Problem 85 - Thomas Calculus

Question

Solve the initial value problems in Exercises \(71-90\) $$\frac{d^{2} r}{d t^{2}}=\frac{2}{t^{3}} ; \quad\left.\frac{d r}{d t}\right|_{t=1}=1, \quad r(1)=1$$

Step-by-Step Solution

Answer

The solution is \( r(t) = \frac{2}{t} + 3t - 4 \).

Step 1: Integrate the second derivative

To start solving the differential equation, first integrate the given second derivative equation. \[ \frac{d^2 r}{dt^2} = \frac{2}{t^3} \]Integrating once with respect to \(t\), we get the first derivative:\[ \frac{dr}{dt} = \int \frac{2}{t^3} \, dt = -\frac{2}{t^2} + C_1 \] where \(C_1\) is the integration constant.

Step 2: Use the first initial condition

Apply the initial condition \( \left. \frac{dr}{dt} \right|_{t=1} = 1 \).Substitute \(t = 1\) and \(\frac{dr}{dt} = 1\) into the equation from Step 1:\[ 1 = -\frac{2}{1^2} + C_1 \]This simplifies to:\[ 1 = -2 + C_1 \]Solving for \(C_1\), we find:\[ C_1 = 3 \]

Step 3: Integrate again to find r(t)

Now integrate the equation for \( \frac{dr}{dt} \) to find \( r(t) \).\[ \frac{dr}{dt} = -\frac{2}{t^2} + 3 \]Integrating with respect to \(t\), we obtain:\[ r(t) = \int \left(-\frac{2}{t^2} + 3\right) dt = \frac{2}{t} + 3t + C_2 \]where \(C_2\) is another integration constant.

Step 4: Use the second initial condition

Utilize the initial condition \( r(1) = 1 \).Substitute \(t = 1\) and \(r(1) = 1\) into the equation from Step 3:\[ 1 = \frac{2}{1} + 3 \times 1 + C_2 \]This simplifies to:\[ 1 = 2 + 3 + C_2 \]Solve for \(C_2\):\[ C_2 = 1 - 5 = -4 \]

Step 5: Write the final solution

Substitute the constants back into the equation for \( r(t) \):\[ r(t) = \frac{2}{t} + 3t - 4 \]This is the solution to the initial value problem that satisfies both initial conditions.

Key Concepts

Second Order Differential Equations

A **second order differential equation** is an equation involving a function and its derivatives, up to the second derivative. In simpler terms, it tells how a function changes based on its current value and the rate of change of the rate of change! It's like calculating acceleration in physics—where you're interested in how the speed itself changes over time.
For instance, in the exercise, we dealt with the equation \( \frac{d^{2} r}{d t^{2}} = \frac{2}{t^{3}} \). Here, we have a second derivative \( \frac{d^{2} r}{d t^{2}} \), indicating it's a second order differential equation.

• These types of equations are common in physics, engineering, and other scientific fields.
• The solution often aims to model fluctuating systems, such as the motion of waves or objects in motion.

Understanding second order differential equations requires grasping concepts like acceleration affecting movement or forces affecting systems. These equations can sometimes be solved step by step, through a method known as integration, which we will explain next.

Integration

**Integration** is like the reverse process of taking a derivative. When you integrate, you find a new function whose derivative is the original function you started with. In our problem, we begin by integrating the second derivative \( \frac{d^2 r}{dt^2} \) to find the first derivative \( \frac{dr}{dt} \).
When working on \( \frac{d^2 r}{dt^2} = \frac{2}{t^{3}} \), integrating \( \frac{2}{t^3} \) with respect to \( t \) gives us\(\frac{dr}{dt} = \int \frac{2}{t^3} \, dt = -\frac{2}{t^2} + C_1\).Here, \( C_1 \) is the constant of integration, which appears because integration is like "undoing" derivatives.

• Integration is used to accumulate values, such as finding an area under a curve.
• In differential equations, it helps us "trace back" from derivatives to the functions themselves.

The next steps involve applying initial conditions to fine-tune these functions and adjust the constants of integration.

Initial Conditions

**Initial conditions** are values given at the start of a problem to help find a specific solution among many possible ones. They are crucial because, without them, you'd end up with a general solution that includes unknown constants.
For the problem at hand, we have two initial conditions:

• \( \left.\frac{dr}{dt}\right|_{t=1} = 1 \)
• \( r(1) = 1 \)

The first condition \( \left.\frac{dr}{dt}\right|_{t=1} = 1 \) specifies the rate of change at \( t=1 \), allowing us to solve for \( C_1 \) by substituting the values in step 2 of our solution. This pinpoints one aspect of the function's behavior.
The second condition \( r(1) = 1 \) describes the value of the function at the same starting point. By using both initial conditions, we fully determine the constants and thus the specific solution for our initial value problem.

Constants of Integration

**Constants of integration** are values that appear when you integrate, representing all possible vertical shifts of the antiderivative. When you solve a differential equation, these constants allow for the general solution to be as broad as possible, accounting for different starting points and conditions.
In our example, we encountered \( C_1 \) and \( C_2 \).

• \( C_1 \) was determined by the initial condition \( \left.\frac{dr}{dt}\right|_{t=1} = 1 \).
• \( C_2 \) was found using the condition \( r(1) = 1 \).

These constants change based on the information given in the initial conditions. Without such conditions, the equation could satisfy infinitely many solutions, which is why constants of integration are key.
They ensure the final solution fits the unique initial value problem we're solving, leading to the specific answer for scenarios given in real-world applications.