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

Short answer

The solutions are \(y(t)=t^2+1\) and \(y(t)=t^2-1\).

Step-by-step solution

Integrate the Differential Equation

The given differential equation is \( \frac{dy}{dt} = 2t \). To find \( y(t) \), we need to integrate the right-hand side with respect to \( t \). This gives: \[ y(t) = \int 2t \, dt = t^2 + C \] where \( C \) is the constant of integration.

Solve for the Initial Condition \(y(0) = 1\)

Using the initial condition \( y(0) = 1 \), substitute \( t = 0 \) into the equation \( y(t) = t^2 + C \): \[ 1 = 0^2 + C \] Thus, \( C = 1 \). The particular solution is: \[ y(t) = t^2 + 1 \]

Solve for the Initial Condition \(y(0) = -1\)

Using the initial condition \( y(0) = -1 \), substitute \( t = 0 \) into the equation \( y(t) = t^2 + C \): \[ -1 = 0^2 + C \] Thus, \( C = -1 \). The particular solution is: \[ y(t) = t^2 - 1 \]

Graph Both Solutions

Now, with the solutions \( y(t) = t^2 + 1 \) and \( y(t) = t^2 - 1 \), we can graph these functions. The first graph represents a parabola shifted up by 1 unit, and the second represents a parabola shifted down by 1 unit. Plot both equations on the same set of axes to compare their shapes and positions.

Key concepts

Differential Equations

A differential equation is an equation involving an unknown function and its derivatives. In our case, the equation is \( \frac{dy}{dt} = 2t \), which describes the rate of change of \( y \) with respect to \( t \). Differential equations are used to model real-world phenomena where quantities change over time or space, like population growth or motion of objects. Solving a differential equation means finding a function that satisfies the equation. Typically, initial conditions are provided to find a specific solution that fulfills certain criteria at a given point, which we call an initial-value problem.

Integration

Integration is a mathematical process used to find the antiderivative or the original function from its derivative. In solving the differential equation \( \frac{dy}{dt} = 2t \), integration helps us find \( y(t) \). By integrating the function \( 2t \), we obtain \( y(t) = t^2 + C \), where \( C \) is the constant of integration. Integration is crucial in differential equations because it allows us to reconstruct the original function, given its rate of change.

Graphing Solutions

Graphing solutions to differential equations can visually illustrate how the function behaves over time or space. In the exercise, the solutions \( y(t) = t^2 + 1 \) and \( y(t) = t^2 - 1 \) are both parabolas. The first one is a standard upward-opening parabola shifted one unit up, and the second is shifted one unit down.

• By plotting both functions on the same graph, we observe how different initial conditions affect the position of the parabola, while the overall shape remains unchanged.
• This method aids in comparing solutions and understanding the influence of initial conditions in real-world contexts.

Constants of Integration

When integrating the derivative of a function, a constant of integration \( C \) is included reflecting the infinite number of possible antiderivatives. This constant is determined using initial conditions given in the problem.

• In this exercise, applying the initial condition \( y(0) = 1 \) gives \( C = 1 \), leading to the solution \( y(t) = t^2 + 1 \).
• For the initial condition \( y(0) = -1 \), we find \( C = -1 \), resulting in \( y(t) = t^2 - 1 \).

Understanding how \( C \) operates is fundamental as it transforms a general solution into a specific solution that exactly matches the intended scenario.