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Chapter 10: Problem 12

Solve the following differential equations: $$ \left(1+t^{2}\right) y^{\prime}=t y^{2} $$

Short Answer

Expert verified
Separate variables, integrate both sides, solve for y. \( y = \ - \frac{2} {\ln( t^2 + 1) + K} \)

Step by step solution

01

Separate Variables

To solve the differential equation, separate the variables so that all terms involving y are on one side of the equation and all terms involving t are on the other side. Start with: \( (1+t^{2}) y^{\text{'} } = t y^{2} \).Rearrange the terms to get: \( \frac{dy}{y^2} = \frac{t}{1+t^2} dt \).
02

Integrate Both Sides

Integrate both sides of the separated equation. The left-hand side becomes: \( \ \ \int \frac{dy}{y^2} = \ -\frac{1}{y} \ \ + C_1 \ \ \) while the right-hand side becomes: \( \int \frac{t}{1+t^2} dt = \ \frac{1}{2} ln{|\ 1 + t^2 \ |} \ + C_2 \ \).
03

Solve for y

Combine the constants of integration: \( \ -\frac{1}{y} = \ \frac{1}{2} ln{|\ 1 + t^2 \ |} \ + C . \ \). Then solve for y: \( y = \ - \frac{2} {\ln( t^2 + 1) + K} \)

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Key Concepts

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

Separation of Variables
In solving differential equations, 'separation of variables' is a fundamental technique. This method helps transform a tough differential equation into simpler integrable parts. Let's break down this process:
Imagine our initial equation: \rightarrow \( (1+t^{2})\, y^{'} = t\, y^{2} \).To separate the variables, we need a strategy to isolate \( y \) terms on one side and \( t \) terms on the other.
Here's how you do it:
  • Divide both sides by \( y^{2} (1+t^{2}) \)
  • Rearrange to get all \( y \)-terms on one side: \( \frac{dy}{y^2} \).
  • Then get all \( t \)-terms on the other: \( \frac{t}{1+t^2} dt \).
Our equation transforms into: \( \frac{dy}{y^2} = \frac{t}{1+t^2}\, dt \).
This neat separation allows us to integrate both sides independently.
Integration
After separating variables, our next big step is 'integration'. Integrating each side solves for the unknown function. Consider our separated parts:
  • The left side: \( \int \frac{dy}{y^2} \).
  • The right side: \( \int \frac{t}{1+t^2}\, dt \).
First, integrate \( \frac{dy}{y^2} \): \rightarrow This integral equals \( -\frac{1}{y} \) + C1.
Next, integrate \( \frac{t}{1+t^2}\, dt \): \rightarrow This results in \( \frac{1}{2}\ln{| 1 + t^2 |} + C2 Thus, after integration, we get: \( -\frac{1}{y} = \frac{1}{2}\ln{| 1 + t^2 |} + C \). Integration helps us transition from a differential to a functional form, which we then solve for \( y \).
Constants of Integration
After integrating, we see constants appearing in our solution - these are 'constants of integration'.
But why do they show up?
When we integrate, we reverse differentiation, potentially losing constant values. These are 'constants of integration' (\( C \), \( C_1 \), \( C_2 \)).
In our example:
  • \( -\frac{1}{y} = \frac{1}{2}\ln{| 1 + t^2 |} + C \)
The individual constants ( \( C_1 \), \( C_2 \)) merged into a single constant (\( C \)) upon combining terms to simplify to: \( y = -\frac{2}{\ln(t^2 + 1) + K} \).

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Most popular questions from this chapter

One or more initial conditions are given for each differential equation in the following exercises. Use the qualitative theory of autonomous differential equations to sketch the graphs of the corresponding solutions. Include a \(y z\) -graph if one is not already provided. Always indicate the constant solutions on the \(t y\) -graph whether they are mentioned or not. \(y^{\prime}=\frac{1}{2} y^{2}-3 y, y(0)=3, y(0)=6, y(0)=9\)

Review concepts that are important in this section. In each exercise, sketch the graph of a function with the stated properties. Domain: \(0 \leq t \leq 6 ;(0,4)\) is on the graph; the slope is always negative, and the slope becomes more negative.

A model that describes the relationship between the price and the weekly sales of a product might have a form such as $$ \frac{d y}{d p}=-\frac{1}{2}\left(\frac{y}{p+3}\right), $$ where \(y\) is the volume of sales and \(p\) is the price per unit. That is, at any time, the rate of decrease of sales with respect to price is directly proportional to the sales level and inversely proportional to the sales price plus a constant. Solve this differential equation. (Figure 6 shows several typical solutions.)

The Los Angeles Zoo plans to transport a California sea lion to the San Diego 'Loo. 'The animal will be wrapped in a wet blanket during the trip. At any time \(t\), the blanket will lose water (due to evaporation) at a rate proportional to the amount \(f(t)\) of water in the blanket, with constant of proportinality \(k=-.3\). Initially, the blanket will contain 2 gallons of seawater. (a) Set up the differential equation satisfied by \(f(t)\). (b) Use Euler's method with \(n=2\) to estimate the amount of moisture in the blanket after 1 hour. (c) Solve the differential equation in part (a) and compute \(f(1)\). (d) Compare the answers in parts (b) and (c) and approximate the error in using Euler's method.

Let \(t\) represent the total number of hours that a truck driver spends during a year driving on a certain highway connecting two cities, and let \(p(t)\) represent the probability that the driver will have at least one accident during these \(t\) hours. Then, \(0 \leq p(t) \leq 1\), and \(1-p(t)\) represents the probability of not having an accident. Under ordinary conditions, the rate of increase in the probability of an accident (as a function of \(t)\) is proportional to the probability of not having an accident. Construct and solve a differential equation for this situation.
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