Problem 42
Question
Find the function \(f(x)\) satisfying the given conditions. $$f^{\prime \prime}(x)=2 x, f^{\prime}(0)=-3, f(0)=2$$
Step-by-Step Solution
Verified Answer
The function that satisfies the given conditions is \(f(x) = \frac{x^{3}}{3} - 3x + 2\).
1Step 1: Integrate the second derivative
The second derivative of \(f(x)\) is given as \(f^{\prime \prime}(x)=2 x\). So, let's integrate \(2x\) with respect to \(x\) to find the first derivative \(f^{\prime}(x)\): \[ \int f^{\prime \prime}(x) dx = \int 2x dx = x^{2} + C \] where \(C\) is the constant of integration.
2Step 2: Apply the first initial condition
We are given \(f^{\prime}(0)=-3\). So, let's plug \(x = 0\) into our derived function \(f^{\prime}(x) = x^{2} + C\) to find \(C\). Doing so, we get that \[f^{\prime}(0) = 0 + C = -3 \Rightarrow C = -3\] . So, our first derivative \(f^{\prime}(x)\) becomes \(f^{\prime}(x) = x^{2} - 3\).
3Step 3: Integrate the first derivative
Now we need to integrate the derived first derivative \(f^{\prime}(x) = x^{2} - 3\) to find the function \(f(x)\). Doing so, we get: \[ \int f^{\prime}(x) dx = \int (x^{2} - 3) dx = \frac{x^{3}}{3} - 3x + D \] where \(D\) is the constant of integration.
4Step 4: Apply the second initial condition
We are given \(f(0)=2\). So, let's plug \(x = 0\) into the derived function \(f(x) = \frac{x^{3}}{3} - 3x + D\) to find \(D\). Doing so, we get that \[f(0) = 0 + D = 2 \Rightarrow D = 2\] . So, our final function \(f(x)\) becomes \(f(x) = \frac{x^{3}}{3} - 3x + 2\).
Key Concepts
Initial Value ProblemIntegration of PolynomialsConstants of Integration
Initial Value Problem
An initial value problem in calculus encapsulates the process of finding a specific function when its derivative, together with one or more points on the original function, is known. Such problems are fundamental in differential equations, where the primary goal is to reconstruct the unknown function that corresponds to the given rate of change.
In the case of the exercise provided, you're asked to determine the function \(f(x)\) that satisfies both the second derivative \(f^{\text{\prime\prime}}(x) = 2x\) and certain initial conditions. Initial conditions refer to the values of the function and/or its derivatives at a particular point, which for this problem are \(f^{\text{\prime}}(0) = -3\) and \(f(0) = 2\). The solution to this type of problem is unique; that is, there is only one function that will satisfy all given conditions.
In the case of the exercise provided, you're asked to determine the function \(f(x)\) that satisfies both the second derivative \(f^{\text{\prime\prime}}(x) = 2x\) and certain initial conditions. Initial conditions refer to the values of the function and/or its derivatives at a particular point, which for this problem are \(f^{\text{\prime}}(0) = -3\) and \(f(0) = 2\). The solution to this type of problem is unique; that is, there is only one function that will satisfy all given conditions.
Integration of Polynomials
Integration is the reverse process of differentiation. When integrating polynomials, each term of the polynomial is integrated individually and the power of the variable is increased by one. The new exponent is then divided by this increased number.
In our exercise, the integration of the polynomial \(2x\) needs to be carried out to find the first derivative of the function. The integration of \(2x\) with respect to \(x\) results in \(x^{2}\), as the power of \(x\) increases from 1 to 2. Following the rule, we divide by the new exponent, but since it's simply \(x^2\), the division is implied to have a coefficient of 1. This step demonstrates the importance of understanding polynomial integration when solving higher-order differential equations or reconstructing functions from their rates of change.
In our exercise, the integration of the polynomial \(2x\) needs to be carried out to find the first derivative of the function. The integration of \(2x\) with respect to \(x\) results in \(x^{2}\), as the power of \(x\) increases from 1 to 2. Following the rule, we divide by the new exponent, but since it's simply \(x^2\), the division is implied to have a coefficient of 1. This step demonstrates the importance of understanding polynomial integration when solving higher-order differential equations or reconstructing functions from their rates of change.
Constants of Integration
When integrating a function, we include a constant of integration to account for the indefinite nature of the antiderivative. Essentially, there are infinitely many antiderivatives for a given function, each differing by a constant value. The constant of integration represents any and all vertical shifts that the antiderivative can have.
In our step-by-step solution, we addressed the constants of integration by applying the initial conditions. When we first integrated to find the first derivative \(f^{\text{\prime}}(x)\), we added the constant \(C\). Then, we used the condition \(f^{\text{\prime}}(0) = -3\) to solve for \(C\). A similar process was used to find the constant \(D\) when integrating the first derivative to find the function \(f(x)\). By setting \(x = 0\) in \(f(x) = \frac{x^{3}}{3} - 3x + D\) and using the initial condition \(f(0) = 2\), we determined that \(D = 2\). These steps ensured that the final function \(f(x)\) would satisfy both the differential equation and the given initial conditions.
In our step-by-step solution, we addressed the constants of integration by applying the initial conditions. When we first integrated to find the first derivative \(f^{\text{\prime}}(x)\), we added the constant \(C\). Then, we used the condition \(f^{\text{\prime}}(0) = -3\) to solve for \(C\). A similar process was used to find the constant \(D\) when integrating the first derivative to find the function \(f(x)\). By setting \(x = 0\) in \(f(x) = \frac{x^{3}}{3} - 3x + D\) and using the initial condition \(f(0) = 2\), we determined that \(D = 2\). These steps ensured that the final function \(f(x)\) would satisfy both the differential equation and the given initial conditions.
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