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Chapter 9: Problem 16

\(15-20\) Solve the initial-value problem. $$t \frac{d y}{d t}+2 y=t^{3}, \quad t>0, \quad y(1)=0$$

Short Answer

Expert verified
The solution is \( y = \frac{t^3}{5} - \frac{1}{5t^2} \).

Step by step solution

01

Rewrite as standard form

First, let's rewrite the given differential equation in its standard linear form. The given equation is \( t \frac{d y}{d t} + 2y = t^3 \). Divide the entire equation by \( t \) to get \( \frac{d y}{d t} + \frac{2}{t}y = t^2 \). This form makes it easier to identify the functions for the integrating factor method.
02

Find the integrating factor

The standard form of a linear first-order differential equation is \( \frac{d y}{d t} + P(t)y = Q(t) \). We identify \( P(t) = \frac{2}{t} \) and \( Q(t) = t^2 \). The integrating factor \( \mu(t) \) is given by \( e^{\int P(t) \, dt} = e^{\int \frac{2}{t} \, dt} \). Calculating this, we get \( \mu(t) = t^2 \).
03

Multiply the equation by the integrating factor

Multiply the entire differential equation by the integrating factor \( t^2 \), resulting in \( t^2 \frac{d y}{d t} + 2ty = t^4 \). Recognize that the left-hand side is now the derivative of the product of the integrating factor and \( y \): \( \frac{d}{dt} (t^2 y) = t^4 \).
04

Integrate both sides

Integrate both sides with respect to \( t \). For the left-hand side, integration gives \( t^2 y \). For the right-hand side, the integral is \( \int t^4 \, dt = \frac{t^5}{5} + C \), where \( C \) is the constant of integration. Thus, we have \( t^2 y = \frac{t^5}{5} + C \).
05

Solve for y

To find \( y \), divide both sides of the equation \( t^2 y = \frac{t^5}{5} + C \) by \( t^2 \): \( y = \frac{t^3}{5} + \frac{C}{t^2} \).
06

Apply the initial condition

Use the initial condition \( y(1) = 0 \) to find \( C \). Substitute \( y = 0 \) and \( t = 1 \) into the equation \( y = \frac{t^3}{5} + \frac{C}{t^2} \). This results in \( 0 = \frac{1^3}{5} + \frac{C}{1^2} \), leading to \( C = -\frac{1}{5} \).
07

Write the final solution

Substitute \( C = -\frac{1}{5} \) back into the equation \( y = \frac{t^3}{5} + \frac{C}{t^2} \). Thus, the final solution is \( y = \frac{t^3}{5} - \frac{1}{5t^2} \).

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Key Concepts

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

Differential Equation
Differential equations are mathematical equations that relate a function with its derivatives. They describe various phenomena such as change, motion, growth, or decay in a wide range of disciplines. In simple terms, they help us understand how something changes over time or space.

For instance, in the given problem, we have a differential equation of the form:
  • Original form: \( t \frac{dy}{dt} + 2y = t^3 \)
This equation connects the unknown function \( y \) and its derivative \( \frac{dy}{dt} \), involving the variable \( t \). Such equations are crucial in modeling real-world situations, as they offer insights on dynamic systems behavior.
Integrating Factor
The integrating factor is a vital technique utilized for solving linear first-order differential equations. It simplifies the process by transforming the equation into a form that's easier to integrate straightforwardly.

In our example, the integrating factor \( \mu(t) \) is calculated using the following formula:
  • \( \mu(t) = e^{\int P(t) \, dt} \)
  • Where \( P(t) = \frac{2}{t} \)
This results in an integrating factor of \( t^2 \). When you multiply the entire differential equation by \( t^2 \), it aligns the terms such that the left-hand side becomes the derivative of a product, simplifying the integration process.
Linear First-Order Differential Equation
A linear first-order differential equation has the standard form \( \frac{dy}{dt} + P(t)y = Q(t) \). In such equations, the highest derivative is of the first order, and the equation is linear in function \( y \).

For our given equation, after simplifying, it becomes:
  • \( \frac{dy}{dt} + \frac{2}{t}y = t^2 \)
This puts it into a format where methods like the integrating factor technique can be employed efficiently. It's crucial to rewrite the equation in this standardized form to apply such systematic approaches to solve it.
Initial Condition
Initial conditions are specific values that the solution of a differential equation must satisfy. They enable us to find a particular solution from a family of possible solutions.

In the problem at hand, the initial condition is \( y(1) = 0 \). This means, when \( t = 1 \), the value of \( y \) must be 0.

Using this information, we determine the constant \( C \) after integrating our equation:
  • Substitute \( t = 1 \) and \( y = 0 \) into \( y = \frac{t^3}{5} + \frac{C}{t^2} \)
This step is crucial to adjust our general solution such that it meets real-world criteria for specific scenarios.

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Most popular questions from this chapter

\(15-20\) Solve the initial-value problem. $$x y^{\prime}=y+x^{2} \sin x, \quad y(\pi)=0$$

Each system of differential equations is a model for two species that either compete for the same resources or cooperate for mutual benefit (flowering plants and insect pollinators, for instance). Decide whether each system describes competition or cooperation and explain why it is a reasonable model. (Ask yourself what effect an increase in one species has on the growth rate of the other.) $$\begin{aligned} \text { (a) } \frac{d x}{d t} &=0.12 x-0.0006 x^{2}+0.00001 x y \\ \frac{d y}{d t} &=0.08 x+0.00004 x y \end{aligned}$$ $$\begin{aligned} \text { (b) } \frac{d x}{d t} &=0.15 x-0.0002 x^{2}-0.0006 x y \\ \frac{d y}{d t} &=0.2 y-0.00008 y^{2}-0.0002 x y \end{aligned}$$

Solve the equation \(y^{\prime}=x \sqrt{x^{2}+1} /\left(y e^{y}\right)\) and graph several members of the family of solutions (if your CAS does implicit plots). How does the solution curve change as the constant \(C\) varies?

Suppose that a population grows according to a logistic model with carrying capacity 6000 and \(k=0.0015\) per year. (a) Write the logistic differential equation for these data. (b) Draw a direction field (either by hand or with a computer algebra system). What does it tell you about the solution curves? (c) Use the direction field to sketch the solution curves for initial populations of \(1000,2000,4000,\) and \(8000 .\) What can you say about the concavity of these curves? What is the significance of the inflection points? (d) Program a calculator or computer to use Euler's method with step size \(h=1\) to estimate the population after 50 years if the initial population is \(1000 .\) (e) If the initial population is \(1000,\) write a formula for the population after \(t\) years. Use it to find the population after 50 years and compare with your estimate in part (d). (f) Graph the solution in part (e) and compare with the solu- tion curve you sketched in part (c).

\(11-14\) Sketch the direction field of the differential equation. Then use it to sketch a solution curve that passes through the given point. $$y^{\prime}=1-x y, \quad(0,0)$$
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