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Chapter 12: Problem 45

Solve the vector initial-value problem for \(\mathbf{y}(t)\) by integrating and using the initial conditions to find the constants of integration. $$ \mathbf{y}^{\prime}(t)=2 t \mathbf{i}+3 t^{2} \mathbf{j}, \mathbf{y}(0)=\mathbf{i}-\mathbf{j} $$

Short Answer

Expert verified
\( \mathbf{y}(t) = (t^2 + 1) \mathbf{i} + (t^3 - 1) \mathbf{j} \)

Step by step solution

01

Identify the given vector differential equation

The differential equation given is \( \mathbf{y}'(t) = 2t \mathbf{i} + 3t^2 \mathbf{j} \). We need to find the vector function \( \mathbf{y}(t) \) that satisfies this equation.
02

Integrate the vector component-wise

To find \( \mathbf{y}(t) \), we integrate each component of the vector separately. The first component is \( 2t \mathbf{i} \), and the second component is \( 3t^2 \mathbf{j} \).1. Integrate \( 2t \) with respect to \( t \):\[ \int 2t \ dt = t^2 + C_1 \] So, the \( \mathbf{i} \) component is \( t^2 \mathbf{i} + C_1 \mathbf{i} \).2. Integrate \( 3t^2 \) with respect to \( t \):\[ \int 3t^2 \ dt = t^3 + C_2 \] So, the \( \mathbf{j} \) component is \( t^3 \mathbf{j} + C_2 \mathbf{j} \).Therefore, \( \mathbf{y}(t) = (t^2 + C_1) \mathbf{i} + (t^3 + C_2) \mathbf{j} \).
03

Apply initial conditions to solve for constants

We use the initial condition \( \mathbf{y}(0) = \mathbf{i} - \mathbf{j} \) to find the constants \( C_1 \) and \( C_2 \).Substitute \( t = 0 \) into \( \mathbf{y}(t) \):\[ \mathbf{y}(0) = (0^2 + C_1) \mathbf{i} + (0^3 + C_2) \mathbf{j} = C_1 \mathbf{i} + C_2 \mathbf{j} \]Since \( \mathbf{y}(0) = \mathbf{i} - \mathbf{j} \), we equate the components to find:- \( C_1 = 1 \) (from the \( \mathbf{i} \) component).- \( C_2 = -1 \) (from the \( \mathbf{j} \) component).
04

Write the final vector function

Substitute \( C_1 = 1 \) and \( C_2 = -1 \) back into \( \mathbf{y}(t) \):\[ \mathbf{y}(t) = (t^2 + 1) \mathbf{i} + (t^3 - 1) \mathbf{j} \]This is the solution for the initial-value problem given.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Initial-Value Problem
An initial-value problem involves finding a function that satisfies a given differential equation and also fulfills specified conditions at the initial point. Let's consider the initial-value problem from the exercise with its vector equation:
  • The differential equation is \( \mathbf{y}'(t) = 2t \mathbf{i} + 3t^2 \mathbf{j} \).
  • The initial condition, provided as \( \mathbf{y}(0) = \mathbf{i} - \mathbf{j} \), is crucial. It tells us the specific value of the function at \( t = 0 \).

The objective is to determine the exact form of \( \mathbf{y}(t) \) that meets both the differential equation and initial condition. It's like solving a puzzle. By using the given initial condition, we can solve for any constants after integrating.
Indeed, this step ensures that the solution aligns with the pre-defined start settings of the problem. This alignment is why we call it an 'initial'-value problem.

Vector Integration
Vector integration is the process of integrating vector-valued functions. It combines the rules of traditional integration and vector calculus. In our exercise, we dealt with the vector function: \( \mathbf{y}'(t) = 2t \mathbf{i} + 3t^2 \mathbf{j} \). To find \( \mathbf{y}(t) \), we integrate each component separately:
  • For the \( \mathbf{i} \)-component: Integrate \( 2t \) with respect to \( t \), leading to \( \int 2t \, dt = t^2 + C_1 \).
  • For the \( \mathbf{j} \)-component: Integrate \( 3t^2 \) with respect to \( t \), resulting in \( \int 3t^2 \, dt = t^3 + C_2 \).

The constants \( C_1 \) and \( C_2 \) represent integration constants that we need to determine. Ensuring these components align with the initial conditions will give us the correct form of \( \mathbf{y}(t) \). Vector integration is thus a systematic approach to unravel each part of the vector function while accommodating the initial puzzle pieces.

Differential Equations
A differential equation contains derivatives of a function and helps describe various phenomena in mathematics and physics. Vector differential equations involve vector functions and their derivatives. In this exercise, \( \mathbf{y}'(t) = 2t \mathbf{i} + 3t^2 \mathbf{j} \) is our differential equation, representing a vector rate of change:
  • The vectors \( \mathbf{i} \) and \( \mathbf{j} \) give us direction, while the expressions \( 2t \) and \( 3t^2 \) provide the rate of change.
  • Solving these involves finding a function \( \mathbf{y}(t) \) that returns the initial state \( \mathbf{y}(0) = \mathbf{i} - \mathbf{j} \) when substituted back into the equation.

The integration generates a general solution, but the initial conditions guide us to a specific solution, ensuring the characteristics of both time and starting point are honored. By solving the intricacies of differential equations, we piece together how processes and functions behave over time.

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