Problem 60 - 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. $$ \frac{d y}{d t}=\frac{t+1}{t v+t v^{3}} \text { where } y(1)=1 \text { . } $$

Step-by-Step Solution

Answer

The implicit solution is \( \frac{y^2}{2} + \frac{y^4}{4} = t + \ln|t| - \frac{3}{4} \).

Step 1: Rewrite the given differential equation

Given the differential equation \( \frac{dy}{dt} = \frac{t+1}{t y + t y^3} \). We aim to separate the variables \( y \) and \( t \) to their respective sides.

Step 2: Separate the variables

Rearrange the equation to \( (t y + t y^3) dy = (t+1) dt \). This results from multiplying both sides of the differential equation by \((t y + t y^3)\) and by \(dt\).

Step 3: Integrate both sides

Integrate each side of the equation. The left side becomes \( \int (y + y^3) dy \) after factoring \( t \) from the left-hand side, and the right side is \( \int \frac{t+1}{t} dt \).

Step 4: Evaluate the integrals

For the left-hand side, \( \int (y + y^3) dy = \frac{y^2}{2} + \frac{y^4}{4} + C_1 \). For the right-hand side, \( \int \left(1 + \frac{1}{t}\right) dt = t + \ln|t| + C_2 \).

Step 5: Combine and simplify the solution

Equate the evaluated integrals: \( \frac{y^2}{2} + \frac{y^4}{4} + C_1 = t + \ln|t| + C_2 \). This is the implicit solution for the differential equation.

Step 6: Apply the initial condition to find constants

Given \( y(1) = 1 \), substitute \( t = 1, y = 1 \) into the equation: \( \frac{1^2}{2} + \frac{1^4}{4} = 1 + \ln(1) + C_2 - C_1 \). Since \( \ln(1) = 0 \), solve for \( C_2 - C_1 \).

Step 7: Simplify the result after applying the initial condition

From the above step, compute any remaining constants using the values found from the initial condition, if possible. Here, \( C_2 - C_1 = -\frac{3}{4} \). Now rewrite the solution accordingly.

Key Concepts

Separation of Variables

Separation of variables is a common method to solve differential equations, especially when the equation is given in a form where the rate of change of a variable (like \( \frac{dy}{dt} \)) can be expressed as a function of two distinct variables. In our exercise, the goal is to manipulate the given equation \( \frac{dy}{dt} = \frac{t+1}{ty + ty^3} \) so that all terms involving the variable \( y \) are moved to one side of the equation, and terms involving \( t \) are moved to the other side.

• First, multiply both sides by \( (ty + ty^3) \) and \( dt \) to "untangle" the two variables.
• The result is \( (ty + ty^3) dy = (t+1) dt \), creating two separate integrals that can be solved individually.

By isolating the variables in this way, you can integrate each side separately, making what seems like a complex problem manageable by breaking it down into simpler steps.

Implicit Functions

An implicit function arises when a differential equation is solved but an explicit solution \( y = f(t) \) is not found. Instead, the relationship between \( y \) and \( t \) is expressed as an equation linking the two variables without plainly solving for one in terms of the other.
In our specific solution, after integrating both sides, we get \[ \frac{y^2}{2} + \frac{y^4}{4} + C_1 = t + \ln|t| + C_2 \]This equation remains implicit because \( y \) is not isolated on one side. Instead of solving algebraically for an explicit function, the equation describes the relationship implicitly.

• Sometimes implicit solutions offer a more general representation of the solution to a differential equation.
• Implicit solutions are especially helpful when isolating the dependent variable in terms of the independent one is algebraically complex or impossible.

Understanding that such representations, while not always providing a direct answer, capture the behavior of solutions over their domain is crucial.

Initial Conditions

Initial conditions are specific values for the variables of a differential equation that allow us to find the particular solution from the general solution. In our exercise, we are given the condition \( y(1) = 1 \), meaning at \( t = 1 \), \( y \) equals 1.
To apply this, substitute \( t = 1 \) and \( y = 1 \) into the implicit solution:\[ \frac{1^2}{2} + \frac{1^4}{4} = 1 + \ln(1) + C_2 - C_1 \]Given that \( \ln(1) = 0 \), this simplifies to finding \( C_2 - C_1 = -\frac{3}{4} \), allowing us to further refine our solution.

• Initial conditions help convert an implicit form into something more specific to the given problem.
• They provide a means to evaluate the constant terms that arise from integration processes.

By understanding initial conditions, you can see how initial setup affects outcomes, effectively tailoring the solution to meet specific scenarios.