Problem 58
Question
Find \(f(x)\) by solving the initial value problem. $$f^{\prime}(x)=1+e^{x}+\frac{1}{x} ; f(1)=3+e$$
Step-by-Step Solution
Verified Answer
The final solution for the initial value problem is \(f(x) = x + e^x + ln |x| + 2\).
1Step 1: Rewrite the given expression
To begin, rewrite the given expression as \(f'(x) = 1 + e^x + \frac{1}{x}\).
2Step 2: Integrate the expression
Now, we need to find the antiderivative (indefinite integral) of the given expression with respect to x. To do this, we integrate the expression term-by-term:
$$
\int f'(x) dx = \int (1 + e^x + \frac{1}{x}) dx
$$
When performing the integration, the antiderivative of each term is:
$$
\int 1 dx = x ;
$$
$$
\int e^x dx = e^x ;
$$
$$
\int \frac{1}{x} dx = ln |x| .
$$
Now we can rewrite and sum up these terms:
$$
f(x) = x + e^x + ln |x| + C
$$
where C is the constant of integration.
3Step 3: Apply the initial condition
We are given an initial condition, which is \(f(1) = 3+ e\). Substitute this value into the general solution for \(f(x)\):
$$
3 + e = 1 + e^1 + ln(1) + C
$$
4Step 4: Solve for the constant of integration
From the previous step, \(C\) is now the only unknown. Simplify the equation:
$$
3 + e = 1 + e + 0 + C \Rightarrow C = 2
$$
5Step 5: Write the final solution
Plug the value of \(C\) back into the general solution for \(f(x)\) and obtain the final solution:
$$
f(x) = x + e^x + ln |x| + 2
$$
Key Concepts
AntiderivativeIndefinite IntegralConstant of IntegrationInitial Condition
Antiderivative
Understanding the concept of an antiderivative is crucial when solving initial value problems in calculus. An antiderivative of a function is another function whose derivative is the original function. For instance, if you have a function denoted by F(x) and its derivative, represented as F'(x), then F(x) is an antiderivative of F'(x). In the context of the given exercise, finding the antiderivative of the function represented by the derivative expression, 1 + e^x + 1/x, is the first step to solving the initial value problem.
It's important to remember that for any function there are infinitely many antiderivatives, each differing by a constant. This is because the derivative of a constant is zero, so adding or subtracting a constant from a function does not change its derivative. In the given solution, the antiderivatives of 1, e^x, and 1/x are x, e^x, and ln|x|, respectively.
It's important to remember that for any function there are infinitely many antiderivatives, each differing by a constant. This is because the derivative of a constant is zero, so adding or subtracting a constant from a function does not change its derivative. In the given solution, the antiderivatives of 1, e^x, and 1/x are x, e^x, and ln|x|, respectively.
Indefinite Integral
The term 'indefinite integral' is often used interchangeably with 'antiderivative', and it represents the family of functions that are all antiderivatives of a given function. When you see the symbol \(\int\underline{\phantom{xxx}}f(x)dx\), it represents the indefinite integral of f(x) with respect to x. The process of finding this integral is called integration. Unlike the definite integral, which calculates the net area under the curve and has limits of integration, the indefinite integral does not have bounds and includes an arbitrary constant of integration, commonly denoted as C.
During integration, we often use known integration rules for common functions. For example, in the exercise, the antiderivative of 1 with respect to x is x, and the antiderivative of e^x is e^x itself, as these follow directly from standard rules of integration.
During integration, we often use known integration rules for common functions. For example, in the exercise, the antiderivative of 1 with respect to x is x, and the antiderivative of e^x is e^x itself, as these follow directly from standard rules of integration.
Constant of Integration
The constant of integration, represented by the letter C, is an essential part of the solution to an indefinite integral. It accounts for the fact that when you differentiate a constant, the result is zero; therefore, when you take an antiderivative, there could have been any constant added to that function without affecting the derivative. In the exercise's solution, after integrating each term of the function separately, we add the constant of integration to the antiderivative.
Since the derivative of a constant is zero, any value of C is technically correct in the context of antiderivatives. However, when solving an initial value problem, you use the given initial condition to find the specific value of C that makes the solution particular to the problem at hand. This is what turns the general solution into a specific solution for the given problem.
Since the derivative of a constant is zero, any value of C is technically correct in the context of antiderivatives. However, when solving an initial value problem, you use the given initial condition to find the specific value of C that makes the solution particular to the problem at hand. This is what turns the general solution into a specific solution for the given problem.
Initial Condition
An initial condition is a known value at a specific point that allows us to solve for the constant of integration, making our general equation a precise solution. In the context of the given exercise, the initial condition provided is f(1) = 3 + e. This piece of information is used in conjunction with the general antiderivative found through integration to solve for the specific value of the constant C.
By plugging the initial condition into the general solution, we get an equation solely in terms of C. Solving this equation yields the value of C that makes our function satisfy the initial condition. For instance, by substituting x = 1 into f(x) and setting it equal to 3 + e, we can solve for the value of C, which turns out to be 2. This value is then used to write the specific solution to the initial value problem, making the process complete.
By plugging the initial condition into the general solution, we get an equation solely in terms of C. Solving this equation yields the value of C that makes our function satisfy the initial condition. For instance, by substituting x = 1 into f(x) and setting it equal to 3 + e, we can solve for the value of C, which turns out to be 2. This value is then used to write the specific solution to the initial value problem, making the process complete.
Other exercises in this chapter
Problem 58
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