Problem 52 - Calculus for Biology and Medicine

Question

In Problems, solve each differential equation with the given initial condition.$$ \frac{d u}{d t}=\frac{\sin t}{u+1}, \text { with } u(0)=3 $$

Step-by-Step Solution

Answer

The implicit solution for the differential equation is \( \frac{u^2}{2} + u = -\cos t + \frac{17}{2} \), with initial condition \( u(0) = 3 \).

Step 1: Identify the Type of Differential Equation

The given differential equation is \( \frac{d u}{d t}=\frac{\sin t}{u+1} \). This is a first-order, non-linear differential equation in the form of a separable differential equation.

Step 2: Separate Variables

To separate variables, rewrite the given equation by moving all terms involving \( u \) to one side and all terms involving \( t \) to the other side:\[ (u+1) du = \sin t \, dt \].

Step 3: Integrate Both Sides

Integrate both sides: \[ \int (u+1) \, du = \int \sin t \, dt \].The left side integrates to \( \frac{u^2}{2} + u + C_1 \), and the right side integrates to \( -\cos t + C_2 \). Equating both sides gives:\[ \frac{u^2}{2} + u = -\cos t + C \].

Step 4: Apply the Initial Condition

We use the initial condition \( u(0) = 3 \) to find the constant \( C \). Substituting \( t = 0 \) and \( u = 3 \) into the equation:\[ \frac{3^2}{2} + 3 = -\cos(0) + C \].This simplifies to \( \frac{9}{2} + 3 = -1 + C \). Solving for \( C \), we have:\[ C = \frac{9}{2} + 3 + 1 = \frac{17}{2} \].

Step 5: Write the General Solution with the Constant

Substitute \( C = \frac{17}{2} \) back into the integral solution:\[ \frac{u^2}{2} + u = -\cos t + \frac{17}{2} \].This equation implicitly defines \( u(t) \) for the given initial condition.

Key Concepts

Separable Differential Equations

Separable differential equations are a special kind of differential equation where you can separate the variables on either side of the equation. These types of equations often appear in simple dynamical systems and allow us to solve problems by getting rid of the derivative  essentially breaking the relationship into two separate integrals. In our example, the equation \( \frac{d u}{d t} = \frac{\sin t}{u+1} \) is indeed separable. This means you can manipulate it to have all terms involving \( u \) on one side and all terms involving \( t \) on another side.

Once separated, you can integrate each side with respect to its respective variable. It often leads to an expression that relates \( u \) and \( t \), which you can then solve for one in terms of the other. This method can sometimes result in an implicit solution depending on the context and complexity of the equation.

• This process forms the core of solving separable differential equations and simplifies the process of working with derivatives.
• Practice makes perfect in recognizing and applying separable differential equations  they are an excellent entry point into the broader topic of differential equations.

Initial Conditions

Initial conditions provide a way to find the specific solution from the general solution of a differential equation. When solving differential equations, especially separable ones, integration gives us a family of solutions. These solutions include an arbitrary constant \( C \).

In our specific problem, the initial condition \( u(0) = 3 \) was vital in determining the value of this integration constant. Here's the process for applying initial conditions:

• After integrating, you end up with an equation that includes a constant \( C \).
• Substitute the initial condition values into the integrated equation.
• Solve for \( C \) using these values, which results in a unique solution for the differential equation.

Without initial conditions, the solution would remain general. Initial conditions are crucial for real-world applications where specific solutions are required, as they provide the context needed to move from theoretical to practical applications in modeling.

Integration Techniques

Integration is a fashionable yet powerful tool in solving separable differential equations, and mastering it is essential. An integral represents the accumulation of quantities, which is a concept used extensively in mathematics to represent various real-world processes such as accumulation of growth, area under a curve, etc.

In our exercise, integrating the expression \( \int (u+1) \, du = \int \sin t \, dt \) involves basic integration techniques. For instance,

• The left side, \( \int (u+1) \, du \), involves the power rule of integration, resulting in \( \frac{u^2}{2} + u \) plus a constant.
• On the right side, integrating \( \int \sin t \, dt \) results in \( -\cos t \) plus a constant. This is due to the fact that the derivative of \(-\cos t\) is \( \sin t \).

Both require you to think carefully about the rules of integration you have learned, from simple to more complex functions. Sometimes, integration techniques may also include substitution, integration by parts, or partial fraction decomposition, depending on the complexity of the function. Practicing these techniques within context helps to improve problem-solving skills.