/*! 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 4 \(2 t+\left(1+y^{3}\right) y^{\p... [FREE SOLUTION] | 91Ó°ÊÓ

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\(2 t+\left(1+y^{3}\right) y^{\prime}=0, \quad y(1)=1\)

Short Answer

Expert verified
Answer: The general solution of the given differential equation is \(y(t) = \tan\left(3t^2 -3+\pi y\right)\).

Step by step solution

01

Rewrite the differential equation in terms of \(y\) and \(t\)

Divide both sides by \((1+y^3)\) to isolate \(y'\) and obtain \(y'=-\frac{2t}{1+y^3}\).
02

Separate the variables

Rewrite the equation to have \(y\) terms on one side and \(t\) terms on the other side: \(-\frac{1}{1+y^3} dy = 2t dt\).
03

Integrate both sides

Integrate \(-\frac{1}{1+y^3} dy = 2t dt\) with respect to \(y\): \(\int -\frac{1}{1+y^3} dy = \int 2t dt\).
04

Solve the integrals

Compute the integrals: \(-\frac{1}{3}\arctan(y) = t^2 + C\).
05

Use the initial condition to find the value of the constant, \(C\)

Plug the initial condition \(y(1) = 1\) into the equation: \(-\frac{1}{3}\arctan(1) = 1^2 + C\). Solving for \(C\), we get \(C = -\frac{1}{3}+\frac{\pi}{12}\).
06

Write the general solution

The general solution of the differential equation is given by: \(y(t) = \tan\left(3t^2 -3+\pi y\right)\).

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Key Concepts

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

Separable Differential Equations
Separable differential equations are a class of ordinary differential equations (ODEs) that can be solved by separation of variables. This method involves rearranging the equation so that all the terms involving one variable are on one side of the equation, and all the terms involving the other variable are on the other side.

In the given example, the differential equation is written as \(2t + (1+y^3)y' = 0\). To solve this, we separate the variables \(y\) and \(t\), rewriting the equation as \(y'=-\frac{2t}{1+y^3}\). Here's how we do it step by step:
  • Rewrite the equation to isolate one of the differentials, such as \(dy\) or \(dt\).
  • Move all terms involving \(y\) to one side and all terms involving \(t\) to the other side, resulting in something akin to \(g(y)dy = f(t)dt\).
  • Integrate both sides to find a solution in terms of \(y\) and \(t\).

Separating variables is powerful because it transforms the problem into simpler integrals that are easier to evaluate. However, this method only works if the equation can be manipulated into a separable form. Not all differential equations are separable, but for those that are, this technique is a beautiful simplification.
Initial Value Problems
Initial value problems (IVPs) are a subset of differential equations where the solution is required to satisfy a specific condition at a given point. This initial condition allows us to determine the exact solution from a family of potential solutions.

In our case, the condition \(y(1)=1\) is provided. Here's the process of using this initial condition:
  • Integrate to find a general solution that includes a constant of integration, \(C\).
  • Apply the initial condition to determine the specific value of \(C\).
  • Substitute the value of \(C\) back into the general solution to obtain the particular solution that satisfies the IVP.

This targeted approach narrows down the infinite solutions to the one solution that passes through the point given by the initial condition, which is crucial for problems with practical applications where specific solutions are required.
Integrating Factors
The concept of integrating factors is a method used to solve certain types of non-separable linear first-order differential equations. An integrating factor is a function that is multiplied by a differential equation to make it easier to solve.

Although the differential equation in the provided exercise is already separable and does not require an integrating factor, the method is important to understand. Here's a general outline for when and how to use integrating factors:
  • Given a linear first-order ODE in the form \(y' + P(t)y = Q(t)\), find an integrating factor \(\mu(t)\) that depends on \(t\).
  • The integrating factor is generally \(\mu(t) = e^{\int P(t)dt}\). Multiplying the entire differential equation by this factor allows us to solve the equation.
  • After applying the integrating factor, the left side of the equation often becomes the derivative of a product, simplifying the equation significantly and allowing integration.

Understanding integrating factors is particularly valuable when dealing with equations that do not readily separate, as they offer another tool for finding solutions to complex problems.

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Most popular questions from this chapter

An auditorium is \(100 \mathrm{~m}\) in length, \(70 \mathrm{~m}\) in width, and \(20 \mathrm{~m}\) in height. It is ventilated by a system that feeds in fresh air and draws out air at the same rate. Assume that airborne impurities form a well-stirred mixture. The ventilation system is required to reduce air pollutants present at any instant to \(1 \%\) of their original concentration in \(30 \mathrm{~min}\). What inflow (and outflow) rate is required? What fraction of the total auditorium air volume must be vented per minute?

Let \(S(t)\) represent the amount of a chemical reactant present at time \(t, t \geq 0\). Assume that \(S(t)\) can be determined by solving the initial value problem $$ S^{\prime}=-\frac{\alpha S}{K+S}, \quad S(0)=S_{0}, $$ where \(\alpha, K\), and \(S_{0}\) are positive constants. Obtain an implicit solution of the initial value problem. (The differential equation, often referred to as the Michaelis-Menten equation, arises in the study of biochemical reactions.)

Use the ideas of Exercise 32 to solve the given initial value problem. Obtain an explicit solution if possible. $$ y^{\prime}=2 t+y+\frac{1}{2 t+y}, \quad y(1)=1 $$

Oscillating Inflow Concentration A tank initially contains \(10 \mathrm{lb}\) of salt dissolved in 200 gal of water. Assume that a salt solution flows into the tank at a rate of \(3 \mathrm{gal} / \mathrm{min}\) and the well-stirred mixture flows out at the same rate. Assume that the inflow concentration oscillates in time, however, and is given by \(c_{i}(t)=0.2(1+\sin t) \mathrm{lb}\) of salt per gallon. Thus, as time evolves, the concentration oscillates back and forth between 0 and \(0.4 \mathrm{lb}\) of salt per gallon. (a) Make a conjecture, on the basis of physical reasoning, as to whether or not you expect the amount of salt in the tank to reach a constant equilibrium value as time increases. In other words, will \(\lim _{t \rightarrow \infty} Q(t)\) exist? (b) Formulate the corresponding initial value problem. (c) Solve the initial value problem. (d) Plot \(Q(t)\) versus \(t\). How does the amount of salt in the tank vary as time becomes increasingly large? Is this behavior consistent with your intuition?

(a) Obtain an implicit solution and, if possible, an explicit solution of the initial value problem. (b) If you can find an explicit solution of the problem, determine the \(t\)-interval of existence. $$ (\ln y) y^{\prime}+t=1, \quad y(3)=e $$

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