Question

Solve the following initial-value problems starting from $y(t=0)=1$ and $y(t=0)=-1.$ Draw both solutions on the same graph$\frac{dy}{dt}=-y$

Short answer

Two solutions: $y(t)=e^{-t}$ and $y(t)=-e^{-t}$.

Step-by-step solution

Identify the Type of Differential Equation

The differential equation given is $\frac{dy}{dt}=-y$. This is a first-order linear ordinary differential equation with a constant coefficient. The solution approach is to find the general solution to this type of equation.

Solve the Differential Equation General Form

For a differential equation $\frac{dy}{dt}=-y$, the general solution can be found using separation of variables or recognizing the form. The solution is given by $y(t)=C{e}^{-t}$, where $C$ is an integration constant to be determined from initial conditions.

Apply Initial Condition for $y(t=0)=1$

Using the initial condition $y(0)=1$, substitute into the general solution: $1=C{e}^0$, which simplifies to $C=1$. Therefore, the particular solution for this initial condition is $y(t)=e^{-t}$.

Apply Initial Condition for $y(t=0)=-1$

Using the initial condition $y(0)=-1$, substitute into the general solution: $-1=C{e}^0$, which simplifies to $C=-1$. Therefore, the particular solution for this initial condition is $y(t)=-e^{-t}$.

Plot Both Solutions on the Same Graph

For visualization, plot the two solutions on the same graph. The function $y(t)=e^{-t}$ will be a decaying exponential starting at 1, and the function $y(t)=-e^{-t}$ will be a decaying exponential starting at -1, both decreasing towards zero as $t$ increases.

Key concepts

Initial-Value Problems

An initial-value problem involves solving a differential equation subject to a specific condition at the start, also known as the initial condition. This is typically given as the value of the solution at a particular point in time.
In the context of differential equations, an initial-value problem helps us find a unique solution that satisfies the given condition. For example, if we have a differential equation $\frac{dy}{dt}=-y$, the initial condition could be stated as $y(0)=1$ or $y(0)=-1$.
Such conditions allow us to determine the constant in the general solution, thus providing a particular solution that conforms to the initial requirements.

• General Solution: This is the solution of a differential equation before applying initial conditions; it often contains an arbitrary constant.
• Particular Solution: By applying the initial condition, we determine this specific solution that fits the starting criteria.

In our example, the initial-value problems help us determine the specific exponential functions that start at 1 and -1 respectively.

First-Order Linear Differential Equations

First-order linear differential equations have a simple form: $\frac{dy}{dt}=ay+b$. The equation $\frac{dy}{dt}=-y$ is an example of this, where $a=-1$ and $b=0$.
This type of differential equation is powerful due to its straightforward nature, making the solving process easy once you understand the structure.

• Characteristics: The key characteristic of first-order differential equations is that they involve only the first derivative of the unknown function. They are said to be linear if they can be arranged into the form $a(t)y+b(t)$.
• Solving Approach: The typical method for solving them involves integrating factor, separation of variables, or recognizing patterns, like in our example.

The specific equation $\frac{dy}{dt}=-y$ has a general solution $y(t)=C{e}^{-t}$, indicating that the solution decays exponentially over time. This pattern aligns with real-world phenomena like radioactive decay.

Exponential Functions

Exponential functions are expressed in the form $y(t)=C{e}^{kt}$. They appear in the solutions of many first-order linear differential equations.
In our differential equation $\frac{dy}{dt}=-y$, the general solution was found to be $y(t)=C{e}^{-t}$, an exponential decay function.

• Decay: Exponential decay functions are characterized by a constant within the exponential term being negative, leading the function to decrease over time.
• Growth: When the constant is positive, the function represents exponential growth, common in modeling processes like population growth.

For the particular solutions, $y(t)=e^{-t}$ when $y(0)=1$ and $y(t)=-e^{-t}$ when $y(0)=-1$, you can picture both solutions diminishing towards zero as $t$ increases. This illustrates how exponential functions can model both positive and negative scenarios effectively.