Question

Finding a Particular Solution Using Separation of Variables In Exercises \(15-24\) , find the particular solution that satisfies the initial condition. $$ \frac{d r}{d s}=e^{r-2 s} \quad r(0)=0 $$

Short answer

The particular solution of the given differential equation is \(-e^{-r} = -\frac{1}{2}e^{-2s} -\frac{1}{2}\).

Step-by-step solution

Separate the Variables

Start by rewriting the differential equation so that all terms involving \(r\) are on one side and all terms involving \(s\) are on the other. That means \(\frac{d r}{e^r} = e^{-2s} ds\).

Integrate Both Sides

The next step is to integrate both sides of the equation with respect to their own variable. The integration is performed as follows: \(\int \frac{d r}{e^r} = \int e^{-2s} ds\). Solving these, we get that the integral \(-e^{-r} = -\frac{1}{2}e^{-2s} + C\).

Apply Initial Condition

At this point, apply the initial condition given by the problem, which is \(r(0) = 0\). In other words, when \(s = 0\), \(r = 0\). Substituting these values into the above result, we find that \(C = -\frac{1}{2}\). Thus, our particular solution becomes \(-e^{-r} = -\frac{1}{2}e^{-2s} -\frac{1}{2}\).

Key concepts

Differential Equations

Differential equations are fundamental tools in mathematics that involve equations containing derivatives of a function. These equations describe how a particular quantity changes concerning another. In real-world scenarios, they can model various phenomena, such as population growth, heat distribution, and motion.In our exercise, we encountered a first-order differential equation, \( \frac{dr}{ds} = e^{r-2s} \), which tells us how the variable \(r\) changes with respect to \(s\). The goal is to find a specific function \(r(s)\) that satisfies this equation along with a given initial condition. Understanding differential equations helps us predict and explain these changes mathematically.

Initial Conditions

When solving differential equations, we often need more than just the general solution; we require a specific solution that satisfies particular constraints. These constraints are called initial conditions. They help us define a unique solution that fits the scenario we're studying.In the exercise, the initial condition is \(r(0) = 0\), meaning when \(s = 0\), the function \(r\) should equal 0. This condition is essential because it helps us find the constant \(C\) after integration. Without initial conditions, the differential equation's solution could represent a family of curves, not just the specific one we need.

Integration Techniques

Integration is a crucial technique in finding solutions to differential equations, especially when using the separation of variables method. This method involves separating the variables into two sides of the equation, where each side contains only one of the variables. Then, we integrate each side individually.In the given problem, we separate the equation, resulting in \(\frac{dr}{e^r} = e^{-2s} ds\). Integrating both sides yields:

• \(\int \frac{dr}{e^r} = -e^{-r}\)
• \(\int e^{-2s} ds = -\frac{1}{2}e^{-2s}\)

These results combine to give us the expression \(-e^{-r} = -\frac{1}{2}e^{-2s} + C\). Mastery of integration techniques is necessary for solving such equations correctly.

Particular Solutions

A particular solution to a differential equation is a single out of possibly many that satisfies the equation as well as any imposed initial conditions. It's the unique solution tailored to fit the specific situation or condition given in the problem.After integrating and simplifying, we arrive at the expression \(-e^{-r} = -\frac{1}{2}e^{-2s} + C\). By applying the initial condition \(r(0) = 0\), we can solve for the constant \(C\), finding it to be \(-\frac{1}{2}\). Substituting \(C\) back gives us the particular solution: \[-e^{-r} = -\frac{1}{2}e^{-2s} - \frac{1}{2}\]This equation satisfies both the differential equation and the initial condition, ensuring it is the correct and specific solution we need.