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\(\frac{d y}{d t}=2 y\)

Short answer

The solutions are \( y(t) = e^{2t} \) and \( y(t) = -e^{2t} \).

Step-by-step solution

Identify the Type of Differential Equation

The given problem is a first-order linear ordinary differential equation (ODE) of the form \( \frac{dy}{dt} = ky \) where \( k = 2 \). This is a separable differential equation.

Separate Variables

To solve the equation \( \frac{dy}{dt} = 2y \), we first separate the variables by dividing both sides by \( y \), which gives us \( \frac{1}{y} \frac{dy}{dt} = 2 \). Then, multiply both sides by \( dt \) to achieve separation: \( \frac{1}{y} dy = 2 dt \).

Integrate Both Sides

Integrate both sides of the equation: \( \int \frac{1}{y} dy = \int 2 dt \). This results in \( \ln |y| = 2t + C \), where \( C \) is the constant of integration.

Solve for y

Exponentiate both sides to solve for \( y \): \( e^{\ln |y|} = e^{2t + C} \), so \( |y| = e^{2t + C} \). Simplify to get \( y = Ce^{2t} \), where \( C = e^C \), an arbitrary positive constant or negative when considering the magnitude.

Apply Initial Conditions

For the initial condition \( y(0) = 1 \), substitute into the equation: \( 1 = Ce^{0} \) implying \( C = 1 \). Thus, the solution is \( y(t) = e^{2t} \). For the initial condition \( y(0) = -1 \), substitute into the equation: \( -1 = Ce^{0} \), implying \( C = -1 \). Thus, the solution is \( y(t) = -e^{2t} \).

Draw the Graphs

Plot both solutions on the same graph. The first solution \( y(t) = e^{2t} \) is an exponentially growing curve starting from 1 when \( t=0 \). The second solution \( y(t) = -e^{2t} \) is an exponential curve that starts at -1 and grows negatively.

Key concepts

Initial Value Problem

In the world of differential equations, an initial value problem (IVP) is a type of problem where one not only has to find a function, which fits a given differential equation, but also fits specific initial conditions. These initial conditions are values that the solution needs to satisfy at a specific point. In this particular case, we have the initial conditions of the form \(y(t=0)=1\) and \(y(t=0)=-1\).

The goal in solving an IVP is to find the unique solution that meets both the differential equation and these initial conditions. Why is this important? Without initial conditions, a differential equation typically has many solutions. By providing an initial condition, you help "anchor" the solution, effectively narrowing down the possibilities to one specific solution.

For example, in the given problem, solving the differential equation \(\frac{dy}{dt} = 2y\) results in the general solution \(y = Ce^{2t}\). However, by applying initial conditions \(y(0) = 1\) and \(y(0) = -1\), we obtain two specific solutions: \(y(t) = e^{2t}\) and \(y(t) = -e^{2t}\), respectively.

Separable Differential Equations

Separable differential equations are a class of equations that can be solved by the method of separation of variables. This technique involves rearranging the differential equation to allow each variable to be isolated on opposite sides of the equation. Once done, each side can be integrated separately to find the solution.

Consider the given equation \(\frac{dy}{dt} = 2y\). First, we separate the variables, bringing all terms involving \(y\) to one side and all terms involving \(t\) to the other, resulting in \(\frac{1}{y} dy = 2 dt\). This is the hallmark of a separable equation: isolating \(dy\) and \(dt\) so they can be independently integrated.

Next, by integrating both sides, \(\int \frac{1}{y} dy = \int 2 dt\), we find \(\ln |y| = 2t + C\). This integration process is crucial in finding the general solution for \(y\).

Finally, solve for \(y\) by exponentiating both sides, resulting in \(|y| = e^{2t+C}\), which simplifies based on the initial conditions to give us specific solutions.

Graphing Solutions of Differential Equations

Graphing solutions to differential equations provides a visual representation that can help in understanding the behavior of the solutions over time. In this example, we have obtained two solutions based on different initial conditions: \(y(t) = e^{2t}\) and \(y(t) = -e^{2t}\), which both arise from the same differential equation \(\frac{dy}{dt} = 2y\).

When you graph these solutions, it's essential to observe their starting points and the trajectory of the curves. The solution \(y(t) = e^{2t}\) begins at \(y = 1\) when \(t = 0\) and depicts an exponentially growing curve. Conversely, \(y(t) = -e^{2t}\) starts at \(y = -1\) and grows negatively or downwards exponentially.

Plotting these functions simultaneously on the same graph allows you to compare how changes in initial values impact the behavior of the solution over time. This process can illustrate key attributes like growth directions and symmetry that might not be obvious from the equations alone. Such visual tools are invaluable for both learning and analyzing differential equations.