/*! 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 8 Solve the following differential... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 8

Solve the following differential equations: $$y^{\prime}=e^{4 y} t^{3}-e^{4 y}$$

Short Answer

Expert verified
The solution is \( y = -\frac{1}{4} \ln(-t^4 + 4t - C') \).

Step by step solution

01

Identify the type of differential equation

The given differential equation is \( y^{\prime}=e^{4 y} t^{3}-e^{4 y} \). It is a first-order ordinary differential equation.
02

Separate variables

Rewrite the equation to separate the variables y and t: \( y' = e^{4y}(t^{3} - 1) \). This can be rewritten as \( \frac{dy}{dt} = e^{4y}(t^{3} - 1) \). Divide both sides by \( e^{4y} \) and multiply by dt: \( \frac{dy}{e^{4y}} = (t^3 - 1) dt \).
03

Integrate both sides

Integrate the separated variables: \(\int \frac{dy}{e^{4y}} = \int (t^3 - 1) dt \). For the left side, use a substitution \( u = 4y \), then \( du = 4dy \), thus \( \frac{1}{4} \int e^{-u} du = \int (t^3 - 1) dt \). This simplifies to \( -\frac{1}{4} e^{-4y} = \frac{t^4}{4} - t + C \).
04

Solve for y

Rewrite the integrated form and solve for y: \( -\frac{1}{4} e^{-4y} = \frac{t^4}{4} - t + C \). Multiply both sides by -4: \( e^{-4y} = - t^4 + 4t - C' \) where \( C' = -4C \). Finally, take the natural logarithm on both sides to solve for y: \( -4y = \ln(-t^4 + 4t - C') \), and then \( y = -\frac{1}{4} \ln(-t^4 + 4t - C') \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

First-Order Differential Equations
First-order differential equations involve derivatives of the first degree. These equations take the form \(y' = f(t, y)\). They describe how the function y changes with respect to t and are foundational because they are often simpler to solve compared to higher-order equations.
For example, our problem starts with \(y' = e^{4y} t^{3} - e^{4y}\). Here, \(y'\) represents the first derivative of y with respect to t.
Separation of Variables
Separation of variables is a method used to solve a first-order differential equation. In this method, we attempt to rewrite the equation so that each variable appears on different sides of the equation. To achieve this, we express the differential equation in the form \(g(y) dy = h(t) dt\).
Our equation \(y' = e^{4y} t^{3} - e^{4y}\) is rewritten as \(\frac{dy}{dt} = e^{4y} (t^{3} - 1)\). By dividing both sides by \(e^{4y}\) and multiplying by \(dt\), we get: \(\frac{dy}{e^{4y}} = (t^3 - 1) dt\). This separation allows us to integrate each side with respect to their own variable.
Integration
Integration is used to solve the separated variables. We integrate both sides of the equation to find \(y\) as an explicit function of \(t\). Integration involves finding the antiderivative of a function.
Here, after separating variables, we integrate: \[ \int \frac{dy}{e^{4y}} = \int (t^3 - 1) dt \].
On the left, we use substitution to simplify the integral: let \(u = 4y\), then \(du = 4dy\), leading to \(\frac{1}{4} \int e^{-u} du\). This helps us solve the integral, simplifying the equation and making it easier to find y.
Natural Logarithm
The natural logarithm, denoted as \(\ln(x)\), is the inverse function of the exponential function with base \(e\). It's used to solve for a variable raised within an exponent in a differential equation.
After integrating our differential equation, we reach the form: \[-\frac{1}{4} e^{-4y} = \frac{t^4}{4} - t + C \].
To solve for \(y\), we multiply by -4, resulting in \( e^{-4y} = -t^4 + 4t - C'\). We then take the natural logarithm on both sides, which helps isolate \(y\): \(-4y = \ln(-t^4 + 4t - C')\).
Finally, we solve for \(y\): \(y = -\frac{1}{4} \ln(-t^4 + 4t - C')\). Using the natural logarithm simplifies solving equations with exponential components.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In an autocatalytic reaction, one substance is converted into a second substance in such a way that the second substance catalyzes its own formation. This is the process by which trypsinogen is converted into the enzyme trypsin. The reaction starts only in the presence of some trypsin, and each molecule of trypsinogen yields 1 molecule of trypsin. The rate of formation of trypsin is proportional to the product of the amounts of the two substances present. Set up the differential equation that is satisfied by \(y=f(t)\) the amount (number of molecules) of trypsin present at time \(t\) Sketch the solution. For what value of \(y\) is the reaction proceeding the fastest? [Note: Letting \(M\) be the total amount of the two substances, the amount of trypsinogen present at time \(t \text { is } M-f(t) .\)]

Solve the given equation using an integrating factor. Take \(t>0\). $$y^{\prime}+\frac{y}{10+t}=0$$

Solve the following differential equations with the given initial conditions. $$\frac{d y}{d t}=\left(\frac{1+t}{1+y}\right)^{2}, y(0)=2$$

Solve the initial-value problem. $$y^{\prime}=2(10-y), y(0)=1$$

The fish population in a pond with carrying capacity 1000 is modeled by the logistic equation $$ \frac{d N}{d t}=\frac{.4}{1000} N(1000-N) $$ Here, \(N(t)\) denotes the number of fish at time \(t\) in years. When the number of fish reached \(275,\) the owner of the pond decided to remove 75 fish per year. (a) Modify the differential equation to model the population of fish from the time it reached \(275 .\) (b) Plot several solution curves of the new equation, including the solution curve with \(N(0)=275.\) (c) Is the practice of catching 75 fish per year sustainable, or will it deplete the fish population in the pond? Will the size of the fish population ever come close to the carrying capacity of the pond?
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Statistics

Read Explanation

Logic and Functions

Read Explanation

Calculus

Read Explanation

Geometry

Read Explanation

Decision Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.