Question

Solve the given initial value problem. \(\frac{d y}{d t}=t * t, \quad y(0)=1\)

Short answer

Question: Solve the initial value problem (IVP) for the first-order ordinary differential equation (ODE) \(\frac{dy}{dt} = t^2\) with the initial condition \(y(0) = 1\). Answer: The particular solution to the given initial value problem is \(y(t) = \frac{1}{3}t^3 + 1\).

Step-by-step solution

Solve the ODE

We are given the following first-order ODE: \(\frac{dy}{dt} = t^2\) To find the function y(t), we need to integrate both sides of the equation with respect to t: \(dy = t^2 dt\) Now, integrate both sides: \(\int dy = \int t^2 dt\) After integrating, we get: \(y(t) = \frac{1}{3}t^3 + C\) where C is the integration constant.

Apply the initial condition

We are given the initial condition \(y(0) = 1\). Apply the initial condition to find the particular solution: \(y(0) = \frac{1}{3}(0)^3 + C = 1\) Since \((0)^3\) is equal to 0, we can simplify the equation to: \(C = 1\) Now, substitute the value of C back into the general solution y(t) equation: \(y(t) = \frac{1}{3}t^3 + 1\) This is the particular solution to the given initial value problem.

Key concepts

Ordinary Differential Equation

An ordinary differential equation (ODE) is a mathematical expression that describes a relation involving a function and its derivatives. It's 'ordinary' because it deals with functions of one variable, in contrast to partial differential equations which involve multiple variables.

In the context of the exercise, we are presented with an ODE \( \frac{dy}{dt} = t^2 \). Our goal is to find a function \( y(t) \) whose derivative with respect to \( t \) is \( t^2 \)—essentially, we are trying to reverse the process of differentiation, which is called integration. Since the ODE is first-order, meaning it involves the first derivative of \( y \), the solution will typically involve an integration constant, which we'll address next.

Integration Constant

The integration constant, typically represented as \( C \), arises when we integrate a differential equation. This constant is a crucial part of finding the general solution to an ODE because it represents the indefinite number of possibilities that arise from anti-differentiation.

For example, upon integrating \( \frac{dy}{dt} = t^2 \) in our exercise, we derive the general solution \( y(t) = \frac{1}{3}t^3 + C \). The \( + C \) part is the integration constant, which we must determine using additional information, such as an initial condition. This constant makes the solution a 'family' of functions, and to identify a 'particular' member of this family, we must specify some conditions, which is where the initial conditions come into play.

Initial Condition

An initial condition in the context of differential equations is a value that allows us to find a specific, or 'particular,' solution to the ODE. It is a way to determine the correct integration constant that will satisfy the problem's constraints.

In the given exercise, the initial condition is \( y(0) = 1 \), meaning that when \( t = 0 \), the function \( y \) must equal 1. Introducing this condition to our general solution \( y(t) = \frac{1}{3}t^3 + C \), allows us to calculate the exact value of \( C \) which is necessary for the problem's particular solution. This step is vital for the function to match the given physical, geometric, or other scientific conditions at a specific point.

Particular Solution

A particular solution to an ODE is a solution that not only satisfies the general equation but also fulfills the given initial or boundary conditions. It is the one specific function from the general solution set that adheres to the conditions imposed by the problem.

In our settled example, the initial condition was utilized to find the value of the integration constant, resulting in the equation \( y(t) = \frac{1}{3}t^3 + 1 \). This expression is the particular solution to the initial value problem given it articulates the exact function \( y(t) \) that not only derives to \( t^2 \) but also confirms \( y(0) = 1 \). It represents the unique answer to our problem, incorporating all provided data.