Problem 16 - Calculus: Early Transcendentals

Question

A function \(y(t)\) is a solution of \(y^{\prime}+t^{k} y=0 .\) Suppose that \(y(0)=1\) and \(y(1)=e^{-13}\). Find \(k\) and find \(y(t)\).

Step-by-Step Solution

Answer

The value of \( k \) is 12, and the function \( y(t) = e^{-\frac{t^{13}}{13}} \).

Step 1: Recognize the Differential Equation Type

The given differential equation is \( y^{\prime} + t^k y = 0 \), which is a first-order linear homogeneous differential equation. This type can often be solved through an integrating factor approach.

Step 2: Identify the Integrating Factor

To solve \( y^{\prime} + t^k y = 0 \) using the integrating factor method, we need to determine the integrating factor \( \mu(t) \) that satisfies \( \mu(t) = e^{\int t^k \ dt} \). Calculate this integral.

Step 3: Solve the Integrating Factor

The integrating factor is given by \( \mu(t) = e^{\int t^k \ dt} = e^{\frac{t^{k+1}}{k+1}} \). We will use this factor to solve the equation.

Step 4: Solve for General Solution

The general solution of the differential equation is given by \( y(t) = \frac{C}{e^{\frac{t^{k+1}}{k+1}}} \) due to the nature of exponential integrals and their solutions. We arrived at this form by multiplying the entire differential equation by \( \mu(t) \), transforming the left side into the derivative of \( \mu(t) y(t) \), then integrating.

Step 5: Apply Initial Conditions

Use the initial condition \( y(0) = 1 \) to find the constant \( C \). When \( t = 0 \), \( y(0) = \frac{C}{e^0} = C \). This implies that \( C = 1 \). Our solution now is \( y(t) = \frac{1}{e^{\frac{t^{k+1}}{k+1}}} \).

Step 6: Apply Second Condition to Find \( k \)

Use the second condition \( y(1) = e^{-13} \) to find \( k \). Substitute \( t = 1 \) into \( y(1) = \frac{1}{e^{\frac{1^{k+1}}{k+1}}} = e^{-13} \), which simplifies to \( \frac{1}{e^{\frac{1}{k+1}}} = e^{-13} \). Equate exponents to obtain \( \frac{1}{k+1} = 13 \), leading to \( k+1 = \frac{1}{13} \), hence \( k = 12 \).

Step 7: Final Solution for \( y(t) \)

Substitute \( k = 12 \) into the general solution \( y(t) = \frac{1}{e^{\frac{t^{12+1}}{12+1}}} = \frac{1}{e^{\frac{t^{13}}{13}}} \). Thus, the final solution is \( y(t) = e^{-\frac{t^{13}}{13}} \).

Key Concepts

Integrating Factor Method

The integrating factor method is a powerful technique to solve first-order linear differential equations. These equations typically appear in the form \( y' + P(t)y = Q(t) \). The key idea is to multiply the entire equation by a function, called the integrating factor \( \mu(t) \), that makes the equation easier to solve.

The integrating factor \( \mu(t) \) is determined by the expression \( \mu(t) = e^{\int P(t) \, dt} \). This method transforms the left side of the differential equation to the derivative of \( \mu(t)y(t) \). Hence, the equation becomes easier to integrate.

In the given exercise, the differential equation \( y' + t^k y = 0 \) uses \( P(t) = t^k \). Thus, the integrating factor is computed as \( \mu(t) = e^{\int t^k \, dt} = e^{\frac{t^{k+1}}{k+1}} \). After multiplying the original equation by this factor, it becomes possible to integrate and find the general solution.

Initial Value Problems

Initial value problems (IVPs) are a type of differential equation problem where the solution is needed to satisfy specific conditions at a certain point. Typically, these conditions are given at \( t = t_0 \) with a function value \( y(t_0) = y_0 \).

In the current example, the initial conditions include \( y(0) = 1 \) and \( y(1) = e^{-13} \). These conditions help in determining the arbitrary constants that appear once we solve the differential equation.

By applying \( y(0) = 1 \) in the general solution, we determined the constant \( C = 1 \) as shown. The second condition \( y(1) = e^{-13} \) provides a way to solve for the unknown value \( k \), as it allowed us to equate exponentials and solve for the needed constant \( k = 12 \). Initial value problems ensure the solution fits within specified parameters, greatly helping to narrow down the correct, unique solution.

Solutions to Differential Equations

Solving differential equations, especially linear ones, often involves finding a general formula that applies to all possible solutions. This solution typically contains arbitrary constants that are adjusted based on initial or boundary conditions.

For the equation in this exercise \( y' + t^k y = 0 \), the general solution was determined to be \( y(t) = \frac{C}{e^{\frac{t^{k+1}}{k+1}}} \). Solving differential equations often involves steps like recognizing the equation form, finding necessary factors or transformations, and finally applying conditions to find specific solutions.

The process involves systematic methods such as using integrating factors, separation of variables, or exact equations. Once the general solution is obtained, applying boundary or initial conditions allows refining the solution to fit specific problem requirements, thus ensuring the solution is applicable to the real-world scenario or context given in the problem.