Question

Solving initial value problems Solve the following initial value problems. $$u^{\prime}(x)=\frac{1}{x^{2}+16}-4, u(0)=2$$

Short answer

The specific solution to the given initial value problem is \(u(x) = \frac{1}{2}\ln\left(x^2+16\right) - 4x +2-2\ln 2\).

Step-by-step solution

Integrate the ODE

Integrate both sides of the equation with respect to \(x\) to find the general solution: $$ \int u^{\prime}(x) \, dx = \int \left( \frac{1}{x^2+16} - 4 \right) \, dx. $$ Using the properties of integrals and antiderivatives, we can split the integral into two parts: $$ \int u^{\prime}(x) \, dx = \int \frac{1}{x^2 + 16} \, dx - \int 4 \, dx. $$

Evaluate Integrals

Evaluate the integrals separately. For the first integral, use substitution: let \(t=x^2+16\), then \(dt=2x\,dx\). We'll also need to divide by \(2\) to balance the \(2x\) we introduced: $$ \frac{1}{2}\int \frac{1}{t} \, dt = \frac{1}{2}\ln|t|+C_1=\frac{1}{2}\ln|x^2+16|+C_1. $$ For the second integral: $$ \int 4 \, dx = 4x+C_2. $$ Thus, we have: $$ u(x) = \frac{1}{2}\ln\left(x^2+16\right) + C_1 - 4x + C_2. $$ Take \(C = C_1 + C_2\) as the constant of integration, the general solution can be written as: $$ u(x) = \frac{1}{2}\ln\left(x^2+16\right) - 4x + C. $$

Solve for the Constant C

Apply the initial condition \(u(0) = 2\) to find the constant \(C\): $$ 2 = \frac{1}{2}\ln\left(0^2+16\right) - 4(0) + C. $$ Solve for \(C\): $$ C=2-\frac{1}{2}\ln 16 = 2-\frac{1}{2}\ln\left(2^4\right) = 2-2\ln 2. $$

Write the Specific Solution

Substitute the value of \(C\) back into the general solution to obtain the specific solution: $$ u(x) = \frac{1}{2}\ln\left(x^2+16\right) - 4x +2-2\ln 2. $$ This is the specific solution to the given initial value problem.

Key concepts

Integration

Integration is a fundamental concept in calculus used to find the area under a curve or the accumulation of quantities. It is the reverse operation of differentiation.
In this exercise, integration is applied to solve an ordinary differential equation (ODE).
This involves calculating the integral of given terms to find a general solution. The process of integrating involves focusing on antiderivatives, which are functions that reverse the derivative process.
Key points about integration include:

• It involves the symbol \( \int \), followed by the function and the differential of the variable \( dx \).
• Integration can be indefinite (without bounds) or definite (with bounds).
• Breaking down complex integrals into simpler parts helps in solving them more easily.

In this problem, integration is split into two parts, making it easier to manage and solve.

Ordinary Differential Equation

An Ordinary Differential Equation (ODE) is an equation involving functions and their derivatives.
Solving an ODE often requires finding a function that satisfies the given derivative equation.
In initial value problems, you are also given specific conditions, known as initial conditions, to find the particular solution.
Important characteristics of ODEs include:

• They involve derivatives with respect to a single variable.
• Initial value problems require solving the ODE and using given conditions to find specific solutions.
• Simplifying an ODE may involve techniques such as substitution or integration of its terms.

In this exercise, the ODE is solved by integration and applying the initial condition \( u(0) = 2 \) to find the specific solution.

Antiderivatives

Antiderivatives are functions whose derivatives yield the original function.
Finding antiderivatives is essential in integration, as they form the general solution of differential equations.
Once an antiderivative is found, it is adjusted using constants to fit any initial conditions given in a problem.
Some essential points to understand about antiderivatives include:

• They are not unique; different constants can yield different particular solutions.
• The general solution of an ODE includes arbitrary constants that are later determined by initial conditions.
• The notation \( F'(x) = f(x) \) indicates that \( F(x) \) is an antiderivative of \( f(x) \).

In our example, finding the antiderivatives of the component parts of the ODE helps in constructing the solution.

Substitution Method

The substitution method is a common technique used in integration to simplify expressions.
It involves changing variables to make integration more straightforward, often turning a complex integral into a more manageable form.
The steps typically include selecting a substitution, transforming the integral, and then solving it.
Here are some key points about the substitution method:

• Selecting a proper substitution is crucial for simplifying the calculation process.
• It requires adjusting the differential \( dx \) using the derivative of the new variable.
• After integrating, the original variable is usually restored to fit the problem context.

In this example, the substitution \( t=x^2+16 \) was used, which simplified the process of finding the antiderivative for part of the ODE, leading to an easier integration process.