Question

Find the function \(F\) that satisfies the following differential equations and initial conditions. $$F^{\prime \prime}(x)=1 ; F^{\prime}(0)=3, F(0)=4$$

Short answer

Question: Given the second-order differential equation \(F^{\prime \prime}(x)=1\) with initial conditions \(F^{\prime}(0) = 3\) and \(F(0) = 4\), find the function F(x). Answer: The function F(x) satisfying the given differential equation and initial conditions is \(F(x) = \frac{1}{2}x^2 + 3x + 4\).

Step-by-step solution

Integrate the differential equation once

Since \(F^{\prime \prime}(x)=1\), we will integrate the equation once concerning x to find \(F^{\prime}(x)\). Here, suppose the integration yields a constant \(c_1\) : $$F^{\prime}(x) = \int 1 \, dx = x + c_1$$

Apply the initial condition for F'(0) and solve for c1

Using the initial condition of \(F^{\prime}(0) = 3\), we will substitute this value into our \(F^{\prime}(x)\) expression and solve for constant \(c_1\) : $$3 = 0 + c_1$$ So, we got \(c_1 = 3\), therefore our \(F^{\prime}(x)\) is now: $$F^{\prime}(x) = x + 3$$

Integrate F'(x) to find F(x)

Integrate \(F^{\prime}(x)\) once more to find our function F(x), suppose it adds another constant, \(c_2\) : $$F(x) = \int (x + 3) \, dx = \frac{1}{2}x^2 + 3x + c_2$$

Apply the initial condition for F(0) and solve for c2

Use the initial condition of \(F(0) = 4\) and put the value into the F(x) expression to solve for constant \(c_2\) : $$4 = \frac{1}{2}(0)^2 + 3(0) + c_2$$ So, we got \(c_2 = 4\). Now, we have our final equation for the function F(x): $$F(x) = \frac{1}{2}x^2 + 3x + 4$$

Key concepts

Integration of Functions

Understanding how to integrate functions is a foundational skill in calculus used to find the antiderivative or the area under a curve of a given function. Essentially, when we integrate a function, we are finding the function whose derivative is the given function. This process can be thought of as the inverse operation to taking a derivative. For instance, if we are given the second derivative of a function, such as in the exercise where we know that \(F''(x) = 1\), our goal is to work backwards to find the original function, \(F(x)\).

The method to carry out this operation generally involves calculating the indefinite integral, adding a constant of integration since the integration process itself can lead to multiple functions. These constants reflect the family of all possible antiderivatives. For example, when integrating 1 with respect to x, the resulting function is \(x + c\), where \(c\) is the integration constant.

Initial Value Problems

An initial value problem is a differential equation paired with a specific value, called an initial condition, which the solution to the equation must satisfy. These problems are fundamental in determining a unique solution to a differential equation. Without initial conditions, only a family of solutions can be found due to the integration constants involved.

In the context of our exercise, we are given the initial conditions \(F'(0) = 3\) and \(F(0) = 4\). These conditions enable us to find the specific values of the integration constants \(c_1\) and \(c_2\), respectively. By substituting these initial values into the integrated functions, we can solve for the constants and, consequently, determine the exact form of our particular solution, \(F(x)\).

Second Derivative

The second derivative of a function, denoted as \(F''(x)\), is the derivative of the derivative of that function. It provides information about the curvature of the function's graph and can tell us whether the function is concave up or concave down at a given point. It is used in various applications such as physics, engineering, and economics to describe acceleration, concavity, and inflection points.

In our exercise, we are given that \(F''(x) = 1\), which means the slope of the graph's tangent line is increasing at a constant rate, indicating the function is linear in its curvature. This information is the starting point to work our way back through integration to find the first derivative and eventually the original function, \(F(x)\).

Integrating Constant

When integrating a function, we always include an arbitrary constant, known as the integration constant. This constant arises because when we differentiate a constant, the result is zero. Therefore, when we integrate, there could be an 'unknown' constant that we had lost in the differentiation process.

In the step-by-step solution of our exercise, after each integration, we account for this constant: after the first integration, we added \(c_1\), and after the second integration, we added \(c_2\). These constants are crucial in solving initial value problems because they are what make the antiderivative we find satisfy the specific initial conditions given, leading to a unique solution of the differential equation.