Question

Follow the instructions for Problem 1 for the following initial value problems: $$ \begin{array}{ll}{\text { (a) } d y / d t=y-5,} & {y(0)=y_{0}} \\ {\text { (b) } d y / d t=2 y-5,} & {y(0)=y_{0}} \\ {\text { (c) } d y / d t=2 y-10,} & {y(0)=y_{0}}\end{array} $$

Short answer

Question: Find the particular solutions for the following initial value problems: (a) $\frac{dy}{dt} = y - 5$, $y(0) = y_0$ Answer: $$y(t) = 5 + (y_0 - 5)e^t$$ (b) $\frac{dy}{dt} = 2y - 5$, $y(0) = y_0$ Answer: $$y(t) = \frac{5}{2} + (y_0 - \frac{5}{2})e^{2t}$$ (c) $\frac{dy}{dt} = y - \frac{5}{2}$, $y(0) = y_0$ Answer: $$y(t) = 5 + \frac{(y_0 - 5)e^{2t}}{2}$$

Step-by-step solution

Problem (a): Integrating the equation

First, let's rewrite the given differential equation and initial condition: $$\frac{dy}{dt} = y - 5, \quad y(0) = y_0$$ To solve this first-order differential equation, we will need to separate variables and integrate both sides: $$\frac{dy}{y - 5} = dt$$ Now let's integrate both sides: $$\int\frac{1}{y-5}dy = \int1dt$$.

Problem (a): Solving for \(y(t)\)

By integrating, we get: $$\ln|y-5| = t + C_1$$ Now let's isolate \(y\): $$y = 5 + e^{t + C_1}$$ We can rewrite this as: $$y = 5 + e^t e^{C_1}$$ For convenience, let's replace the constant term \(e^{C_1}\) with another constant \(C_2\): $$y = 5 + C_2 e^t$$.

Problem (a): Applying the initial condition

Now we will use the initial condition \(y(0) = y_0\) to find the value of \(C_2\). Substitute \(t=0\) and \(y=y_0\) into the equation: $$y_0 = 5 + C_2 e^0$$ This gives us: $$C_2 = y_0 - 5$$.

Problem (a): Final solution

Now we can write the final solution for problem (a) by substituting \(C_2\) back into the equation: $$y(t) = 5 + (y_0 - 5)e^t$$ Follow the same steps for problems (b) and (c).

Problems (b) and (c): Final solutions

After solving for the constants and applying the initial conditions, we can write the final solutions for problems (b) and (c): Problem (b): $$y(t) = \frac{5}{2} + (y_0 - \frac{5}{2})e^{2t}$$ Problem (c): $$y(t) = 5 + \frac{(y_0 - 5)e^{2t}}{2}$$

Key concepts

Initial Value Problems

Initial value problems involve solving differential equations with specified conditions at a particular point, often referred to as the initial condition. In these problems, you are given a differential equation along with the value of the unknown function at a particular point. This point is usually where the independent variable is zero, which helps determine any arbitrary constants that arise during integration.

For example, in problem (a), the initial condition is given by \( y(0) = y_0 \), meaning that when the independent variable \( t \) is zero, the dependent variable \( y \) equals \( y_0 \). This condition is critical as it allows us to find the specific solution from the general solution of the differential equation. Without it, you would end up with a family of solutions instead of one definitive answer.

Using initial conditions simplifies complex problems into more manageable tasks and is key to solving real-world problems where a specific outcome is desired.

First-Order Differential Equations

A first-order differential equation is any equation that involves the first derivative of an unknown function. Essentially, it relates the rate of change of the function to the function itself. The 'first-order' designation means that the highest derivative in the equation is the first derivative.

For example, the equation \( \frac{dy}{dt} = y - 5 \) in the exercise is a first-order differential equation. Here, \( \frac{dy}{dt} \) represents the rate of change of \( y \) with respect to \( t \).

First-order equations appear ubiquitously in various fields such as physics, engineering, and economics. They can describe many real-world phenomena, like exponential growth, cooling processes, and more. Understanding these equations opens the door to analyzing and predicting outcomes in systems that change over time.

Separation of Variables

Separation of variables is a powerful technique used to solve first-order differential equations. The core idea is to rearrange the equation so that all terms involving the dependent variable \( y \) are on one side of the equation, and all the terms involving the independent variable \( t \) are on the other.

In problem (a), the equation \( \frac{dy}{dt} = y - 5 \) is restructured as \( \frac{dy}{y-5} = dt \). This separation allows for both sides to be integrated individually, simplifying the process of finding a solution.

Once separated, you integrate each side with respect to its particular variable. This technique assumes that each side of the equation can be expressed as a function of a single variable, allowing the solution to be constructed piece by piece.

Integration

Integration is the process of finding the integral of a function, and it is an essential step in solving differential equations, especially when using methods like separation of variables. In essence, integration is the reverse process of differentiation.

In problem (a), once the variables are separated, integration is applied: \( \int \frac{1}{y-5} dy = \int 1 dt \). Integrating gives \( \ln|y-5| = t + C_1 \), where \( C_1 \) is an integration constant.

Integration is vital as it allows us to transform differential equations into algebraic ones. By finding the anti-derivative of both sides of a separated equation, we proceed closer to finding the explicit solution for the differential equation.

Exponential Functions

Exponential functions frequently appear in the solutions to differential equations, particularly those describing growth or decay processes. An exponential function is characterized by a constant base raised to a variable exponent, often seen as \( e^x \) where \( e \) is Euler's number.

In problem (a), after integrating and applying the initial condition, the solution takes the form \( y(t) = 5 + (y_0 - 5) e^t \). The exponential term \( e^t \) signifies the dynamic change in \( y \) with respect to \( t \).

Because of their properties, exponential functions are ideal for modeling situations involving compound growth, decay, and other processes that are proportional to their current state. They provide a simple and precise description of the phenomena that change exponentially over time.