Problem 25

Question

Determine all functions $f$ satisfying the given conditions. $$ f^{\prime \prime}(x)=\sin x, f^{\prime}(\pi)=-2, f(0)=4 $$

Step-by-Step Solution

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Answer

The function is $f(x) = -\sin x - 3x + 4$.

1Step 1: Solve the Second Derivative

We start by solving the second derivative equation. We know that $f''(x) = \sin x$. To find $f'(x)$, we integrate $f''(x)$ with respect to $x$. The indefinite integral of $\sin x$ is $-\cos x$. Thus, $f'(x) = -\cos x + C_1$, where $C_1$ is an integration constant.

2Step 2: Apply the First Derivative Condition

Next, we use the given condition $f'(\pi) = -2$. From Step 1, we have the expression $f'(x) = -\cos x + C_1$. Plugging in $x = \pi$, we get $f'(\pi) = -\cos \pi + C_1 = -(-1) + C_1 = 1 + C_1 = -2$. Solving for $C_1$, we find $C_1 = -3$. Thus, $f'(x) = -\cos x - 3$.

3Step 3: Solve for the Original Function

Now, we integrate $f'(x) = -\cos x - 3$ with respect to $x$ to find $f(x)$. The integral of $-\cos x$ is $-\sin x$ and the integral of $-3$ is $-3x$. Thus, $f(x) = -\sin x - 3x + C_2$, where $C_2$ is another integration constant.

4Step 4: Apply the Function Condition

We use the initial condition $f(0) = 4$ to find $C_2$. From Step 3, we have $f(x) = -\sin x - 3x + C_2$. Plugging in $x = 0$, we get $f(0) = -\sin 0 - 3(0) + C_2 = 0 + C_2 = 4$. Thus, $C_2 = 4$. Hence, the original function is $f(x) = -\sin x - 3x + 4$.

Key Concepts

Integration TechniquesInitial ConditionsIntegration Constants

Integration Techniques

Integration techniques are crucial when solving differential equations, as they allow us to find a function from its derivative. In our problem, we start with a second derivative, and our goal is to work our way back to the original function. This is achieved through integration.

Here are some key points about the integration process:

The second derivative given is $f''(x) = \sin x$. To find the first derivative $f'(x)$, we integrate $\sin x$ with respect to $x$. The integral of $\sin x$ is $-\cos x$.
After integrating $\sin x$, we introduce an integration constant, resulting in $f'(x) = -\cos x + C_1$.
Next, to find the original function $f(x)$, we need to integrate $f'(x) = -\cos x - 3$, leading to $f(x) = -\sin x - 3x + C_2$.

Integration not only helps in finding the primitive function but also involves understanding the role of integration constants, which lead us to the complete general solution.

Initial Conditions

Initial conditions are specific values that help to determine the integration constants in a differential equation solution. They are given parts of the problem that indicate the behavior or value of the function or its derivatives at specific points.

In the given exercise, there are two initial conditions:

$f'(\pi) = -2$
$f(0) = 4$

We use these conditions to find the specific values of the integration constants $C_1$ and $C_2$. By substituting $x = \pi$ into the expression for $f'(x)$, we solve for $C_1$, ensuring the solution satisfies $f'(\pi) = -2$. Similarly, substituting $x = 0$ into the expression for $f(x)$ allows us to find $C_2$, ensuring $f(0) = 4$.

These conditions uniquely define the particular solution from the infinite possibilities offered by the general solution.

Integration Constants

Integration constants are arbitrary constants that arise when taking the indefinite integral of a function. They are necessary because integration is the inverse operation of differentiation, and differentiating a constant results in zero. Hence, there is an infinite number of possible antiderivatives for any given function.

In this problem, as we integrate twice, we get two integration constants, $C_1$ and $C_2$. Initially, when integrating the second derivative $f''(x) = \sin x$, we introduce $C_1$, resulting in the expression $f'(x) = -\cos x + C_1$.

Then, as we integrate this first derivative to find $f(x)$, we introduce another constant $C_2$, giving the expression $f(x) = -\sin x - 3x + C_2$. These constants $C_1$ and $C_2$ are later determined using the initial conditions provided, which gives us a complete, particular solution rather than a general one.

$C_1$ adjusts the slope of $f'(x)$
$C_2$ influences the vertical shift of $f(x)$

Recognizing the role of integration constants is essential in accurately describing the behavior of functions derived from differential equations.

Problem 25

Problem 26

Other exercises in this chapter

Problem 25

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Problem 25

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Problem 26

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Problem 26

Find all inflection points (if any) of the graph of the function. Then sketch the graph of the function. $$ g(x)=x^{4}-2 x^{3} $$

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