Problem 59 - 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. $$ \frac{d y}{d t}=\sqrt{\frac{t+1}{y+1}} \text { where } y(0)=1 $$

Step-by-Step Solution

Answer

The implicit solution is \((y+1)^{3/2} = (t+1)^{3/2} + 1\).

Step 1: Separate the Variables

We begin by separating the variables to isolate terms involving \( y \) on one side and terms involving \( t \) on the other. The differential equation is given by \( \frac{dy}{dt} = \sqrt{\frac{t+1}{y+1}} \). To separate the variables, we rewrite this as \( (y+1)^{1/2} \, dy = (t+1)^{1/2} \, dt \).

Step 2: Integrate Both Sides

Next, integrate both sides of the equation with respect to their respective variables. The left side becomes \( \int (y+1)^{1/2} \, dy \) and the right side becomes \( \int (t+1)^{1/2} \, dt \).

Step 3: Solve the Integrals

Solve each integral separately. The integral \( \int (y+1)^{1/2} \, dy \) can be solved using the substitution \( u = y + 1 \) which yields \( \frac{2}{3}(y+1)^{3/2} + C_1 \). The integral \( \int (t+1)^{1/2} \, dt \) solves similarly with substitution \( v = t + 1 \), giving \( \frac{2}{3}(t+1)^{3/2} + C_2 \).

Step 4: Combine and Simplify the Results

Combine the results of your integrations to get: \( \frac{2}{3}(y+1)^{3/2} + C_1 = \frac{2}{3}(t+1)^{3/2} + C_2 \). Since \( C_1 - C_2 \) can be a single constant \( C \), rewrite the equation as \( \frac{2}{3}(y+1)^{3/2} = \frac{2}{3}(t+1)^{3/2} + C \).

Step 5: Solve for Integration Constant Using Initial Condition

To find the constant \( C \), use the initial condition \( y(0) = 1 \). Substitute \( t = 0 \) and \( y = 1 \) into the equation: \( \frac{2}{3}(1+1)^{3/2} = \frac{2}{3}(0+1)^{3/2} + C \). This simplifies to \( \frac{2}{3} \times 2^{3/2} = \frac{2}{3} \times 1^{3/2} + C \), resulting in \( C = \frac{4}{3} - \frac{2}{3} = \frac{2}{3} \).

Step 6: Write the Final Implicit Solution

Insert the constant \( C = \frac{2}{3} \) back into the equation from Step 4 to get the final implicit solution: \( \frac{2}{3}(y+1)^{3/2} = \frac{2}{3}(t+1)^{3/2} + \frac{2}{3} \). This simplifies to \( (y+1)^{3/2} = (t+1)^{3/2} + 1 \).

Key Concepts

Separation of Variables

The separation of variables is a powerful method for solving ordinary differential equations. This technique works well when an equation can be rearranged so that each variable is on a different side of the equation. In this exercise, we started with the differential equation \( \frac{dy}{dt} = \sqrt{\frac{t+1}{y+1}} \). By multiplying both sides to swap and isolate the terms involving different variables, we expressed it as \((y+1)^{1/2} \, dy = (t+1)^{1/2} \, dt\). This step is crucial because it paves the way for individual integration of each side.

• The main goal is to separate the variables so each appears only on one side in a product with its differential.
• Once the variables are separated, we can integrate each side with respect to its variable.

The flexibility of this method allows us to handle a wide range of differential equations, making it an essential technique in mathematical problem-solving.

Implicit Function

Sometimes, solving a differential equation might lead to a solution in the form of an implicit function. This means the solution relates the variables indirectly, unlike explicit solutions where one variable is directly solved in terms of another. Our problem arrives at such an implicit form: \((y+1)^{3/2} = (t+1)^{3/2} + 1\).

• An implicit function is not easily solved for a single variable but still conveys the relationship between the variables.
• Implicit solutions are often acceptable when explicit solutions are complex or not plausible.

In this case, although \(y\) isn't isolated, the solution is still valid as it defines the relationship between \(y\) and \(t\). Learning to work with implicit functions expands your ability to interpret and solve real-world problems with more complex dynamics.

Initial Condition

Initial conditions are essential for determining a unique solution to a differential equation. Essentially, they provide specific values for the variables at a given point, which we use to calculate any unknown constants. In this exercise, the initial condition is given as \(y(0) = 1\). This means that when \(t = 0\), \(y = 1\).

• By substituting the initial condition into the solution equation, we solve for the integration constant \(C\).
• This process is vital for arriving at a unique solution that matches the conditions of the problem.

Initial conditions are crucial in many practical applications, as they help tailor general solutions to specific situations or points of interest.