/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 Solve the following differential... [FREE SOLUTION] | 91Ó°ÊÓ

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Solve the following differential equations: $$ \frac{d y}{d t}=\frac{t^{2} y^{2}}{t^{3}+8} $$

Short Answer

Expert verified
y = -\frac{1}{\frac{1}{3} \ln |t^3 + 8| + C}

Step by step solution

01

- Identify the type of differential equation

The given differential equation is \( \frac{dy}{dt} = \frac{t^2 y^2}{t^3 + 8} \). It is a first-order, non-linear differential equation.
02

- Separate the variables

Rewrite the equation to separate the variables y and t: \[ \frac{1}{y^2} dy = \frac{t^2}{t^3 + 8} dt \]
03

- Integrate both sides

Integrate both sides of the equation: \[ \int \frac{1}{y^2} dy = \int \frac{t^2}{t^3 + 8} dt \]
04

- Solve the left side integral

The left side integral simplifies to: \[ \int y^{-2} dy = -y^{-1} = -\frac{1}{y} \]
05

- Solve the right side integral

For the right side, use substitution. Let \(u = t^3 + 8 \) so \( du = 3t^2 dt \) and the integral becomes: \[ \int \frac{t^2}{t^3 + 8} dt = \int \frac{1}{3} \frac{1}{u} du = \frac{1}{3} \int \frac{1}{u} du = \frac{1}{3} \ln |u| = \frac{1}{3} \ln |t^3 + 8| \]
06

- Combine and solve for y

Combining both integrals, we get: \[ -\frac{1}{y} = \frac{1}{3} \ln |t^3 + 8| + C \] where C is the integration constant. Rearranging for y gives: \[ y = -\frac{1}{\frac{1}{3} \ln |t^3 + 8| + C} \]

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Key Concepts

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

Separation of Variables
Separation of variables is a method used to solve first-order differential equations. The key idea is to rewrite the equation so that each variable and its derivative are on separate sides of the equation. This technique works best when the differential equation is in a form where the variables can be separated easily.
For example, in the given equation \( \frac{d y}{d t}=\frac{t^{2} y^{2}}{t^{3}+8} \), we start by rewriting it as: \[ \frac{1}{y^2} dy = \frac{t^2}{t^3 + 8} dt \]
In this form, all terms involving y are on one side, and all terms involving t are on the other side of the equation. This separation allows us to integrate each side independently. Always ensure that both sides of the equation are separable before proceeding. You can then solve each part separately to find expressions for y and t.
Integration by Substitution
Integration by substitution is often necessary when dealing with integrals that are not straightforward. The idea is to make a substitution that simplifies the integral into a more manageable form.
In our example, after separating the variables, we arrived at two integrals: \[ \int \frac{1}{y^2} dy = \int \frac{t^2}{t^3 + 8} dt \]
While the integral on the left is straightforward, the right side requires substitution. Let \ u = t^3 + 8 \; then \ du = 3t^2 dt \.Rewriting the integral in terms of u gives us: \[ \int \frac{t^2}{t^3 + 8} dt = \int \frac{1}{3} \frac{1}{u} du \]
This substitution simplifies the integral substantially, making it possible to integrate: \[ \int \frac{1}{u} du = \ln |u| = \ln |t^3 + 8| \]
Remembering to account for the constant of integration, we ensure to substitute back any terms to return to the original variable for the final solution.
Non-linear Differential Equations
Non-linear differential equations are equations where the unknown function and its derivatives appear to powers greater than one or in products. These can be more complicated to solve compared to linear differential equations because standard methods for linear equations do not typically apply.
In our example, \( \frac{d y}{d t}=\frac{t^{2} y^{2}}{t^{3}+8} \) is non-linear because the term \ y^2 \ appears. Despite this complexity, separating variables or other techniques like substitution can still work effectively.
When you encounter a non-linear differential equation, look for ways to simplify it. Techniques like separation of variables, as shown here, or more advanced methods like integrating factors for non-linear equations, might be required.
These methods often transform the non-linear equation into a form that is easier to integrate or solve. Each problem may require a different approach, so being familiar with various methods is useful.

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Most popular questions from this chapter

Solve the initial-value problem. $$ y^{\prime}+2 y \cos (2 t)=2 \cos (2 t), y\left(\frac{\pi}{2}\right)=0 $$

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.)

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. $$ y^{\prime}=y^{3}, y(0)=-1, y(0)=1 $$

Solve the given equation using an integrating factor. Take \(t>0\). $$ y^{\prime}-2 t y=-4 t $$

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. \(y^{\prime}=\frac{1}{2} y^{2}-3 y, y(0)=3, y(0)=6, y(0)=9\)

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