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Chapter 5: Problem 51

Find the general solution of the differential equation. $$ \frac{d y}{d t}=t(1-t), t \geq 0 $$

Short Answer

Expert verified
The general solution is \( y(t) = \frac{t^2}{2} - \frac{t^3}{3} + C \).

Step by step solution

01

Recognize the Equation Type

Identify the given differential equation as a first-order ordinary differential equation (ODE). It is in the standard form \( \frac{d y}{d t} = f(t) \), where \( f(t) = t(1-t) \). This indicates that we can solve it using integration.
02

Separate Variables

Since the equation is already in the form \( \frac{d y}{d t} = f(t) \), it can be directly integrated with respect to \( t \). Thus, we will integrate both sides: \(\int d y = \int t(1-t) \, dt \).
03

Integrate the Right Side

Expand the function to integrate: \( t(1-t) = t - t^2 \). Now integrate term-by-term: \[ \int (t - t^2) \, dt = \int t \, dt - \int t^2 \, dt. \]
04

Solve the Integrals

Compute each integral separately: - \( \int t \, dt = \frac{t^2}{2} \) - \( \int t^2 \, dt = \frac{t^3}{3} \) Combine these results: \[ \int (t - t^2) \, dt = \frac{t^2}{2} - \frac{t^3}{3}. \]
05

Write the General Solution

Include the constant of integration \( C \) to express the general solution: \[ y(t) = \frac{t^2}{2} - \frac{t^3}{3} + C. \] This is the general solution to the differential equation.

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Key Concepts

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

First-order Differential Equations
Ordinary differential equations (ODEs) form the foundation for describing many natural phenomena in fields like physics and engineering. A first-order differential equation is arguably the simplest of these kinds of equations. It involves derivatives and functions of only one variable, typically written in the form:
\[ \frac{d y}{d t} = f(t, y) \]
  1. "\( y \)" is the dependent variable.
  2. "\( t \)" is the independent variable.
  3. "\( f(t, y) \)" is a function of the independent variable and possibly the function \( y \) itself.
In our specific example, the equation \( \frac{d y}{d t} = t(1-t) \) is identified as a first-order ODE without a \( y \) dependency in \( f(t, y) \). This makes it a simpler version known as a 'separable' differential equation, allowing us to implement the method of integration directly.
Integration
Integration is a central operation when dealing with first-order ODEs. The process involves finding an antiderivative or integrating both sides of the equation to find the solution. Understanding the following points is key:
  • The primary goal is to "reverse" differentiation, moving from a rate of change (a derivative) back to a function.
  • In the context of differential equations, integrating provides us with a function \( y(t) \) that satisfies the relationship given by the derivative \( \frac{d y}{d t} \).
In the case of \( \frac{d y}{d t} = t(1-t) \), we expanded \( t(1-t) \) to obtain \( t - t^2 \). We then integrate both components separately:
  • \( \int t \, dt = \frac{t^2}{2} \)
  • \( \int t^2 \, dt = \frac{t^3}{3} \)
By combining these results, we form the antiderivative which is crucial for constructing the general solution.
General Solution
The outcome of solving a differential equation is often a 'general solution,' which captures all possible solutions using a constant of integration. This highlights:
  • The general solution represents an infinite family of curves, depending on the constant of integration \( C \).
  • Substituting different values for \( C \) results in different specific solutions of the differential equation.
For our equation, integrating gives us:\[ y(t) = \frac{t^2}{2} - \frac{t^3}{3} + C \]This formula represents the general solution. Each choice of \( C \) corresponds to a distinct function \( y(t) \) that satisfies the original equation at various initial conditions. Initial conditions are often set, in real-world scenarios, to pinpoint one specific solution from this family of solutions, offering valuable insights into specific behaviors modeled by the original differential equation.

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Most popular questions from this chapter

The equation $$ x^{2}-5=0 $$ has two solutions. Use the Newton-Raphson method to approximate the two solutions to four decimal places.

The growth of a particular population is described by a power law model, in which the rate of growth is given by a function: $$ r(t)=\frac{A}{(t+a)^{m}} $$ where \(A, m\), and \(a\) are all unknown constants. Given the following data for the size of the population, calculate the value for these constants that would fit the model to the data: $$ \begin{array}{ll} \hline \boldsymbol{t} & \boldsymbol{r}(\boldsymbol{t}) \\ \hline 0 & 1.89 \\ 1 & 1.31 \\ 3 & 0.988 \\ \hline \end{array} $$ Hint: Eliminate \(A\) first. It may help to then take logarithms of the equations that you derive after eliminating \(A\).

Find the general antiderivative of the given function. $$ f(x)=-3 \sin \left(\frac{\pi}{3} x\right)+4 \cos \left(-\frac{\pi}{4} x\right) $$

Find the general antiderivative of the given function. $$ f(x)=\sec ^{2}(2 x) $$

Solve the initial-value problem. $$ \frac{d y}{d x}=\frac{x^{2}}{3}, \text { for } x \geq 0 \text { with } y(0)=2 $$
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