/*! 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 19 Find the general solution of eac... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the general solution of each differential equation. Use \(C, C_{1}, C_{2}, \ldots .\) to denote arbitrary constants. $$y^{\prime \prime}(t)=60 t^{4}-4+12 t^{-3}$$

Short Answer

Expert verified
Question: Find the general solution of the second-order differential equation: \(y^{\prime \prime}(t)=60 t^{4}-4+12 t^{-3}\). Answer: The general solution of the given differential equation is \(y(t) = 2t^6-2t^2+\frac{C_1}{3}t^3+Ct+C_2\), where \(C_1\) and \(C_2\) are arbitrary constants.

Step by step solution

01

Identify the given differential equation

The given differential equation is: $$y^{\prime \prime}(t)=60 t^{4}-4+12 t^{-3}$$
02

Integrate the first time

Integrate the differential equation with respect to \(t\): $$ \int y^{\prime \prime}(t) \, dt= \int (60 t^{4}-4+12 t^{-3}) \, dt$$ $$y'(t) = \int 60t^4 \, dt - \int 4 \, dt + \int 12t^{-3} \, dt$$ Now integrate each term separately: $$y'(t) = 12t^5 - 4t + C_1t^2 + C$$ Where \(C_1\) and \(C\) are constants.
03

Integrate the second time

Integrate \(y'(t)\) with respect to \(t\): $$ y(t) = \int y'(t) \, dt = \int (12t^5 - 4t + C_1t^2 + C) \, dt$$ Now integrate each term separately: $$ y(t) = \frac{12}{6}t^6 - \frac{4}{2}t^2 + \frac{C_1}{3}t^3+ Ct + C_2$$ Where \(C_2\) is an arbitrary constant.
04

Write down the general solution

The general solution of the given differential equation is: $$y(t) = 2t^6-2t^2+\frac{C_1}{3}t^3+Ct+C_2$$

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 91Ó°ÊÓ!

Key Concepts

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

Integration
When solving differential equations, integration is a key mathematical tool. It allows us to reverse the process of differentiation and find the original function given its derivative. In the exercise, we had the second derivative of a function, denoted as \(y''(t)\). To find the original function \(y(t)\), we performed integration twice. First, we integrated \(y''(t)\) to get \(y'(t)\).This involved integrating each term separately, as shown below:
  • The term \(60t^4\) became \(12t^5\).
  • The constant \(-4\) became \(-4t\) since the integral of a constant \(a\) with respect to \(t\) is \(at\).
  • The term \(12t^{-3}\) became \(-6t^{-2}\) since the integral of \(t^n\) is \(\frac{t^{n+1}}{n+1}\), assuming \(n eq -1\).
Once we found \(y'(t)\), we integrated it again to find \(y(t)\), the general solution. Each term was integrated independently:
  • \(12t^5\) became \(2t^6\).
  • \(-4t\) became \(-2t^2\).
  • \(C_1t^2\) involved a simple power rule integration.
  • \(C\) became \(Ct\).
Arbitrary Constants
Arbitrary constants play a crucial role when dealing with differential equations. They arise from the integration processes. Every time you integrate, you introduce a new arbitrary constant because integration is the inverse operation of differentiation, which destroys specific constant values.In the original equation, upon the first integration of \(y''(t)\), we introduce an arbitrary constant \(C\). Upon the second integration of \(y'(t)\), we introduce another arbitrary constant \(C_2\). These constants are named arbitrarily as \(C_1, C_2, \ldots\), and they reflect unknown initial conditions or particular solutions that will be determined if additional information is provided, such as initial values or boundary conditions.
General Solution
The general solution of a differential equation incorporates all possible solutions, represented with arbitrary constants. This gives the solution flexibility to fit different conditions.For example, in the exercise, the general solution is expressed as:\[ y(t) = 2t^6 - 2t^2 + \frac{C_1}{3}t^3 + Ct + C_2 \]Here, \(C_1\), \(C\), and \(C_2\) are arbitrary constants. The structure of the solution means it can be tailored to meet specific criteria or initial conditions by determining particular values for these constants.
  • It reflects the capability to fit various scenarios by achieving different particular solutions.
  • If we had specific initial conditions like \(y(0) = 5\), we would substitute \(t = 0\) in the general solution, set it equal to 5, and solve for one of the arbitrary constants.
This flexibility makes the general solution extremely valuable in both theoretical and practical scenarios, wherever modeling of real-world situations is necessary.

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

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. If the growth rate function for a population model is positive, then the population is increasing. b. The solution of a stirred tank initial value problem always approaches a constant as \(t \rightarrow \infty\) c. In the predator-prey models discussed in this section, if the initial predator population is zero and the initial prey population is positive, then the prey population increases without bound.

For each of the following stirred tank reactions, carry out the following analysis. a. Write an initial value problem for the mass of the substance. b. Solve the initial value problem and graph the solution to be sure that \(m(0)\) and \(\lim _{t \rightarrow \infty} m(t)\) are correct. A \(1500-L\) tank is initially filled with a solution that contains \(3000 \mathrm{g}\) of salt. A salt solution with a concentration of \(20 \mathrm{g} / \mathrm{L}\) flows into the tank at a rate of \(3 \mathrm{L} / \mathrm{min.}\) The thoroughly mixed solution is drained from the tank at a rate of \(3 \mathrm{L} / \mathrm{min}\)

Solve the following initial value problems and leave the solution in implicit form. Use graphing software to plot the solution. If the implicit solution describes more than one curve, be sure to indicate which curve corresponds to the solution of the initial value problem. $$y y^{\prime}(x)=\frac{2 x}{\left(2+y^{2}\right)^{2}}, y(1)=-1$$

Solve the differential equation for Newton's Law of Cooling to find the temperature in the following cases. Then answer any additional questions. A glass of milk is moved from a refrigerator with a temperature of \(5^{\circ} \mathrm{C}\) to a room with a temperature of \(20^{\circ} \mathrm{C}\). One minute later the milk has warmed to a temperature of \(7^{\circ} \mathrm{C}\). After how many minutes does the milk have a temperature that is \(90 \%\) of the ambient temperature?

A differential equation of the form \(y^{\prime}(t)=f(y)\) is said to be autonomous (the function \(f\) depends only on \(y\) ). The constant function \(y=y_{0}\) is an equilibrium solution of the equation provided \(f\left(y_{0}\right)=0\) (because then \(y^{\prime}(t)=0\) and the solution remains constant for all \(t\) ). Note that equilibrium solutions correspond to horizontal lines in the direction field. Note also that for autonomous equations, the direction field is independent of t. Carry out the following analysis on the given equations. a. Find the equilibrium solutions. b. Sketch the direction field, for \(t \geq 0\). c. Sketch the solution curve that corresponds to the initial condition \(y(0)=1\). $$y^{\prime}(t)=y(y-3)(y+2)$$

See all solutions

Recommended explanations on Math Textbooks

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.