Problem 8 - Calculus for Biology and Medicine

Question

Find the equilibria of the following differential equations. $$ \frac{d y}{d t}=y^{1 / 3}-1 $$

Step-by-Step Solution

Answer

The equilibrium is at \( y = 1 \).

Step 1: Set the Differential Equation to Zero

To find the equilibria of a differential equation, set the derivative to zero. Thus, we solve the equation: \[\frac{d y}{d t} = y^{1 / 3} - 1 = 0\]

Step 2: Solve for the State Variable, y

Rearrange the equation from Step 1 to isolate the term involving y:\[y^{1/3} = 1\]Next, cube both sides to solve for y:\[y = 1^3\]

Step 3: Simplify the Solution

Since cubing 1 gives 1, the resulting solution for y is:\[y = 1\]

Step 4: Verify Equilibrium Condition

To verify, substitute y = 1 back into the original equation:\[\frac{d y}{d t} = 1^{1/3} - 1 = 0\]Since this holds true, y = 1 is indeed an equilibrium.

Key Concepts

Equilibria

An equilibrium in a differential equation is a state where the system doesn't change over time. This means that at equilibrium, the derivative (rate of change of the system) is zero. Equilibria are important because they show us stable points in a system's behavior.

• To find the equilibrium, set the differential equation to zero.
• This helps us find values where the state of the system doesn't change, indicating balance or stability.
• In the given problem, setting \( \frac{dy}{dt} = y^{1/3} - 1 = 0 \) led us to determine the equilibrium point.

Understanding equilibria allows one to analyze whether a system will settle down into a steady state over time, or if it will keep changing indefinitely.

State Variable

The state variable in a differential equation is a variable that represents the state of the system at any given time. It’s what changes as time progresses. In our given exercise, the state variable is \( y \).

• The value of the state variable \( y \) represents a specific condition of the system.
• Solving for \( y \) when \( \frac{dy}{dt} = 0 \) helps identify key points such as equilibria.
• Here, solving \( y^{1/3} = 1 \) helps us understand the system's behavior when it is not changing.

Grasping the concept of state variables is crucial in modeling dynamic systems, as they are the variables we need to understand to predict future states of the system.

Cube Root

The cube root is an operation that finds a number which, when multiplied by itself three times, gives the original number. In the given differential equation, the cube root is essential in manipulating the expression \( y^{1/3} \).

• Taking the cube root of a number can help simplify the expression into a more solvable form.
• In our step-by-step solution, we used the cube root property to isolate \( y \), leading to solving \( y^{1/3} = 1 \).
• Cubing both sides derived that \( y = 1^3 \), which equals 1.

Understanding how to solve cube roots aids greatly in rewriting and simplifying differential equations.

Solutions Verification

Verifying solutions in differential equations is a critical step to ensure the solutions are correct and applicable. Once a potential solution is determined, substitute it back into the original equation to see if it satisfies all conditions.

• Verification involves checking consistency with the original differential equation.
• This step ensures no mistakes were made during the process of finding solutions.
• In our problem, substituting \( y = 1 \) back into \( \frac{dy}{dt} = y^{1/3} - 1 \) resulted in zero, confirming it as an equilibrium point.

Always verify your solutions to decisively conclude their validity and better understand their meaning in the context of the equation.