Question

Solve the following initial-value problems starting from \(y_{0}=10 .\) At what time does \(y\) increase to 100 or drop to \(1 ?$$\frac{d y}{d t}=4 t\)

Short answer

The function reaches 100 at approximately \( t = 6.71 \). It never reaches 1.

Step-by-step solution

Understand the Differential Equation

The differential equation given is \( \frac{dy}{dt} = 4t \). This is a first-order linear differential equation. Our goal is to find a function \( y(t) \) such that when we differentiate it with respect to \( t \), it equals \( 4t \).

Integrate to Find y(t)

To solve this equation, we need to integrate with respect to \( t \). The integral of \( 4t \) with respect to \( t \) is:\[y(t) = \int 4t \, dt = 2t^2 + C\]where \( C \) is the constant of integration.

Apply Initial Condition

Using the initial condition \( y(0) = 10 \), we substitute \( t = 0 \) and \( y = 10 \) into the equation \( y(t) = 2t^2 + C \):\[10 = 2(0)^2 + C \,\Rightarrow C = 10\]Thus, \( y(t) = 2t^2 + 10 \).

Solve for When y(t) = 100

To find when \( y(t) = 100 \), set \( 2t^2 + 10 = 100 \) and solve for \( t \):\[2t^2 + 10 = 100 \,\Rightarrow 2t^2 = 90 \,\Rightarrow t^2 = 45 \,\Rightarrow t = \sqrt{45} \,\approx 6.71\]

Solve for When y(t) = 1

To find when \( y(t) = 1 \), set \( 2t^2 + 10 = 1 \) and solve for \( t \):\[2t^2 + 10 = 1 \,\Rightarrow 2t^2 = -9\]This is impossible since \( t^2 \) cannot be negative, thus \( y(t) \) never decreases to \( 1 \).

Key concepts

Initial Value Problem

The concept of an initial value problem in differential equations refers to finding a specific solution to a differential equation that satisfies given conditions. These conditions are known as the initial conditions. In this exercise, the condition is set at the start of the scenario, which is time zero. We are told that at time zero, the value of the function is 10, noted as \( y(0) = 10 \).
This initial condition is crucial because it allows us to determine the constant of integration, helping us identify a unique solution among the infinite possibilities for a given differential equation.
Understanding this allows you to solve questions about how a system evolves over time and predict its behavior under specific circumstances.

Integration

Integration is a fundamental concept in calculus that involves finding the original function given its derivative. In solving our first-order differential equation \( \frac{dy}{dt} = 4t \), integration is used to determine the function \( y(t) \).
The process involves integrating the right-hand side of \( \frac{dy}{dt} = 4t \), leading us to the function \( y(t) = 2t^2 + C \). Here, \( \int 4t \, dt \) was evaluated, adding the constant \( C \) since integration introduces an arbitrary constant.
This step transforms a rate of change (the derivative) back into an actual function that describes the system being modeled.

Constant of Integration

The constant of integration \( C \) appears in solutions to differential equations whenever you perform indefinite integration. It represents an unknown constant because the act of differentiating loses this information.
By using the initial condition \( y(0) = 10 \), we substituted into the equation \( y(t) = 2t^2 + C \) to solve for \( C \).
This is done by setting \( 2(0)^2 + C = 10 \) and solving for \( C \), yielding \( C = 10 \).
Incorporating the constant of integration is a key step in finding particular solutions to differential equations that satisfy initial conditions.

First-Order Linear Differential Equation

A first-order linear differential equation is one of the simplest types, involving the first derivative of the unknown function. In the given problem, \( \frac{dy}{dt} = 4t \) is classified as first-order because it includes \( dy/dt \).
Such equations are linear because they can be written in the standard form \( a(t)y' + b(t)y = c(t) \), though in this case it simplifies to \( y' = 4t \) since \( b(t) = 0 \).
The simplicity of first-order equations makes them great stepping stones for more complex differential equations, building foundational skills needed for manipulation and solution.
Recognizing the type of differential equation you are dealing with is essential as it dictates the method of solution and affects the overall approach.