Question

Find a solution to the initial value problem that is continuous on the given interval \([a, b]\). \(y^{\prime}+\frac{1}{t} y=g(t), \quad y(1)=1 ; \quad g(t)=\left\\{\begin{array}{ll}3 t, & 1 \leq t \leq 2 \\ 0, & 2

Short answer

Answer: The solution to the initial value problem is \(y(t) = \left\{\begin{array}{ll}t^{2}, & 1 \leq t \leq 2 \\ 8/t, & 2<t \leq 3\end{array}\right.\).

Step-by-step solution

Separate cases according to g(t)

We'll analyze the differential equation in two separate cases, according to the defined behavior of g(t). Case 1: \(1 \leq t \leq 2\) In this interval, g(t) is defined as \(g(t) = 3t\). Case 2: \(2 < t \leq 3\) In this interval, g(t) is defined as \(g(t) = 0\).

Find the general solutions of the differential equations for each case

For each case, we will find the general solution of the given differential equation. Case 1: \(y^{\prime}+\frac{1}{t} y=3t\) Use an integrating factor, which is the exponent of the integral of the coefficients of y. Integrating factor: \(e^{\int \frac{1}{t} dt} = e^{\ln(t)} = t\) Now, multiply the differential equation by the integrating factor. \(t y^{\prime}+y=3t^{2}\) Now, notice that the left-hand side of the equation is the derivative of the product of y and the integrating factor: \((ty)^{\prime} = 3t^{2}\). Integrate to find the function \(ty\). \(\int (ty)^{\prime} dt = \int 3t^{2} dt \Rightarrow ty = t^{3} + C_{1} \Rightarrow y(t) = t^{2} + \frac{C_{1}}{t}\) Case 2: \(y^{\prime}+\frac{1}{t} y=0\) Integrating factor: \(e^{\int \frac{1}{t} dt} = e^{\ln(t)} = t\) \(t y^{\prime}+y=0\) \((ty)^{\prime} = 0\) Integrate to find the function \(ty\). \(\int (ty)^{\prime} dt = \int 0 dt \Rightarrow ty = C_{2} \Rightarrow y(t) = \frac{C_{2}}{t}\)

Use the initial condition and continuity

Now, we'll use the initial condition \(y(1) = 1\) to find the specific solutions. For case 1: \(y(1) = 1^{2} + \frac{C_{1}}{1} \Rightarrow C_{1} = 0\). So, \(y(t) = t^{2}\) in the interval \(1 \le t \le 2\). For continuity at t = 2, we make sure the value of the function at t = 2 is the same for both cases. \(y(2) = 2^{2} = 4\) for the first case, so the second case must have the same value. \(y(2) = \frac{C_{2}}{2} \Rightarrow C_{2} = 8\). So, \(y(t) = \frac{8}{t}\) in the interval \(2 < t \le 3\). Thus, the solution to the problem is: \(y(t) = \left\{\begin{array}{ll}t^{2}, & 1 \leq t \leq 2 \\ 8/t, & 2<t \leq 3\end{array}\right.\)

Key concepts

Initial Value Problem

An initial value problem is a type of differential equation that comes with specific conditions at a given point. These conditions help determine a unique solution to the differential equation. In the exercise, we are given the differential equation \( y^{\prime}+\frac{1}{t} y=g(t) \) with the initial condition \( y(1)=1 \). The goal is to find a solution that satisfies this condition at \( t=1 \).

Initial value problems are pivotal because they provide the necessary constraints to pinpoint a particular solution among infinitely many possibilities. Without an initial condition, we might only determine a family of solutions, but not the specific one needed.

These problems often involve functions that describe physical phenomena, requiring the solution to adhere to real-world conditions.

Integrating Factor

The integrating factor is a crucial technique for solving linear first-order differential equations, such as the one posed in the exercise. It transforms the equation into a form where the left-hand side is easily integrable. This simplifies the process of finding solutions.

In our example, the integrating factor is calculated as follows:

• Start with the equation: \( y^{\prime}+\frac{1}{t} y = g(t) \).
• Find the integrating factor: \( \mu(t) = e^{\int \frac{1}{t} \, dt} = e^{\ln(t)} = t \).
• Multiply the entire differential equation by \( t \) to obtain a new equation that can be integrated directly: \( (ty)^{\prime} = tg(t) \).

This conversion allows us to simplify the solution-finding process, highlighting the power of the integrating factor as a mathematical tool.

Piecewise Function

A piecewise function is used to define functions with different expressions over various intervals. This concept is relevant in our exercise because the function \( g(t) \) is defined piecewise:

• For \( 1 \leq t \leq 2 \), \( g(t) = 3t \).
• For \( 2 < t \leq 3 \), \( g(t) = 0 \).

This kind of function is common in real-world applications where conditions or rules change across intervals.

Solving differential equations with piecewise functions involves solving them separately in each interval and then ensuring they align at the boundaries to achieve continuity. The challenge arises when attempting to ensure that the solutions are continuously connected across these intervals.

Continuity in Differential Equations

Continuity is a fundamental concept when solving differential equations, especially those involving piecewise functions. It requires that solutions not only exist over each interval but also match at the boundaries to avoid abrupt changes.

In the exercise, we ensured continuity by matching the values of the solution at \( t=2 \), where the behavior of \( g(t) \) changes. We found solutions for each piecewise segment:

• For \( 1 \leq t \leq 2 \), the solution was \( y(t) = t^{2} \).
• For \( 2 < t \leq 3 \), the solution needed to match at \( t=2 \). Thus, \( y(2) = \frac{C_{2}}{2} = 4 \) ensured \( C_{2} = 8 \), giving us \( y(t) = \frac{8}{t} \).

This careful matching highlights the importance of continuity and its role in creating accurate and realistic models in calculus and differential equations.