Problem 58 - Calculus for Biology and Medicine

Question

In Problems you will need to solve differential equations by separation of variables. In these problems it will not always be possible to solve explicitly for \(y\) in terms of \(t ;\) instead your solution may take the form of an implicit function relating the two variables. $$ \frac{d y}{d t}=\frac{t^{2}+1}{\cos y+\sin y} \text { where } y(0)=0 $$

Step-by-Step Solution

Answer

Implicit solution is \( \sin y - \cos y = \frac{t^3}{3} + t - 1 \).

Step 1: Separate Variables

To separate variables, we need all terms involving \( t \) on one side and all terms involving \( y \) on the other. Starting with the given equation: \( \frac{d y}{d t} = \frac{t^2 + 1}{\cos y + \sin y} \), multiply both sides by \( \cos y + \sin y \) to get \( (\cos y + \sin y) d y = (t^2 + 1) d t \). This allows us to integrate both sides with respect to their respective variables.

Step 2: Integrate Both Sides

Integrate both sides of the equation separately. The left-hand side becomes \( \int (\cos y + \sin y) \, d y \) and the right-hand side becomes \( \int (t^2 + 1) \, d t \). After integration, we have: \( \int \cos y \, dy + \int \sin y \, dy = \int t^2 \, dt + \int 1 \, dt \). This leads to \( \sin y - \cos y = \frac{t^3}{3} + t + C \), where \( C \) is the constant of integration.

Step 3: Apply Initial Condition

Use the initial condition \( y(0) = 0 \) to find the value of \( C \). Substitute \( t = 0 \) and \( y = 0 \) into \( \sin y - \cos y = \frac{t^3}{3} + t + C \), we get \( \sin(0) - \cos(0) = 0 + 0 + C \). Simplifying, \( -1 = C \). Therefore, the solution becomes \( \sin y - \cos y = \frac{t^3}{3} + t - 1 \).

Step 4: Confirm General Form

The solution in implicit form is \( \sin y - \cos y = \frac{t^3}{3} + t - 1 \). Solving explicitly for \( y \) can be complex or not possible with elementary functions. For this problem, the implicit form is acceptable as the relationship between \( y \) and \( t \).

Key Concepts

Separation of Variables

Separation of variables is a method used to solve differential equations, which allows us to simplify the problem by separating the variables involved. The goal is to have one side of the equation with terms involving only one variable, and the other side with terms involving the other variable. This makes the equation integrable with respect to each variable separately.
In the original exercise, the differential equation given is \( \frac{dy}{dt} = \frac{t^2 + 1}{\cos y + \sin y} \). To separate the variables, we rearrange the terms to get all \( t \)-related terms on one side and all \( y \)-related terms on the other. Here’s how that looks like:

• Multiply both sides by \( \cos y + \sin y \) to transfer all \( y \)-related terms on one side: \( (\cos y + \sin y) dy = (t^2 + 1) dt \).
• Now, each side of the equation is prepared for integration with respect to its variable: integrate \( dy \) with \( y \)-terms, and \( dt \) with \( t \)-terms.

Separation of variables essentially transforms the differential equation into two integrals that are easier to handle. This procedure is a fundamental technique for dealing with equations of this type.

Initial Condition

When solving differential equations, the initial condition is crucial for determining a specific solution rather than a general form with a constant of integration. The initial condition provides specific values of the variables that satisfy the differential equation at a particular point.
In the given exercise, the initial condition is \( y(0) = 0 \). This means when \( t = 0 \), then \( y \) is also zero. After integrating both sides and introducing an integration constant \( C \), the specific initial condition helps us find the exact value of this constant. For this problem, substituting \( t = 0 \) and \( y = 0 \) into the equation \( \sin y - \cos y = \frac{t^3}{3} + t + C \) leads to \( -1 = C \).
This step is essential to find the particular solution where the condition reflects physical reality or specific requirements of the problem, making it distinct from only having a family of general solutions.

Implicit Function

An implicit function is a type of function where the relationship between the variables isn’t expressed as one variable being written explicitly in terms of another. In contrast, an explicit function provides the direct equation of one variable in terms of others.
In some cases, like this exercise, solving the equation explicitly for one attribute like \( y \) might be too complicated or not possible with elementary functions. Instead, the solution remains an implicit equation: \( \sin y - \cos y = \frac{t^3}{3} + t - 1 \).
This implicit relation still fully describes the connection between \( y \) and \( t \), even though it’s not rearranged into an explicit format. Many real-world problems use implicit functions, as they perfectly capture the essential relations without needing simplification that might not be feasible.