\), so \(A=2\). When \(x=-1\), we get the equation \(2 = -B\), so \(B=-2\). Thus, we have the partial fractions as $$\frac{2}{x^2 + x} = \frac{2}{x} - \frac{2}{x+1}.$$

Integrate the partial fractions

Now integrate the right-side partial fractions: $$\int p'(x) \, dx = \int \left(\frac{2}{x} - \frac{2}{x+1}\right) \, dx = 2\int\frac{1}{x} \, dx - 2\int\frac{1}{x+1} \, dx.$$ Apply the integral of the natural logarithm: $$p(x) = 2\ln|x| - 2\ln|x+1| + C,$$ where C is the integration constant.

Use the initial condition to find the particular solution

Now, we will use the initial condition \(p(1)=0\) to find the value of C: $$0 = 2\ln|1| - 2\ln|1+1| + C.$$ Since \(\ln|1|=0\) and \(\ln|2|\neq 0\), we get: $$0 = -2\ln|2| + C \Rightarrow C = 2\ln|2|.$$

Write the final particular solution

Combine the general solution and the value of C to get the particular solution for the initial value problem: $$p(x) = 2\ln|x| - 2\ln|x+1| + 2\ln|2|.$$ This is the final solution to the given initial value problem.

Key concepts

Differential Equations

Differential equations are like equations in math that involve functions and their derivatives. Essentially, they show the relationship between a function and the rates at which it changes. These equations come in handy when trying to model complex systems in fields like physics, engineering, and economics. There are different types of differential equations, such as ordinary (ODEs) and partial (PDEs), based on the number of independent variables they involve. In our case, the differential equation we are working with is an ordinary differential equation because it involves a single independent variable, \( x \). By solving a differential equation, we seek a function that satisfies the equation's constraints, which often include an initial condition too. This initial condition, provided at a specific point, allows us to find a unique solution that otherwise would include an unknown constant.

Partial Fractions

Partial fractions make life easier when we need to integrate complex rational expressions, especially those with polynomial denominators. Instead of dealing directly with a complicated fraction, we break it down into simpler parts called partial fractions. This process, known as decomposition, expresses a complex fraction as the sum of simpler fractions.

• The target is to have terms like \( \frac{A}{x} \) and \( \frac{B}{x+1} \), which are easier to integrate.
• We begin by identifying the factors of the polynomial in the denominator.
• Next, we express the given fraction as a sum based on these factors.
• Finally, we solve for the constants \( A \) and \( B \) using algebraic methods.

For instance, in the given problem, the expression \( \frac{2}{x^2 + x} \) was decomposed into \( \frac{2}{x} - \frac{2}{x+1} \). This made the integration process straightforward.

Integration

Integration is the reverse process of differentiation. It allows us to find the function from its derivative. When dealing with integrals, especially indefinite ones, we're always searching for a family of functions that can represent the anti-derivative.

• In the context of solving differential equations, integration helps us move from a derivative back to the original function.
• For rational expressions like those obtained from partial fraction decomposition, integration often involves natural logarithms.

In the example problem, integrating \( \frac{2}{x} - \frac{2}{x+1} \) leads to the expression \( 2\ln|x| - 2\ln|x+1| + C \). Here, \( C \) is an arbitrary constant, indicating an entire set of functions that can be solutions. It's this process that restores the function from its rate of change.

Initial Condition

Initial conditions are critical in determining a specific solution from the endless possibilities present in the general solution of a differential equation. When integrating, we introduce an arbitrary constant, \( C \). This constant is adjusted based on the initial condition to find the particular solution relevant to the problem at hand.

• An initial value problem provides an extra piece of information, like \( p(1) = 0 \), which fixes \( C \).
• This condition ensures the solution curve passes through a specified point.
• By plugging values into the general solution, we can solve for \( C \) and determine the exact form of our function.

Using the initial condition \( p(1) = 0 \), we computed \( C \) to be \( 2\ln|2| \). This adjusted our solution to \( p(x) = 2\ln|x| - 2\ln|x+1| + 2\ln|2| \), completing our task of identifying the exact behavior of the system described by the differential equation.