Question

Solve the initial value problem $$ y^{\prime \prime}=f(t), \quad y(0)=0, \quad y^{\prime}(0)=0 $$ where $$ f(t)=m+1, \quad m \leq t

Short answer

Question: Determine the solution to the initial value problem of the second-order differential equation \(y^{\prime \prime}=f(t)\), with initial conditions \(y(0)=0\) and \(y^{\prime}(0)=0\), where \(f(t)=m+1\) for \(m \leq t<m+1\) with \(m=0,1,2,\ldots\). Answer: The solution to this initial value problem for the given range of t is \(y(t) = \frac{1}{2}(m+1)t^2\) for \(m \leq t<m+1\) with \(m=0,1,2,\ldots\).

Step-by-step solution

Integrate the function f(t) for each interval

Notice that \(f(t)=m+1\) for \(m \leq t<m+1\) where \(m=0,1,2, \ldots\). This means we will have different equations for different intervals of t. Integrate \(f(t)\) with respect to t to get \(y^{\prime}(t)\): $$ y^{\prime}(t) = \int f(t) \, dt = \int (m+1) \, dt = (m+1)t + C_1, $$ where \(C_1\) is a constant of integration.

Integrate y'(t) to obtain y(t)

Integrate \(y^{\prime}(t)\), as found in Step 1, to find \(y(t)\): $$ y(t) = \int y^{\prime}(t) \, dt = \int[(m+1)t + C_1] \, dt = \frac{1}{2}(m+1)t^2 + C_1t + C_2. $$ Here, \(C_2\) is another constant of integration.

Apply the initial conditions

We have the initial conditions \(y(0)=0\) and \(y^{\prime}(0)=0\). Apply them to the solution we found in Steps 1 and 2. For \(y(0)=0\): $$ 0 = \frac{1}{2}(m+1)(0)^2 + C_1(0) + C_2 \quad \Rightarrow C_2 = 0. $$ For \(y^{\prime}(0)=0\): $$ 0 = (m+1)(0) + C_1 \quad \Rightarrow C_1 = 0. $$

Write the final solution

Substituting \(C_1=0\) and \(C_2=0\) back into the expression for \(y(t)\), we get the final solution: $$ y(t) = \frac{1}{2}(m+1)t^2 $$ for \(m \leq t<m+1\) with \(m=0,1,2,\ldots\). This is the solution to the given initial value problem.

Key concepts

Differential Equations

Differential equations are a type of equation that involves unknown functions and their derivatives. These equations are fundamental in describing various physical phenomena such as motion, heat, and sound. They allow us to understand how changes in one variable can affect another. In the exercise, we tackle a differential equation where the second derivative of a function, noted as \(y''\), is given by another function, \(f(t)\). The process of solving such equations usually involves finding the original function \(y(t)\) that satisfies both the equation and any initial or boundary conditions specified. This involves integrating the function step-by-step to backtrack from derivatives to the original function.

Integration

Integration is the reverse process of differentiation and is used to find the original function when its derivative is known. In the given problem, integration is used twice. First, we integrate \(f(t)=m+1\) to find the first derivative of \(y(t)\), denoted as \(y'(t)\). The result is \(y'(t) = (m+1)t + C_1\), where \(C_1\) is the constant of integration obtained.
Next, the process is repeated for \(y'(t)\) to get \(y(t)\). The result, \(y(t) = \frac{1}{2}(m+1)t^2 + C_1t + C_2\), includes another constant of integration, \(C_2\). These constants are crucial as they adjust the function to align with given initial conditions.

Boundary Conditions

Boundary conditions are given at the start or end values of the problem and provide the necessary information to solve for the constants of integration. In this exercise, the boundary conditions given are \(y(0)=0\) and \(y'(0)=0\).
Applying these conditions helps to calculate the values of \(C_1\) and \(C_2\). For \(y(0)=0\), substituting \(t = 0\) into \(y(t)\) gives \(C_2 = 0\). Similarly, applying \(y'(0)=0\) gives \(C_1 = 0\). Without these conditions, the integration process would yield multiple solutions; however, they ensure we find the specific solution that meets the initial setup of the problem.

Constant of Integration

The constant of integration arises whenever you perform indefinite integration. It represents an unknown constant that must be determined using additional information, typically initial or boundary conditions. In this problem, during the integration of \(f(t)\) and subsequently \(y'(t)\), \(C_1\) and \(C_2\) emerge as constants of integration.

• \(C_1\) appears from the integration of the differential equation to find \(y'(t)\).
• \(C_2\) is added upon integrating \(y'(t)\) to find \(y(t)\).

Using the given initial conditions, we set both \(C_1\) and \(C_2\) to zero. This adjustment tunes the general solution provided by integration to the particular solution required by the problem's conditions.