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. \(f^{\prime}(x)=3 x+\sin x ; f(0)=3\)

Short answer

Answer: The particular solution to the given differential equation and initial condition is \(f(x) = \frac{3}{2}x^2 - \cos(x) + 4\).

Step-by-step solution

Integrate the given differential equation

To find the integral of the given differential equation \(f'(x) = 3x + \sin(x)\), we can integrate each term separately: $$f(x) = \int (3x + \sin(x)) dx = \int 3x dx + \int \sin(x) dx$$ Now, integrate each term: $$f(x) = \frac{3}{2}x^2 - \cos(x) + C$$ where \(C\) is the constant of integration.

Find the particular value of the constant of integration

To find the particular value of the constant \(C\), we will use the initial condition \(f(0) = 3\). Plug in \(x = 0\) and \(f(0) = 3\) in the function \(f(x)\) obtained in Step 1: $$3 = \frac{3}{2}(0)^2 - \cos(0) + C$$ Solve for \(C\): $$C = 3 + 1 = 4$$ Now we have the particular solution as: $$f(x) = \frac{3}{2}x^2 - \cos(x) + 4$$

Graph the functions

To graph the general solution (family of functions), you would plot the function \(f(x) = \frac{3}{2}x^2 - \cos(x) + C\) for several different values of \(C\). To graph the particular solution, you would plot the function with the specific \(C\) found in Step 2, which is \(f(x) = \frac{3}{2}x^2 - \cos(x) + 4\). Note that it is not possible to directly provide visual graphs in this response, but there are numerous online tools and software available (such as Desmos, Geogebra, or WolframAlpha) for creating these plots.

Key concepts

Integrating Differential Equations

Understanding how to integrate differential equations is an essential skill for students studying calculus or working in fields which involve mathematical modeling. Differential equations represent the relationship between a function and its derivatives, and integrating these equations yields information about the original function before it was differentiated.

For example, consider the differential equation from the exercise: \(f'(x) = 3x + \sin(x)\). To 'integrate' this equation means we look for an antiderivative or a function whose derivative gives us the right-hand side of our equation.

Differential equations can often be solved by separating the equation into manageable parts, such as polynomials and trigonometric functions, which can be integrated individually. After performing the integration of \(3x\) and \(\sin(x)\), we obtain a function in terms of \(x\) plus an unknown constant \(C\), which represents the general solution to the differential equation. The reason an unknown constant appears is that derivatives lose constant information—the reason one must consider initial conditions to find particular solutions.

Initial Value Problems

Once the general solution to a differential equation is found, we can address an 'initial value problem' (IVP), which is a situation where we know the value of the function or its derivatives at a particular point. IVPs are crucial because they allow us to find a unique solution by determining the constant of integration \(C\).

In our exercise, the initial condition is \(f(0) = 3\). This means when \(x = 0\), the value of the function \(f\) is 3. By plugging \(x = 0\) into our general solution \(\frac{3}{2}x^2 - \cos(x) + C\) and setting \(f(0)\) equal to 3, we get a simple equation that allows us to solve for \(C\), thus converting our general solution into a particular solution.

Initial value problems are frequently encountered in the real world where a specific circumstance dictates one set of conditions—like the initial speed of a car or the initial population of a species in a biological model.

Antiderivative

An antiderivative is a function that reverses the process of differentiation. For any given function, an antiderivative is a function whose derivative is the original function. Often in calculus, finding an antiderivative is referred to as 'integration', and it's most commonly expressed with the integral symbol.

The antiderivative isn't unique; because differentiation strips away constant information, any constant could be added to an antiderivative and it would still work. For the exercise at hand, \(f'(x) = 3x + \sin(x)\), we seek the antiderivative to determine the original function \(f(x)\). This leads us to the general solution \(f(x) = \frac{3}{2}x^2 - \cos(x) + C\), where \(C\) represents an infinite number of possible antiderivatives.

Understanding antiderivatives, integration, and the property of lacking uniqueness due to the constant \(C\) is pivotal in grasping calculus concepts and solving more complex differential equations.