Problem 2 - Calculus for the Life Sciences: A Modeling Approach Volume II

Question

Which of the following possible solutions satisfies the initial condition and the differential equation. Possible Solution \(\quad\) Initial Condition \(\quad\) Differential Equation a. \(\quad y(t)=e^{2 t}+2 e^{t} \quad y(0)=2\) \(y^{\prime}(t)-y(t)=e^{2 t}\) b. \(\quad y(t)=e^{2 t}+2 e^{t} \quad y(0)=3 \quad y^{\prime}(t)-y(t)=e^{2 t}\) \(\begin{array}{lll}\text { c. } y(t)=\sqrt{t+1} & y(0)=1 & y(t) \times y^{\prime}(t)=1\end{array}\) d. \(y(t)=t^{3}\) \(y(1)=1 \quad y^{\prime}(t) / y(t)=3 t\) e. \(\quad y(t)=\frac{1}{t}\) f. \(y(t)=t^{3}\) \(y(1)=1 \quad y^{\prime}(t) / y(t)=3 t\) g. \(\quad y(t)=t e^{t}\) \(y(0)=0 \quad y^{\prime}(t)-y(t)=e^{t}\)

Step-by-Step Solution

Answer

Option g satisfies both the initial condition and the differential equation.

Step 1: Evaluate Initial Condition for Option a

For \(y(t)=e^{2t}+2e^{t}\), substitute \(t=0\) to check if the initial condition \(y(0)=2\) holds. Calculate \(y(0)=e^{0}+2e^{0}=1+2=3\). Therefore, it does not satisfy \(y(0)=2\).

Step 2: Evaluate Initial Condition for Option b

For \(y(t)=e^{2t}+2e^{t}\), substitute \(t=0\) to check the initial condition \(y(0)=3\). Calculate \(y(0)=1+2=3\), which satisfies the initial condition.

Step 3: Verify Differential Equation for Option b

Differentiate \(y(t)=e^{2t}+2e^{t}\) to get \(y'(t)=2e^{2t}+2e^{t}\). Substitute \(y(t)\) and \(y'(t)\) into the differential equation \(y'(t)-y(t)=e^{2t}\) to verify: \[(2e^{2t}+2e^{t})-(e^{2t}+2e^{t})=e^{2t}\],which simplifies to \(e^{2t}=e^{2t}\). The equation is satisfied.

Step 4: Evaluate Initial Condition for Option c

For \(y(t)=\sqrt{t+1}\), substitute \(t=0\) to check if the initial condition \(y(0)=1\) holds. Calculate \(y(0)=\sqrt{1}=1\). The condition is satisfied.

Step 5: Verify Differential Equation for Option c

Differentiate \(y(t)=\sqrt{t+1}\) to find \(y'(t)=\frac{1}{2\sqrt{t+1}}\). The differential equation is \(y(t) \times y'(t) = 1\). Verify:\[\sqrt{t+1} \times \frac{1}{2\sqrt{t+1}} = \frac{1}{2} \, eq 1\].The equation is not satisfied.

Step 6: Evaluate Initial Condition for Option d

For \(y(t)=t^3\), substitute \(t=1\) to check if the initial condition \(y(1)=1\) holds. Calculate \(y(1)=1^3=1\). The condition is satisfied.

Step 7: Verify Differential Equation for Option d

Differentiate \(y(t)=t^3\) to get \(y'(t)=3t^2\). Substitute into \(\frac{y'(t)}{y(t)}=3t\):\[\frac{3t^2}{t^3}=\frac{3}{t}\].The equation is not satisfied.

Step 8: Evaluate Initial Condition and Differential Equation for Option g

For \(y(t) = te^t\), substitute \(t=0\) to check the initial condition \(y(0)=0\). Calculate \(y(0)=0\), which is satisfied. Differentiate to obtain \(y'(t)=e^t+te^t\). Substitute into the equation \(y'(t)-y(t)=e^t\):\[(e^t+te^t)-(te^t)=e^t\].The differential equation is satisfied.

Key Concepts

Initial Conditions in Differential Equations

When solving differential equations, initial conditions are crucial for determining the specific solution that fits a given problem. A differential equation can have many solutions, but initial conditions help pin down which one is appropriate for a specific scenario.

Initial conditions provide the values of the function and often its derivatives at a particular point. They act as a starting point that allows us to solve for unknown constants in the solution. For instance, if we have a differential equation with the general solution \(y(t)=C_1e^{t} + C_2e^{-t}\), and an initial condition such as \(y(0)=2\), we can substitute \(t=0\) into the equation to find the values of \(C_1\) and \(C_2\).

In the exercise, by substituting the value of \(t\) given in the initial conditions, students can test whether each potential solution satisfies those conditions. This is a critical step, as only solutions that satisfy the initial conditions are valid. It's like finding the puzzle piece that fits perfectly into our problem.

Problem-Solving in Differential Equations

Problem-solving is a process that follows identifying the correct mathematical method appropriate for the differential equation at hand. Usually, solving differential equations involves:

• Substituting guessed solutions or applying known techniques (like separation of variables or integrating factors) into the differential equation
• Using initial conditions to solve for unknown constants
• Checking the solution for consistency with any given conditions

In the exercise provided, each possible solution was put through a process to determine if it met the initial conditions and the differential equation. For example, when evaluating option (b) with \(y(t)=e^{2t} + 2e^{t}\) and comparing it to the differential equation \(y'(t)-y(t)=e^{2t}\), calculations were performed by differentiating and substituting back into the equation.

This step-by-step approach is key in mastering problem-solving for differential equations, allowing for the identification of a solution that uniquely fits the initial conditions and the equation.

Verification of Solutions

Once a potential solution is found, the next critical step is verifying its validity. Verification involves checking both the differential equation and the initial conditions to ensure all components align.

Verification steps often include:

• Calculating the derivative(s) of the proposed solution
• Substituting these derivatives and the solution into the original differential equation
• Seeing if the resultant expressions hold true throughout

In the provided solution, we verified option (b) by plugging the derived values of \(y'(t)\) and \(y(t)\) back into the equation to confirm it holds true: \[(2e^{2t} + 2e^{t}) - (e^{2t} + 2e^{t}) = e^{2t}\] simplifies to \(e^{2t} = e^{2t}\). This perfect match confirms the solution is valid.

Verification ensures accuracy and helps build confidence in one's ability to solve differential equations correctly. It's like a double-check in a math problem, ensuring no steps were overlooked or miscalculated.