Problem 11 - Calculus

Question

In Exercises \(2-28,\) use separation of variables to find the solutions to the differential equations subject to the given initial conditions. $$2 \frac{d u}{d t}=u^{2}, u(0)=1$$

Step-by-Step Solution

Answer

The solution is \( u = \frac{2}{2-t} \).

Step 1: Identify Differential Equation and Initial Condition

The given differential equation is \( 2 \frac{du}{dt} = u^2 \). The initial condition is \( u(0) = 1 \). The goal is to use separation of variables to solve this equation and apply the initial condition.

Step 2: Separate Variables

Rearrange the terms to separate \( u \) and \( t \):\[ 2 \frac{du}{dt} = u^2 \]Rewriting gives: \[ \frac{2}{u^2} du = dt \] Now, the variables are separated, with \( du \) on one side and \( dt \) on the other.

Step 3: Integrate Both Sides

Integrate both sides of the equation:\[ \int \frac{2}{u^2} \, du = \int \, dt \]This results in:\[ -\frac{2}{u} = t + C \] where \( C \) is the constant of integration.

Step 4: Solve for the Constant using Initial Conditions

Apply the initial condition \( u(0) = 1 \) to find \( C \):\[ -\frac{2}{1} = 0 + C \] Thus, \( C = -2 \).

Step 5: Express Solution

Substitute \( C = -2 \) back into the integrated equation:\[ -\frac{2}{u} = t - 2 \] Solving for \( u \) gives:\[ u = \frac{2}{2-t} \] This is the solution to the differential equation with the given initial condition.

Key Concepts

Separation of Variables

Separation of Variables is a powerful technique for solving differential equations. It involves rearranging the equation so that each of the variables is isolated on opposite sides of the equation.
To apply separation of variables to the equation given in the exercise, we first recognize the structure of the differential equation: \(2 \frac{du}{dt} = u^2\). The term \(\frac{du}{dt}\) suggests a derivative, indicating that \(u\) and \(t\) are our two important variables.
We begin by rewriting the differential equation to separate these variables:

• Move terms involving \(u\) to one side.
• Move terms involving \(t\) to the other side.

Starting from \(2 \frac{du}{dt} = u^2\), we divide both sides by \(u^2\) and multiply by \(dt\), resulting in \(\frac{2}{u^2} du = dt\).
Now, the variables are separated, with functions of \(u\) on one side and functions of \(t\) on the other side. This sets the stage for integration, our next step.

Initial Conditions

Initial conditions are essential for giving uniqueness to the solutions of differential equations. Without them, we might find a general solution but not a specific one. In our exercise, the initial condition is \(u(0) = 1\).
An initial condition specifies the value of the solution at a particular point, usually denoted in terms of the dependent variable at \(t = 0\) or some other value.
Using the initial condition impacts the integration result by helping us find a specific constant of integration, denoting where the curve crosses the value given by the initial condition.

• In this problem, after integrating, we found \(-\frac{2}{u} = t + C\).
• Applying \(u(0) = 1\) helps us determine \(-\frac{2}{1} = 0 + C\), giving us \(C = -2\).

Integration

Integration is a fundamental concept used in solving separated differential equations. Once you've separated the variables, you can integrate each side with respect to its respective variable.
For our exercise, after separation, the equation becomes \(\int \frac{2}{u^2} \, du = \int \, dt\). Integration requires calculating the antiderivative of each side:

• The left side, \(\int \frac{2}{u^2} \, du\), simplifies to \(-\frac{2}{u} + C\), where \(C\) is the constant of integration from the rule of indefinite integration.
• On the right side, \(\int \, dt\) results simply in \(t\).

The resulting equation, \(-\frac{2}{u} = t + C\), represents a family of curves depending on the value of \(C\). The integration acknowledges the relationship between \(u\) and \(t\) established by the original differential equation.

Constant of Integration

The constant of integration, represented by \(C\), appears in the process of integrating expressions in differential equations. It reflects the notion that there are infinitely many antiderivatives for a function, differing by a constant.
For the differential equation at hand, the integration process produces the result \(-\frac{2}{u} = t + C\). Here, \(C\) is the arbitrary constant added to account for all possible antiderivatives of a function.
The initial condition provided, \(u(0) = 1\), aids in pinpointing this constant. Substituting these values, we found \(C = -2\).

• This specific value adjusts our solution to meet the initial conditions given and rules out all other possible curves that don't pass through \((0,1)\).
• The final solution, \(u = \frac{2}{2-t}\), is thus adjusted specifically to kind of "fit" the initial conditions specified in the problem.

In this way, the constant of integration ensures that our solution to the differential equation is unique and adheres to any initial condition provided.