Problem 12 - Calculus Early Transcendentals: Pearson New International Edition

Question

First find the general solution (involving a constant C) for the given differential equation. Then find the particular solution that satisfies the indicated condition. (See Example 2.) \(\frac{d u}{d t}=u^{3}\left(t^{3}-t\right) ; u=4\) at \(t=0\)

Step-by-Step Solution

Answer

The particular solution is \( u = \sqrt{\frac{1}{2 \left( \frac{t^4}{4} - \frac{t^2}{2} + \frac{1}{32} \right)}} \).

Step 1: Separate the Variables

To solve the differential equation \( \frac{du}{dt} = u^3(t^3-t) \), we first separate the variables by moving all terms involving \(u\) to one side, and all terms involving \(t\) to the other side. This gives us: \[ \frac{du}{u^3} = (t^3 - t) \, dt \]

Step 2: Integrate Both Sides

Now that the variables are separated, integrate both sides. Integrate \(\frac{du}{u^3}\) with respect to \(u\) and \((t^3 - t)\) with respect to \(t\):\[ \int \frac{du}{u^3} = \int (t^3 - t) \, dt \]The integrals compute to:\[ -\frac{1}{2u^2} = \frac{t^4}{4} - \frac{t^2}{2} + C \]

Step 3: Solve for the General Solution

From the integration, we have:\[ -\frac{1}{2u^2} = \frac{t^4}{4} - \frac{t^2}{2} + C \]Solving for \(u^2\), we get:\[ u = \pm \sqrt{\frac{1}{2\left( \frac{t^4}{4} - \frac{t^2}{2} + C \right)}} \]This is the general solution involving the constant \(C\).

Step 4: Apply Initial Condition to Find Particular Solution

Use the initial condition \(u = 4\) at \(t = 0\) to find the particular solution. Substitute these values into the general solution:\[ 4 = \pm \sqrt{\frac{1}{2(0 - 0 + C)}} \]This simplifies to:\[ 16 = \frac{1}{2C} \]Solving for \(C\), we find \(C = \frac{1}{32}\).

Step 5: Write the Particular Solution

Substitute \(C = \frac{1}{32}\) back into the general solution to get the particular solution:\[ u = \pm \sqrt{\frac{1}{2\left( \frac{t^4}{4} - \frac{t^2}{2} + \frac{1}{32} \right)}} \]Considering the initial condition and the practical sense, \(u\) must be positive, so:\[ u = \sqrt{\frac{1}{2 \left( \frac{t^4}{4} - \frac{t^2}{2} + \frac{1}{32} \right)}} \]

Key Concepts

Differential Equations

Differential equations are mathematical equations that involve derivatives of a function. They are essential in describing various phenomena in science and engineering, such as changes in physics, chemistry, and biology. In a differential equation, you're often trying to find a function which satisfies the equation. Here's a quick rundown:

• The goal is to find a function that describes how something changes.
• These equations relate the function and its derivatives.
• They're crucial for modeling how systems evolve over time.

In the problem we're solving, we have the differential equation \( \frac{du}{dt} = u^3(t^3-t) \). This equation tells us how the function \(u\) changes with respect to \(t\). The task involves finding the function \(u(t)\) that satisfies this equation.

Integration

Integration is the reverse process of differentiation, and it's an essential tool in solving differential equations. When you integrate, you are essentially finding a function whose derivative is the given function.

• You integrate with respect to the variable of interest.
• This process allows you to 'undo' differentiation.
• The constant of integration \(C\) appears after integration, representing all possible solutions.

In the exercise, once we separate the variables, we integrate \(\frac{du}{u^3}\) to get a function \(-\frac{1}{2u^2}\) and \((t^3 - t)\) to get \(\frac{t^4}{4} - \frac{t^2}{2} + C\). These integrals are key to finding the general solution of the differential equation.

Initial Value Problems

Initial value problems involve finding a specific solution to a differential equation that satisfies an initial condition. This condition is a specific value of the function at a certain point. It ensures that among many possible solutions, we select the one unique to our problem.

• An initial condition might be the position of a particle at \(t=0\).
• This condition "pins down" the solution.
• Without it, you'd have a family of solutions.

In our case, the initial condition \(u = 4\) when \(t = 0\) is crucial. It allows us to find the particular solution by determining the specific value of the constant \(C\). By substituting these values, we found \(C = \frac{1}{32}\), ensuring that the solution pass through the point \( (0, 4) \).

Separation of Variables

Separation of variables is a technique used to solve differential equations, especially those that can be easily manipulated so that all terms involving one variable are on one side, and all terms involving another variable are on the other.

• This technique simplifies the integration process.
• It works well when the equation can be rearranged into a product of functions of each variable separately.
• It allows us to integrate each side with respect to its own variable.

In the exercise, starting with the equation \( \frac{du}{dt} = u^3(t^3 - t) \), we rearranged it to \( \frac{du}{u^3} = (t^3 - t) dt \). Now, each side contains only one variable, which we can then integrate, leading us towards the general solution. This method is powerful and often the first step in dealing with differential equations.