Question

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

Short answer

Answer: The function F(x) is given by \(F(x) = -\cos x + 3x + (3-3\pi)\).

Step-by-step solution

Integration of the differential equation (First Integration)

First, we need to find the antiderivative of \(F^{\prime \prime}(x)=\cos x\). To do this, we will integrate \(\cos x\) with respect to \(x\): $$\int \cos x \, dx = \sin x + C_1$$ Now, we have that: $$F^{\prime}(x)=\sin x + C_1$$

Integration of the differential equation (Second Integration)

Next, we need to find the antiderivative of the previously found first derivative, \(F^{\prime}(x)=\sin x + C_1\). Integrate again with respect to dx: $$\int (\sin x + C_1) \, dx = -\cos x + C_1x + C_2$$ Now, we have that: $$F(x)=-\cos x + C_1x + C_2$$

Apply the initial conditions

We are given two initial conditions: \(F^{\prime}(0) = 3\) and \(F(\pi) = 4\). We will apply these initial conditions to find \(C_1\) and \(C_2\). For \(F^{\prime}(0) = 3\), substitute \(x=0\) into the first derivative of F: $$3 = \sin(0) + C_1$$ Since \(\sin(0) = 0\), we get \(C_1 = 3\). For \(F(\pi) = 4\), substitute \(x=\pi\) into F: $$4 = -\cos(\pi) + 3\pi + C_2$$ Since \(\cos(\pi) = -1\), we get \(C_2 = 4-3\pi -1 = 3-3\pi\).

Write the final function

Finally, substitute the values of \(C_1\) and \(C_2\) back into the function \(F(x)\): $$F(x) = -\cos x + 3x + (3-3\pi)$$ This is the function that satisfies the given differential equation and initial conditions.

Key concepts

Initial Conditions

In differential equations, initial conditions help specify unique solutions. Imagine initial conditions as instructions that shape how a curve should behave at certain points. For example, when we know that \(F'(0) = 3\) and \(F(\pi) = 4\), we can adjust constants in our function to fulfill these requirements.
Without initial conditions, there could be countless possibilities for our function.

• Your initial conditions lock in specific paths for the function, ensuring a single solution.
• These conditions guide us in determining unknown constants (like \(C_1\) and \(C_2\) from the solution).

To sum up, initial conditions are essential for pinning down definite answers from indefinite functions.

Antiderivatives

Antiderivatives are the opposite of derivatives. They help us retrace our steps from a derivative to the original function. Consider them as a way to decode the changes captured by derivatives back into the function itself.
A single function can have multiple antiderivatives, differing by a constant term. This is why we often add a constant, like \(C\), when finding an antiderivative:

• The antiderivative of \(\cos x\) is \(\sin x + C_1\).
• The antiderivative of \(\sin x + C_1\) is \(-\cos x + C_1 x + C_2\).

These constants are crucial because they adjust the base function to fit specific initial conditions. So, antiderivatives are important for solving differential equations, letting us transform rates of change back into standard functions.

Integration

Integration is the process of finding an antiderivative. Think of it as the reversing of differentiation. When we integrate, we are essentially gathering up small pieces into a whole.
In the given problem, integration is performed twice: once on the second derivative \(F''(x) = \cos x\), and then again on the first derivative \(F'(x) = \sin x + C_1\).

• The first integration turns \(F''(x) = \cos x\) into \(F'(x) = \sin x + C_1\).
• The second integration further breaks down \(F'(x)\) into \(F(x) = -\cos x + C_1x + C_2\).

Both steps involve adding constants of integration \(C_1\) and \(C_2\), representing the indefinite nature of antiderivatives before incorporating initial conditions.

Second Derivative

The second derivative \(F''(x)\) represents the derivative of \(F'(x)\), capturing the rate of change of the rate of change. It serves as a fantastic tool for understanding acceleration or concavity of functions in practical scenarios.
In this exercise, \(F''(x) = \cos x\) gives us key insights:

• The function's second derivative shows periodic behavior since \(\cos x\) varies between -1 and 1.
• Integrating twice brings us from this acceleration view to the original function \(F(x)\).

By systematically integrating the second derivative and applying given initial conditions, we unravel the original function while gaining insights into its motion or curvature.