Problem 56 - 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. $$ \frac{d y}{d t}=\frac{y t}{\ln y} \text { where } y(1)=e $$

Step-by-Step Solution

Answer

The implicit solution to the differential equation is \( (\ln y)^2 = t^2 \).

Step 1: Separate the Variables

First, we need to rewrite the given differential equation, \( \frac{d y}{d t}=\frac{y t}{\ln y} \), to separate the variables \( y \) and \( t \). This can be done by dividing both sides by \( y \) and multiplying both sides by \( \ln y \), yielding: \( \ln y \frac{d y}{y} = t \, dt \).

Step 2: Integrate Both Sides

Next, integrate both sides of the equation from Step 1. The left-hand side becomes \( \int \ln y \frac{d y}{y} \), and the right-hand side becomes \( \int t \, dt \).

Step 3: Solve the Left-Hand Side Integration

For the left-hand side, use substitution. Let \( u = \ln y \), then \( du = \frac{1}{y} \, dy \). The integral becomes \( \int u \, du \), which results in \( \frac{u^2}{2} = \frac{(\ln y)^2}{2} \).

Step 4: Solve the Right-Hand Side Integration

For the right-hand side, integrate with respect to \( t \), thus \( \int t \, dt = \frac{t^2}{2} + C \), where \( C \) is the constant of integration.

Step 5: Write the Combined Solution

Combine the results of Steps 3 and 4: \( \frac{(\ln y)^2}{2} = \frac{t^2}{2} + C \).

Step 6: Apply Initial Condition

Use the initial condition \( y(1) = e \) to find \( C \). Substitute \( t = 1 \) and \( y = e \) into the equation: \( \frac{(\ln e)^2}{2} = \frac{1^2}{2} + C \). Since \( \ln e = 1 \), the equation simplifies to \( \frac{1}{2} = \frac{1}{2} + C \), giving \( C = 0 \).

Step 7: Final Implicit Solution

Substitute \( C = 0 \) back into the combined solution to obtain the implicit solution: \( (\ln y)^2 = t^2 \).

Key Concepts

Separation of Variables

Separation of variables is a powerful technique used to solve differential equations. The aim is to manipulate the equation so that each variable and its differential appear on different sides of the equation. This makes it easier to integrate each side separately. To perform separation of variables in the given differential equation \( \frac{d y}{d t} = \frac{y t}{\ln y} \), we rearrange terms to isolate \( y \) on one side and \( t \) on the other. Here are the steps:

• Divide both sides by \( y \) to get \( \frac{d y}{y} = \frac{t}{\ln y} \, dt \).
• Multiply both sides by \( \ln y \) to get \( \ln y \frac{d y}{y} = t \, dt \).

Now, each side of the equation is ready to be integrated separately. This careful separation sets the stage for the integration step, leading towards finding a solution to the differential equation.

Implicit Function

An implicit function is different from an explicit function in that it doesn't solve explicitly for one variable in terms of the others. Instead, it relates variables within an equation form. For the solved differential equation problem, our solution came out as an implicit function: \( (\ln y)^2 = t^2 \). In this form, neither \( y \) nor \( t \) is expressed solely in terms of the other variable. Solving implicitly is common in differential equations when simplifying the solution to an explicit form isn’t straightforward or necessary. It allows us to express the relationship between the variables correctly without further simplification that could be complex or impossible. This form expresses all possible solutions where the initial condition is also respected.

Initial Condition

An initial condition is a specified value that a solution to a differential equation must satisfy. Using an initial condition is crucial as it helps to find the specific solution from a family of possible solutions. In the problem, the initial condition given is \( y(1) = e \). This means when \( t = 1 \), \( y \) must equal \( e \). To use this condition:

• Substitute \( t = 1 \) and \( y = e \) into the implicit solution \( (\ln y)^2 = t^2 \).
• This gives \( (\ln e)^2 = 1^2 \), simplifying to \( 1 = 1 + C \). Here, we solve for \( C \) which turns out to be \( 0 \).

Applying the initial condition narrows down the infinite set of solutions to the one specific solution that meets the problem's criteria. This makes it unique and ensures it adheres to the initial state described by the problem.