Question

Graph several functions that satisfy the following differential equations. Then find and graph the particular function that satisfies the given initial condition. $$f^{\prime}(x)=2 \cos 2 x ; f(0)=1$$

Short answer

Answer: The amplitude is \(\frac{1}{2}\), the period is \(\pi\), and there is no phase shift.

Step-by-step solution

Integrate the differential equation

In this step, we will integrate the given differential equation. The differential equation is: $$ f^{\prime}(x)=2 \cos 2x $$ To find the function, we must integrate this expression. So we have: $$ f(x) = \int 2 \cos 2x \, dx $$ Recall that the integral of cosine is positive sine. Therefore, we can apply the substitution rule u-substitution method, by setting \(u=2x\). Then, \(\frac{du}{dx} = 2\) and \(\frac{du}{2} = dx\). Our integral then becomes: $$ f(x) = \int \cos u \, \frac{du}{2} $$ $$ f(x) = \frac{1}{2}\int \cos u \, du $$ $$ f(x) = \frac{1}{2} \sin u +C $$ Now let's substitute back \(u=2x\) to obtain: $$ f(x) = \frac{1}{2} \sin{2x} + C $$ where C is the constant of integration.

Find the constant of integration using the initial condition

We are given the initial condition: \(f(0)=1\). Let's use this to find our constant of integration, C. $$ 1 = f(0) = \frac{1}{2} \sin{(0)} +C $$ Since, \(\sin{0} =0\), we have: $$ 1=C $$ Thus, the particular function is: $$ f(x) = \frac{1}{2} \sin{2x} + 1 $$

Graph the function

To graph the function, we need to find the amplitude, period and phase shift of the sine function. Amplitude: \(\frac{1}{2}\) Period: Recall that the period of a sine function in the form of \(\sin{Bx}\) is given by: $$ T = \frac{2\pi}{B} $$ In our case, \(B=2\), so: $$ T = \frac{2\pi}{2} = \pi $$ Phase shift: There is no phase shift in this function. Now, we can plot the function \(f(x) = \frac{1}{2} \sin{2x} + 1\). It is a sine wave with an amplitude of \(\frac{1}{2}\), a period of \(\pi\) and no phase shift. The graph will oscillate between \(0.5\) and \(1.5\).

Key concepts

Initial Conditions

When solving differential equations, initial conditions are extremely important to determine a specific solution. A differential equation provides a family of solutions or functions, which all satisfy the equation. However, when an initial condition is specified—as it is here with \(f(0) = 1\)—it pinpoints exactly one function from that family.- The initial condition \(f(0) = 1\) tells us the specific value of the function \(f(x)\) at \(x = 0\). - By applying this condition, we can solve for the constant of integration \(C\), as it caters the function specifically to the initial value.- In practice, this means our solution isn't just a general equation; it's customized to pass through a specific point on the graph by adjusting \(C\).
Understanding this concept means you can accurately determine exact solutions rather than just general solutions.

Integration

Integration is a process used to find a function from its derivative function, effectively the reverse of differentiation. In this problem, we start with a differential equation: \(f'(x) = 2 \cos 2x\). We want to find \(f(x)\) by integrating the right-hand side.- The integral \(\int 2 \cos 2x \, dx\) tells us we're looking for a function whose derivative is \(2 \cos 2x\). - The substitution method (u-substitution) is crucial here. By setting \(u=2x\), the integration transforms to \(\int \cos u \, \frac{du}{2}\). - After integration, this provides us with \(\frac{1}{2} \sin 2x + C\), a more specific form of the antiderivative.- Remember, the constant \(C\) is a placeholder for any possible shift or translation along the y-axis, which we resolve using initial conditions.

Sine Function

The sine function is instrumental in trigonometry and periodic functions, manifesting in waves that are symmetric and oscillatory. In our scenario, the sine function helps express the solution to the differential equation.- The expression \(f(x) = \frac{1}{2} \sin{2x} + C\) unfolds the nature of the function. The coefficient \(\frac{1}{2}\) denotes the amplitude, representing the height from the centerline of the wave to its peak.- The term \(2x\) inside the sine function impacts the frequency, inversely connected to the function's period. The frequency here implies that the function completes a full cycle over \(\pi\) units.- Adjusting the constant \(C=1\) with the initial condition, shifts the graph vertically, moving the sine wave up by 1 unit.
This adaptability of the sine function makes it perfect for representing various oscillating behaviors.

Graphing Functions

Once the function \(f(x) = \frac{1}{2} \sin{2x} + 1\) is determined, graphing provides a visual understanding, revealing the interaction between mathematical components.- Begin by identifying the key aspects of the function: its amplitude, period, and vertical shift.- With an amplitude of \(\frac{1}{2}\), the wave will rise and fall by this amount from the central line.- The period is \(\pi\), implying the function repeats every \(\pi\) units along the x-axis, unlike standard \(2\pi\) sine functions.- A vertical shift of \(+1\) moves the entire wave upward, meaning the midline is y=1, making the range of the function \([0.5, 1.5]\).- Plotting these points provides clarity on how the function behaves and fits within real-world contexts.
Graphical representation solidifies understanding of theoretical concepts, making abstract functions tangible.