Problem 3 - Calculus for the Life Sciences: A Modeling Approach Volume II

Question

The special case of \(y^{\prime}=f(t, y)\) in which \(f(t, y)=F(t)(f\) is independent of \(y)\) has a familiar solution from the Fundamental Theorem of Calculus I. Check by substitution that $$y(t)=y_{a}+\int_{a}^{t} F(x) d x \quad \text { solves } \quad y(a)=y_{a} \quad \text { and } \quad y^{\prime}(t)=F(t)$$ The differential equation $$y(a)=y_{a} \quad y^{\prime}(t)=F(t)$$ has therefore been completely solved. Henceforth we will consider that \(f\) is dependent on \(y\) and possibly also on \(t\).

Step-by-Step Solution

Answer

The provided integral solves the differential equation and the initial condition is met.

Step 1: Identify the Differential Equation

We are given the initial condition \( y(a) = y_a \) and the differential equation \( y'(t) = F(t) \). This is a first-order differential equation with respect to \( t \).

Step 2: Check Solution by Substitution

We are asked to verify that the solution \( y(t) = y_a + \int_a^t F(x) \, dx \) satisfies these conditions. First, differentiate this expression with respect to \( t \).

Step 3: Differentiate the Integral

We calculate the derivative of \( y(t) = y_a + \int_a^t F(x) \, dx \) using the Fundamental Theorem of Calculus, which gives \( y'(t) = F(t) \). This shows that our proposed solution indeed satisfies the differential equation \( y'(t) = F(t) \).

Step 4: Verify the Initial Condition

Substitute \( t = a \) into the expression for \( y(t) \). We have \( y(a) = y_a + \int_a^a F(x) \ dx = y_a + 0 = y_a \), satisfying the initial condition \( y(a) = y_a \).

Step 5: Conclusion

Since both the differential equation \( y'(t) = F(t) \) and the initial condition \( y(a) = y_a \) are satisfied by \( y(t) = y_a + \int_a^t F(x) \, dx \), this is indeed a solution to the problem.

Key Concepts

initial condition

In the realm of differential equations, an initial condition is a prerequisite value that helps define a specific solution to a differential equation among many potential ones. Consider an equation like \( y'(t) = F(t) \). Without an initial condition, this equation might have numerous solutions depending on the behavior of \( F(t) \). Therefore, the initial condition \( y(a) = y_a \) helps pin down one precise solution.
The initial condition tells us the state of the function at a particular point, usually at the beginning or sometimes at a significant point in time. When solving first-order differential equations, this condition is crucial for ensuring the function meets specific criteria.

• It provides a starting point for solving the equation.
• It guarantees that the solution aligns with the physical, biological, or other real-world scenarios represented by the equation.
• Without it, the solution could be arbitrary, lacking context or specificity.

By incorporating the initial condition, the solution satisfies the differential equation across its domain while honoring that initial state precisely.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus plays a pivotal role in linking differentiation and integration, two cornerstone concepts in calculus. There are two parts to this theorem. Here, we are particularly interested in the first part, which facilitates the solution of our differential equation.
This theorem asserts that if a function \( F \) is continuous over an interval, then if \( f \) is an antiderivative of \( F \) over that interval, the following holds:\[ \int_a^t F(x) \, dx = f(t) - f(a) \]This means that the derivative of an integral of \( F(x) \) leads back to the function \( F \). In the context of solving a differential equation like \( y'(t) = F(t) \), it allows us to express the solution in terms of integrals.
It confirms that the proposed solution \( y(t) = y_a + \int_a^t F(x) \, dx \) satisfies our differential equation by ensuring the derivative \( y'(t) \) calculates simply as \( F(t) \). Instead of solving first-order differential equations directly, this theorem provides a shortcut by facilitating integral computation as a means to derive solutions.

integral solutions

Integral solutions offer a method to solve differential equations by finding a function whose derivative matches a given function. The concept of integral solutions is crucial when the solution to the differential equation can be expressed as an integral.
For \( y'(t) = F(t) \), we're interested in finding \( y(t) \). Here, the integral solution becomes: \[ y(t) = y_a + \int_a^t F(x) \, dx \]This implies that to solve the problem, we calculate the definite integral of \( F(x) \) from \( a \) to \( t \), and adjust it by the initial condition \( y_a \).
Integral solutions are particularly powerful because:

• They convert a differential equation into a more manageable integral form.
• They allow inclusion of an initial condition naturally through integral limits.
• They adhere to the Fundamental Theorem of Calculus, confirming reversibility from integration back to the original derivative.

Thus, integral solutions provide an efficient means to articulate the solution to a variety of differential equations, especially when looking for explicit functions representing real-world situations.