Question

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

Short answer

Question: Find the function \(F(x)\) that satisfies the third-order differential equation \(F^{\prime \prime \prime}(x) = 4x\) and the initial conditions \(F^{\prime \prime}(0) = 0, F^{\prime}(0) = 1, F(0) = 3\). Answer: \(F(x) = \frac{1}{2}x^4 + x + 3\)

Step-by-step solution

Integrate the given differential equation once.

The given differential equation is \(F^{\prime \prime \prime}(x) = 4x\). To find \(F^{\prime \prime}(x)\), we can integrate both sides with respect to x: $$F^{\prime\prime}(x) = \int 4x dx = 2x^2 + C_1$$

Apply the initial condition for \(F^{\prime\prime}(x)\)

We know that \(F^{\prime \prime}(0) = 0\). Therefore, we can substitute this information into our equation for \(F^{\prime\prime}(x)\) and solve for \(C_1\): $$0 = 2(0)^2 + C_1 \Rightarrow C_1 = 0$$ Now, we have: $$F^{\prime\prime}(x) = 2x^2$$

Integrate the equation obtained in Step 2 once.

Now, we integrate \(F^{\prime\prime}(x)\) with respect to x to find the expression for \(F'(x)\): $$F^{\prime}(x) = \int 2x^2 dx = \frac{2}{3}x^3 + C_2$$

Apply the initial condition for \(F^{\prime}(x)\)

We know that \(F^{\prime}(0) = 1\). Therefore, we can substitute this information into our equation for \(F^{\prime}(x)\) and solve for \(C_2\): $$1 = \frac{2}{3}(0)^3 + C_2 \Rightarrow C_2 = 1$$ Now we have: $$F^{\prime}(x) = \frac{2}{3}x^3 + 1$$

Integrate the equation obtained in Step 4 once.

Finally, we integrate \(F^{\prime}(x)\) with respect to x to find the expression for \(F(x)\): $$F(x) = \int\left(\frac{2}{3}x^3 + 1\right) dx = \frac{1}{2}x^4 + x + C_3$$

Apply the initial condition for \(F(x)\)

We know that \(F(0) = 3\). Therefore, we can substitute this information into our equation for \(F(x)\) and solve for \(C_3\): $$3 = \frac{1}{2}(0)^4 + (0) + C_3 \Rightarrow C_3 = 3$$ Now we have our final solution for \(F(x)\): $$F(x) = \frac{1}{2}x^4 + x + 3$$

Key concepts

Initial Value Problems

An initial value problem in calculus is a type of differential equation along with specified values, known as initial conditions, which the solution must satisfy. In the context of our exercise, the differential equation is given as \(F^{\prime \prime \prime}(x)=4x\) with initial conditions \(F^{\prime \prime}(0)=0\), \(F^{\prime}(0)=1\), and \(F(0)=3\). These conditions are essential because without them, the differential equation would have an infinite number of solutions. However, the initial conditions allow us to find a unique solution that exactly fits the criteria established at the outset.

Integration of Functions

Integration is the process of finding the antiderivative, or the function F that represents the accumulated area under the curve of a given function, f. When we integrate a function, we're essentially finding the total quantity if the rate of change is known. For our function, given its third derivative \(F^{\prime \prime \prime}(x)=4x\), we perform successive integrations to walk back the layers of derivatives until we reach the original function F(x).

This sequential integration helps us construct the function from its derivatives, thus building a solution step-by-step for the initial value problem.

Calculus Initial Conditions

The initial conditions of a differential equation are specific values assigned at a certain point, often denoted as a starting point like \(x=0\). These initial conditions are used to determine the constants of integration that result from the indefinite integration process. In the exercise we're examining, the initial conditions give us the information needed to solve for \(C_1\), \(C_2\), and \(C_3\).

Applying each initial condition appropriately ensures that the function and its derivatives line up with the physical or geometric situation described by the problem. The initial conditions act as constraints that tailor the general formula into a specific one.

Antiderivatives

An antiderivative of a function f(x) is a function with a derivative that equates to f(x). It's essentially the reverse operation of differentiation. The concept of antiderivatives is at the core of solving a problem like ours where we integrate to determine the function F from its higher derivatives.

When integrating to find an antiderivative, we often encounter a '+ C', where C represents the constant of integration. Every antiderivative includes this constant because differentiation wipes out constants, leaving an infinite number of possible antiderivatives. The initial conditions provided in a problem help to find the correct constant and thus the specific solution required.