Question

Graph several functions that satisfy each of the following differential equations. Then find and graph he particular fimction that satisfies the given initial condition. $$f^{\prime}(x)=2 x-5 ; f(0)=4$$

Short answer

Question: Solve the first-order differential equation \(f^{\prime}(x)=2x-5\) with the given initial condition \(f(0)=4\), graph some functions that satisfy the differential equation, and find the particular function that also satisfies the initial condition.

Step-by-step solution

1. Integrate the differential equation

Integrate the given differential equation, \(f^{\prime}(x)=2x-5\), to find the general solution: $$f(x)=\int (2x-5) \, dx$$ $$f(x)=2\int x \, dx - 5\int 1 \, dx$$ $$f(x)=x^2 - 5x + C$$ Here, \(C\) is the constant of integration that will determine the particular solution.

2. Apply the initial condition

Use the initial condition \(f(0)=4\) to find the particular solution: $$4=f(0)=0^2 - 5(0) + C$$ $$4=C$$

3. Write the particular solution

Replace \(C\) with the value determined from the initial condition: $$f(x) = x^2 - 5x + 4$$

4. Graph the general solution

Graph a few functions that satisfy the general solution, \(f(x)=x^2 - 5x + C\). You may try changing the value of the constant to observe different functions: - \(C=0\): \(f(x)=x^2 - 5x\) - \(C=2\): \(f(x)=x^2 - 5x + 2\) - \(C=4\): \(f(x)=x^2 - 5x + 4\)

5. Graph the particular solution

Finally, we graph the particular solution that also satisfies the given initial condition, \(f(x) = x^2 - 5x + 4\). Observe how it passes through the point \((0, 4)\), which is the given initial condition. Now, you should see several functions that satisfy the given differential equation, as well as the particular function that satisfies the initial condition. The particular function is just one of the many general functions that satisfy the differential equation, but it's the one that meets the specific initial condition provided.

Key concepts

Initial Value Problems

An Initial Value Problem (IVP) in the context of differential equations involves finding a function that satisfies both the differential equation and a given initial condition. Initial conditions, often denoted as values such as f(0)=4, serve as a starting point for the function and help in determining the unique solution.

When dealing with an IVP, one typically integrates the differential equation to obtain the general solution, which includes a constant of integration. It is the initial condition that allows us to find the specific value of this constant, thus enabling us to find the unique, sole solution that goes through the initial point. This tailored solution is referred to as the particular solution. In the case of the example exercise, the initial value problem is defined by the differential equation f'(x)=2x-5 and the initial condition f(0)=4.

Integration of Functions

Integration is a fundamental operation in calculus, often viewed as the reverse process of differentiation. When you integrate a function, you're essentially looking for the original quantity whose rate of change is described by the given function. This process is invoked in finding the general solution of differential equations.

In the exercise provided, the differential equation f'(x)=2x-5 was integrated to yield f(x)=x^2 - 5x + C. Here, each term was integrated separately: 2x became x^2 after integrating with respect to x, and -5 became -5x. The constant of integration C is crucial, as it represents an infinite number of possible functions that are all solutions to the differential equation.

Graphing Functions

Graphing functions is a visual interpretation of equations. It provides insight into the behavior and characteristics of functions, like upward or downward trends, curves, hills, and valleys. For differential equations, graphing the solutions can show us how the function behaves for different values of the constant of integration.

In our example, several functions satisfying the differential equation were graphed with different constants C. This illustrates how the variation of C affects the position of the graph vertically, resulting in a family of curves that mirror each other along the y-axis. The particular solution, associated with the specific initial condition f(0)=4, is also graphed. This graph passes through the point (0, 4) and provides students with a concrete understanding of how the initial value affects the particular solution.

Particular and General Solutions

A general solution of a differential equation includes a constant of integration C, and it represents a whole set of functions that satisfy the differential equation.

Distinguishing the General and Particular Solutions

The exercise showcases how to find both. The general solution found through integration, f(x)=x^2 - 5x + C, gives us a curve for every real number value of C. However, to find the particular solution, the one that fulfills the initial value condition, we substitute the initial condition into the general solution. In our example, setting C=4 yields f(x)=x^2 - 5x + 4, which is the unique curve that passes through the point (0, 4). Graphically, this specific curve can be identified among the family of curves as the one that intersects the y-axis at y=4.