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Chapter 4: Problem 42

Finding a Particular Solution In Exercises \(37-44,\) find the particular solution of the differential equation that satisfies the initial condition(s). \(f^{\prime \prime}(x)=3 x^{2}, f^{\prime}(-1)=-2, f(2)=3\)

Short Answer

Expert verified
The particular solution of the given differential equation that satisfies the initial conditions is \(f(x)=\frac{1}{4}x^{4} - x + 1\).

Step by step solution

01

Integration

Begin by integrating the differential equation \(f^{\prime \prime}(x)=3x^{2}\) with respect to \(x\).\nAfter integrating, we will get \(f'(x)=x^{3} + C_{1}\), where \(C_{1}\) is the constant of integration.
02

Apply the first initial condition

Now, apply the first initial condition \(f^{\prime}(-1)=-2\) to find the value of \(C_{1}\).\nWhen substituting \(f'(-1) = -2\) into the equation \(f'(x)=x^{3} + C_{1}\), we get -2 = -1 + \(C_{1}\). Solving for \(C_{1}\), we find that \(C_{1}=-1\).
03

Second Integration

Now integrate the result from step 1 again, now with the found constant \(C_{1}\). So, \(f(x)=\frac{1}{4}x^{4} - x + C_{2}\), where \(C_{2}\) is another integration constant.
04

Apply the second initial condition

Now, apply the second initial condition \(f(2)=3\) to find the value of \(C_{2}\).\nSubstituting \(f(2) = 3\) into \(f(x)=\frac{1}{4}x^{4} - x + C_{2}\), we get 3 = 4-2+\(C_{2}\).\nSolving for \(C_{2}\), we find that \(C_{2}=1\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integrating Differential Equations
Integrating a differential equation is a fundamental technique used to find a function when its derivative is given. A differential equation represents a relationship between a function and its derivatives. To solve such an equation, we integrate the relationship to find a more general form of the function.

When asked to integrate a differential equation like the one in our exercise, \( f''(x)=3x^2 \), we integrate step by step. With each integration, we sum up the area under the curve of the rate of change, moving us from acceleration (in this case, \( f''(x) \) which is akin to acceleration in physics) to velocity (\( f'(x) \)) and, with a second integration, to position (\( f(x) \)).

During the first integration, we find the velocity function \( f'(x) \) from the acceleration function \( f''(x) \) and introduce a constant of integration, \( C_1 \), because indefinite integration opens up the possibility to add any constant and still satisfy the differential equation. Think of it as finding all possible functions whose rate of change for a particular interval would result in \( f''(x) \).
Initial Value Problem
An initial value problem involves not only integrating a differential equation but also determining the specific solution that satisfies given initial conditions. In other words, we're finding a particular solution to the equation that passes through a point or set of points.

In our exercise, solving the initial value problem is necessary after the integration steps to identify the precise function \( f(x) \). This is achieved by using the initial values such as \( f'(鈭�1)=鈭�2 \) and \( f(2)=3 \). These initial values represent constraints that the function must satisfy, and when we substitute these specific points, we can solve for the constants of integration we've previously added. This step is crucial because it's what differentiates our particular solution from the general solution which includes the constants. The provided initial values act as 'anchors' that pin down the function in the coordinate space.
Constant of Integration
The constant of integration is introduced when we perform indefinite integration on a differential equation. It is symbolized as \( C \) and can be any real number. Since the derivative of a constant is zero, adding this constant does not affect the validity of the antiderivative.

In the context of our exercise, when we integrate \( f''(x) \) to find \( f'(x) \) and again to find \( f(x) \) we introduce \( C_1 \) and \( C_2 \) respectively. These constants are placeholders until we apply initial conditions that enable us to solve for their specific values. The constants are necessary because without them, we would be missing a family of potential solutions. They represent the infinite number of vertical shifts that our function could undergo and still have the same rate of change with respect to \( x \). The initial conditions in an initial value problem serve to find the exact values of these constants for the particular solution, making the solution unique.

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Most popular questions from this chapter

Modeling Data An experimental vehicle is tested on a straight track. It starts from rest, and its velocity v (in meters per second) is recorded every 10 seconds for 1 minute (see table). $$\begin{array}{|c|c|c|c|c|c|c|}\hline t & {0} & {10} & {20} & {30} & {40} & {50} & {60} \\ \hline v & {0} & {5} & {21} & {40} & {62} & {78} & {83} \\\ \hline\end{array}$$ \begin{equation} \begin{array}{l}{\text { (a) Use a graphing utility to find a model of the form }} \\ {v=a t^{3}+b t^{2}+c t+d \text { for the data. }} \\ {\text { (b) Use a graphing utility to plot the data and graph the model. }} \\ {\text { (c) Approximate the distance traveled by the vehicle during }} \\ {\text { the test. }}\end{array} \end{equation}

Even and Odd Functions In Exercises 79 and 80, write the integral as the sum of the integral of an odd function and the integral of an even function. Use this simplification to evaluate the integral. $$\int_{-3}^{3}\left(x^{3}+4 x^{2}-3 x-6\right) d x$$

Even and Odd Functions In Exercises 73-76, evaluate the integral using the properties of even and odd functions as an aid. $$\int_{-2}^{2} x\left(x^{2}+1\right)^{3} d x$$

True or False? In Exercises \(63-68\) , determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. $$\int_{a}^{b}[f(x)+g(x)] d x=\int_{a}^{b} f(x) d x+\int_{a}^{b} g(x) d x$$

True or False? In Exercises \(63-68\) , determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. $$\int_{a}^{b} f(x) g(x) d x=\left[\int_{a}^{b} f(x) d x\right]\left[\int_{a}^{b} g(x) d x\right]$$
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