Question

In Exercises \(25-29\) solve the initial value problem and sketch the graph of the solution. $$ \begin{array}{|l|l|l|} \hline \mathrm{C} / \mathrm{G} & y^{\prime}+7 y=e^{3 x}, & y(0)=0 \end{array} $$

Short answer

In summary, the given initial value problem is a first-order linear ODE. We used the integrating factor method to solve the ODE, resulting in the particular solution: $$y(x) = e^{-7x}\left(\frac{1}{10}e^{10x} - \frac{1}{10}\right).$$ By analyzing the solution's structure, we can sketch the graph, which displays exponential growth for $$x > 0$$ and exponential decay for $$x < 0.$$ The graph passes through the point $$(0, 0)$$ due to the initial condition.

Step-by-step solution

Identify the type of the ODE

This ODE is a first-order linear ODE since it can be written in the form $$y^\prime + P(x) y = Q(x),$$ where $$P(x) = 7$$ and $$Q(x) = e^{3x}.$$

Rewrite the ODE in a suitable form

Since the ODE is already in a suitable form for finding an integrating factor, we can skip this step.

Apply the appropriate method

We will use the integrating factor method for first-order linear ODEs. The integrating factor $$I(x)$$ can be found as follows: $$ I(x) = e^{\int P(x) \, dx} = e^{\int 7 \, dx} = e^{7x}.$$ Now, multiply the entire ODE with the integrating factor $$I(x)$$: $$ e^{7x}(y^\prime + 7y) = e^{7x} \cdot e^{3x}.$$ This results in: $$ (y e^{7x})^\prime = e^{10x}.$$ Now we integrate both sides with respect to $$x$$: $$\int (y e^{7x})^\prime dx = \int e^{10x} dx.$$ This results in: $$y e^{7x} = \frac{1}{10}e^{10x} + C,$$ where $$C$$ is the constant of integration. Next, we solve for $$y(x)$$ by isolating $$y$$ in the equation above: $$y(x) = e^{-7x}\left(\frac{1}{10}e^{10x} + C\right).$$

Apply the initial condition

Now that we have a general solution, we can apply the initial condition $$y(0) = 0$$: $$y(0) = e^{-7(0)}\left(\frac{1}{10}e^{10(0)} + C\right) = \frac{1}{10} + C.$$ We find that $$C = -\frac{1}{10}$$, and thus, the particular solution is given by: $$y(x) = e^{-7x}\left(\frac{1}{10}e^{10x} - \frac{1}{10}\right).$$

Sketch the graph of the solution

To sketch the graph of the solution, we rely on the properties of the exponential function and the specific structure of the given solution. The solution can be considered as the sum of two functions: $$e^{-7x}(\frac{1}{10}e^{10x}) - e^{-7x}\frac{1}{10}.$$ The first function grows exponentially as $$x$$ increases, and the second function decays exponentially as $$x$$ increases because of the $$e^{-7x}$$ term. This implies that the graph of the solution approaches the first function as $$x$$ increases and approaches the second function as $$x$$ decreases. The sketch of the function should thus show an exponential growth for $$x > 0$$, with the curve going through the point $$(0, 0)$$ due to the initial condition, and an exponential decay for $$x < 0$$.

Key concepts

First-order linear ODE

Understanding what a first-order linear ordinary differential equation (ODE) exactly is, helps in solving these types of problems more effectively. A first-order linear ODE has the general form:\[ y' + P(x) y = Q(x) \] The 'first-order' indicates that the highest derivative present in the equation is the first derivative, which is represented by \(y'\). The term 'linear' refers to the fact that both \(y\) and \(y'\) appear in linear combinations, not powers or other functions.
This ODE directly shows a linear combination of the function \(y\) and its first derivative \(y'\). Recognizing this form is the first step in solving these type of initial value problems.

Integrating Factor Method

To solve a first-order linear ODE, one efficient technique is the integrating factor method. This involves finding an appropriate function, called the integrating factor, which simplifies the equation. Here’s the general approach to using this method:

• First, identify \(P(x)\) from the given ODE.
• Calculate the integrating factor \(I(x) = e^{\int P(x) \, dx}\). This is done by integrating \(P(x)\) with respect to \(x\).
• Multiply the entire ODE by the integrating factor \(I(x)\). This will transform the left-hand side of the equation into the derivative of a product \((y I(x))'\).
• Integrate both sides with respect to \(x\) to find the solution.

In this specific exercise, the integrating factor \(I(x) = e^{7x}\) was used to effectively simplify the original differential equation and facilitate its integration.

Particular Solution

Once we have the general solution, identifying the particular solution forms the next crucial step in solving initial value problems. This involves applying any given initial conditions. In other words, we use the specific conditions provided to find the value of any constant of integration that appeared during the solution process.
In this exercise, the initial condition given was \(y(0) = 0\). By substituting \(x = 0\) and \(y = 0\) into the general solution, we solve for the constant \(C\). This ensures the solution fits exactly with the given initial condition, refining the general solution into a unique particular solution. It’s an essential step to ensure that the solution not only works in theory, but also suits the specific characteristics of the problem presented.

Exponential Growth and Decay

Exponential growth and decay describe how a function increases or decreases in value as the variable \(x\) changes. These concepts are pivotal in understanding the behavior of the solutions to ODEs, particularly when exponential terms are involved.
In the exercise, the given differential equation involved terms that were either exponentially growing or decaying.

• The term \(e^{10x}\) represents exponential growth, as the function increases rapidly for positive \(x\).
• Conversely, \(e^{-7x}\) illustrates exponential decay, where the function diminishes as \(x\) increases.

These exponential behaviors influence how the solution graph appears. For instance, as \(x > 0\), the graph shows exponential growth, while for \(x < 0\), it demonstrates exponential decay. Understanding these effects helps in correctly sketching and interpreting the solution’s graph.