Problem 103 - University Calculus: Early Transcendentals

Question

Solve the initial value problems in Exercises. $$\frac{d v}{d t}=\frac{3}{t \sqrt{t^{2}-1}}, \quad t>1, v(2)=0$$

Step-by-Step Solution

Answer

The solution to the initial value problem is \( v(t) = 3\sqrt{t^2-1} - 3\sqrt{3} \).

Step 1: Understanding the Differential Equation

We are given a first-order differential equation \( \frac{dv}{dt} = \frac{3}{t \sqrt{t^2 - 1}} \) with an initial condition \( v(2) = 0 \). Our goal is to find the function \( v(t) \) that satisfies both the differential equation and the initial condition.

Step 2: Separate Variables

To solve the differential equation, we separate the variables. We rewrite the equation as \( dv = \frac{3}{t \sqrt{t^2 - 1}} dt \). This allows us to integrate both sides separately.

Step 3: Integrate Both Sides

Integrate both sides of the equation. The left-hand side becomes \( \int dv = v + C \). The right-hand side becomes \( \int \frac{3}{t \sqrt{t^2 - 1}} dt \). We can simplify the integral on the right using a substitution method.

Step 4: Use Substitution

Let \( u = t^2 - 1 \). Then \( du = 2t\, dt \) or \( dt = \frac{du}{2t} \). Substitute into the integral to get \( \int \frac{3}{t \sqrt{u}} \cdot \frac{du}{2t} = \frac{3}{2} \int \frac{1}{\sqrt{u}} du \).

Step 5: Solve the New Integral

The integral \( \int \frac{1}{\sqrt{u}} du \) is \( 2\sqrt{u} + C \). Thus, \( \frac{3}{2} \int \frac{1}{\sqrt{u}} du = 3\sqrt{u} + C \). Returning to the original variable \( t \), we have the integral result: \( 3\sqrt{t^2-1} + C \).

Step 6: Write the General Solution

The general solution from integrating both sides is \( v(t) = 3\sqrt{t^2-1} + C \).

Step 7: Apply Initial Condition

Use the initial condition \( v(2) = 0 \) to find \( C \). Substitute \( t = 2 \) into the equation: \( 0 = 3\sqrt{2^2-1} + C \). Thus, \( 0 = 3\sqrt{3} + C \), leading to \( C = -3\sqrt{3} \).

Step 8: Final Solution

Include the constant \( C \) in the solution to get the final answer: \( v(t) = 3\sqrt{t^2-1} - 3\sqrt{3} \). Ensure the solution respects the initial conditions and the domain of \( t > 1 \).

Key Concepts

Differential Equation

A differential equation is a type of equation that relates a function with its derivatives. In simple terms, it involves an unknown function and its rates of change.
It's widely used in mathematics to model how things change over time or space. The particular focus of differential equations is to determine the function itself.
For example, let's consider the differential equation in our exercise:

• \( \frac{dv}{dt} = \frac{3}{t \sqrt{t^2 - 1}} \)
• It is a first-order differential equation because it involves the first derivative, \( \frac{dv}{dt} \).
• It suggests how the variable \( v \) changes with respect to \( t \).

Such equations often require specific conditions, known as initial conditions, to find unique solutions. They ensure the solution meets particular criteria, such as \( v(2) = 0 \) in this scenario.

Separation of Variables

Separation of variables is a widely-used method for solving differential equations. It's especially useful for first-order differential equations like the one we are working on here.
The technique involves rearranging the equation so each variable and its derivative are on opposite sides of the equation.
This allows us to treat each part separately and integrate them independently. Let's see how this works:

• Starting with the equation: \( \frac{dv}{dt} = \frac{3}{t \sqrt{t^2 - 1}} \)
• Rearrange to: \( dv = \frac{3}{t \sqrt{t^2 - 1}} dt \)

Now both sides are set up for integration. One side in terms of \( v \) and the other in terms of \( t \). Once separated, each side can be integrated to advance towards a solution.

Integration by Substitution

Integration by substitution is a powerful technique used to simplify complex integrals. It allows you to make a substitution that turns an intricate integral into a simpler one.
It's similar to reversing the chain rule in differentiation. In this exercise, we applied this to simplify the right-hand integral:

• Substitute \( u = t^2 - 1 \)
• Then, \( du = 2t\, dt \) implies \( dt = \frac{du}{2t} \)
• Substituting turns the integral into: \( \int \frac{3}{t \sqrt{u}} \cdot \frac{du}{2t} = \frac{3}{2} \int \frac{1}{\sqrt{u}} du \)

This simplification allows us to calculate the integral easily, solving it to find \( 3\sqrt{u} + C \). Then, revert to the original variable \( t \) to integrate back into the initial context.

First-Order Differential Equation

First-order differential equations involve derivatives of the first order (the first derivative) and the function itself. They are often written as \( \frac{dy}{dt} = g(t, y) \). In these equations:

• The primary aim is to find the function \( y \) as it changes over time \( t \).
• Our example is \( \frac{dv}{dt} = \frac{3}{t \sqrt{t^2 - 1}} \).
• Such equations are important because they model a wide range of phenomena, from cooling coffee to population growth.

In solving this, one typically applies methods like separation of variables or integrating factors. The solution usually involves an arbitrary constant \( C \), which is determined using the initial condition provided. For our exercise, the initial condition \( v(2) = 0 \) was used to find the specific constant value, ensuring a well-defined solution.