Question

Find the solution to the indicated initial value problem, and use ezplot to plot it. \(y^{\prime}=t(y+1)\) with \(y(0)=1\) over \([0,2]\)

Short answer

The solution is \( y(t) = 2e^{\frac{t^2}{2}} - 1 \).

Step-by-step solution

Understand the Problem

We are given the differential equation \( y' = t(y+1) \) with the initial condition \( y(0) = 1 \). Our objective is to find the solution \( y(t) \) that satisfies both the differential equation and the initial condition. Once we find \( y(t) \), we'll plot it over the interval \([0, 2]\).

Separate Variables

First, we separate the variables in the differential equation. We rewrite it as \( \frac{dy}{dt} = t(y+1) \), which becomes\[ \frac{dy}{y+1} = t \, dt. \]

Integrate Both Sides

Integrate both sides of the separated equation. The left side becomes \( \int \frac{dy}{y+1} = \ln|y+1| \), and the right side becomes \( \int t \, dt = \frac{t^2}{2} + C \), where \( C \) is the constant of integration. Thus, we have\[ \ln|y+1| = \frac{t^2}{2} + C. \]

Solve for the Constant Using the Initial Condition

Apply the initial condition \( y(0) = 1 \) to find \( C \). Substitute \( t = 0 \) and \( y = 1 \) into our integrated equation: \( \ln|1+1| = \ln 2 = 0 + C \). Thus, \( C = \ln 2 \).

Solve for the General Solution

Substitute \( C = \ln 2 \) back into the integrated equation: \[ \ln|y+1| = \frac{t^2}{2} + \ln 2. \] Exponentiate both sides to solve for \( y \):\[ |y+1| = 2e^{\frac{t^2}{2}}. \] Since \( y+1 \) is positive, we have \( y+1 = 2e^{\frac{t^2}{2}} \), yielding the solution\[ y = 2e^{\frac{t^2}{2}} - 1. \]

Plot the Solution Using ezplot

To visualize the solution, use ezplot (or any plotting tool of your choice) to plot \( y(t) = 2e^{\frac{t^2}{2}} - 1 \) over the interval \( [0, 2] \). This verifies the behavior of the solution within the given range.

Key concepts

Initial Value Problem

When it comes to understanding ordinary differential equations (ODEs), initial value problems are extremely important. An initial value problem not only includes a differential equation, but also an initial condition. This initial condition specifies the value of the solution at a particular point, usually where the equation begins to be evaluated. In this exercise, the differential equation given is \( y' = t(y+1) \), accompanied by the initial condition \( y(0) = 1 \). This essentially tells us how the solution starts or initiates its journey at \( t = 0 \), which is a crucial piece of information for solving the equation accurately.

Separation of Variables

The method of separation of variables is a popular technique used to solve differential equations like the one in our problem. This technique involves rearranging the equation such that each variable is on a different side of the equation. For the given equation \( \frac{dy}{dt} = t(y+1) \), we separate the variables to get \( \frac{dy}{y+1} = t \, dt \). This means we are expressing the derivative in terms of \( y \) on one side and \( t \) on the other, which makes it easier for us to integrate and find the solution. It's like dividing responsibilities into more manageable parts, one for each variable.

Integration

Integration is the next step after you have separated the variables in a differential equation. In this exercise, we integrate both sides to find the antiderivatives. For \( \frac{dy}{y+1} \), the integral is \( \ln|y+1| \). On the other side of the equation, integrating \( t \) with respect to \( t \) gives us \( \frac{t^2}{2} \), plus a constant \( C \), which arises because we are working with indefinite integrals. The complete integrated form is \( \ln|y+1| = \frac{t^2}{2} + C \). This integration step helps in transforming our equation into one that is ready to be solved for \( y \).

Function Plotting

Function plotting is an effective way to visually understand the behavior of the solution over a specified interval. After finding the solution \( y(t) = 2e^{\frac{t^2}{2}} - 1 \), we use plotting tools to graph this function over the interval \([0, 2]\). By examining this plot, we can verify our solution and observe its growth trend over the designated range. Seeing the function on paper or screen can give a much clearer understanding of how the mathematical expressions we derived behave in practical terms.

Differential Equation Solution

Finally, the solution to a differential equation is what ties everything together. It is essentially the function that satisfies both the differential equation and its initial condition. For our problem, solving the integrated equation with the initial condition \( y(0) = 1 \), we arrived at the solution \( y = 2e^{\frac{t^2}{2}} - 1 \). This solution offers us a precise formula to predict the behavior of \( y \) based on any time \( t \) within the interval we specified. Thus, solving a differential equation not only involves arithmetic but also a conceptual understanding of the equation, the conditions, and how they interact.