Problem 1 - Calculus for the Life Sciences: A Modeling Approach Volume II

Question

Show that each solution satisfies the initial condition and the differential equation. $$\begin{aligned} &\text { Solution } \quad \text { Initial Condition } \quad \text { Differential Equation }\\\ &\text { a. } \quad y(t)=e^{2 t}+e^{t} \quad y(0)=2 \quad y^{\prime}(t)-y(t)=e^{2 t} \end{aligned}$$ $$\text { b. } \quad y(t)=\frac{1}{3} e^{t}+\frac{2}{3} e^{-2 t}, \quad y(0)=1, \quad y^{\prime}(t)+2 y(t)=e^{t}$$ c. \(\quad y(t)=t e^{t} \quad y(0)=0 \quad y^{\prime}(t)-y(t)=e^{t}\) d. \(\quad y(t)=\frac{t^{2}}{3}+\frac{1}{t}, \quad y(1)=\frac{4}{3}, \quad t \times y^{\prime}(t)+y(t)=t^{2}\) e. \(\quad y(t)=\sqrt{t+1} \quad y(0)=1 \quad y(t) \times y^{\prime}(t)=\frac{1}{2}\) f. \(\quad y(t)=\sqrt{1+t^{2}}, \quad y(0)=1, \quad y(t) \times y^{\prime}(t)=t\) \(\begin{array}{lll}\text { g. } y(t)=\sqrt{4+t^{2}} & y(0)=2 & y(t) \times y^{\prime}(t)=t\end{array}\) \(\begin{array}{lll}\text { Solution } & \text { Initial } & \text { Differential Equation }\end{array}\) Condition h. \(y(t)=\frac{1}{t+1}, \quad y(0)=1, \quad y^{\prime}(t)+(y(t))^{2}=0\) i. \(\quad y(t)=0.5+0.5 e^{-0.2 \sin t} \quad y(0)=1, \quad y^{\prime}(t)+0.2(\cos t) y(t)=0.1 \cos t\) j. \(\quad y(t)=\tan t, \quad y(0)=0, \quad y^{\prime}(t)=1+(y(t))^{2}\) k. \(y(t)=3\) \(y(0)=3, \quad y^{\prime}(t)=(y(t)-1) \times(y(t)-3) \times(y(t)-5)\) l. \(y(t)=5\) \(y(0)=5, \quad y^{\prime}(t)=(y(t)-1) \times(y(t)-3) \times(y(t)-5)\)

Step-by-Step Solution

Answer

Solution a satisfies both the initial condition and the differential equation.

Step 1: Verify Initial Condition a

Substitute \( t = 0 \) into \( y(t) = e^{2t} + e^t \). This gives \( y(0) = e^0 + e^0 = 1 + 1 = 2 \). The initial condition is satisfied.

Step 2: Verify Differential Equation a

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

Key Concepts

Initial Conditions

Initial conditions are a crucial part of solving differential equations. They are the known values of the solution at a specific point in time, typically at the beginning of the problem, i.e., when the independent variable (often time, denoted as \( t \)) is zero. By providing initial conditions, we can pinpoint the exact solution out of potentially infinite solutions to a differential equation.
Consider an example: with the function \( y(t) = e^{2t} + e^{t} \) and the initial condition \( y(0) = 2 \), we can check if this condition is met by substituting \( t = 0 \) into the function. As shown in the step-by-step solution, it calculates to \( e^0 + e^0 = 1 + 1 = 2 \). Since the calculated value matches the given initial condition, it confirms that the function satisfies \( y(0) = 2 \).

• Initial conditions help us to verify if we have the right particular solution for a differential equation.
• They allow us to solve for constants of integration that arise from solving a differential equation.

So, always remember to apply the initial conditions to confirm the correctness of your solution.

Verifying Solutions

Verifying a solution to a differential equation means confirming that the solution satisfies both the differential equation itself and the given initial conditions. Once a potential solution is found, this verification is necessary to ensure the solution is truly valid for the problem conditions.
For instance, after finding the function \( y(t) = e^{2t} + e^t \), we calculated its derivative, which is \( y'(t) = 2e^{2t} + e^t \). This is done because differential equations involve derivatives and they need to hold true when substituted back into the original differential equation.
In our example, substituting both \( y(t) \) and \( y'(t) \) into the equation \( y'(t) - y(t) = e^{2t} \), we get:
\[ 2e^{2t} + e^t - (e^{2t} + e^t) = e^{2t} \]
This simplifies correctly to \( e^{2t} = e^{2t} \), showing that the original differential equation is satisfied.

• Verification is a crucial step in ensuring the proposed solution satisfies all given conditions.
• An unverified solution could lead to inaccuracies and possible errors in the results.

It's an essential step for problem confirmation in differential equations.

First Order Differential Equations

First order differential equations are differential equations of the form \( y'(t) = f(t, y) \), where \( y'(t) \) is the first derivative of \( y(t) \) with respect to \( t \). These equations involve only the first derivative, making them the simplest type of differential equations.
They appear in many scientific fields, modeling processes such as heat conduction, population growth, and electrical circuits. To solve these equations, we often use methods like separation of variables, integrating factors, or exact equations.
For example, one of the problems in our original exercise involves the equation \( y'(t) - y(t) = e^{2t} \). It's a linear first order differential equation. Solving it involves finding a function \( y(t) \) whose derivative and its relation to \( y(t) \) result in the equation being balanced.

• They are crucial for modeling real-world phenomena that change at a rate proportional to a single factor.
• Solving these equations often serve as a foundation for tackling more complex higher-order differential equations.

Understanding first order differential equations is fundamental to mastering more advanced mathematical concepts.