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Chapter 2: Problem 15

Determine the general form of the function \(M(t, y)\) or \(N(t, y)\) that will make the given differential equation exact. $$ N(t, y) y^{\prime}+t^{2}+y^{2} \sin t=0 $$

Short Answer

Expert verified
In summary, the given differential equation would be exact if the function N(t, y) has the general form: $$ N(t, y) = -2y \cos t + f(y) $$ where \(f(y)\) is an arbitrary function of \(y\) only.

Step by step solution

01

Identifying the given differential equation

In this problem, the given differential equation is: $$ N(t, y)y'+t^2+y^2 \sin t = 0 $$ Already, we know that \(M(t, y) = t^2 + y^2 \sin t\) and we are given that \(N(t, y) y' = -M(t, y)\), which implies that finding \(M(t, y)\) is not necessary. Therefore, our task is to find the form of \(N(t, y)\) that makes the given equation exact.
02

Finding partial derivatives

To find the general form for \(N(t, y)\), we will compute the partial derivatives of \(M\) and \(N\) with respect to \(y\) and \(t\), respectively: $$ \frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(t^2 + y^2 \sin t) $$ and $$ \frac{\partial N}{\partial t} = \frac{\partial}{\partial t}(N(t, y)). $$ Now, let's compute the partial derivative of \(M\) with respect to \(y\): $$ \frac{\partial M}{\partial y} = 2y \sin t $$
03

Finding the general form of \(N(t, y)\)

Recall that for the given equation to be exact, we must have \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial t}\). Using the partial derivative of \(M(t, y)\) that we found, we can write the condition as: $$ 2y \sin t = \frac{\partial N}{\partial t} $$ Integrating both sides with respect to \(t\), we get: $$ N(t, y) = \int 2y \sin t \, dt $$ Now, you can integrate the right side: $$ N(t, y) = 2y(-\cos t) + f(y) $$ Here, we have the general form of \(N(t, y)\) that makes the given differential equation exact: $$ N(t, y) = -2y \cos t + f(y) $$ where \(f(y)\) is an arbitrary function of \(y\) only.

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Key Concepts

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

Partial Derivatives
Partial derivatives are a fundamental concept in multivariable calculus. When dealing with functions of several variables, a partial derivative with respect to one variable measures how the function changes as that specific variable is varied, while keeping all other variables constant.
This concept is crucial when working with exact differential equations because:
  • We use partial derivatives to verify if a differential equation is exact.
  • For an equation of the form \(M(t, y)dt + N(t, y)dy = 0\), the partial derivatives \(\frac{\partial M}{\partial y}\) and \(\frac{\partial N}{\partial t}\) must be equal for the equation to be exact.

In our solution, calculating the partial derivative \(\frac{\partial M}{\partial y} = 2y \sin t\) helped establish the condition needed for exactness.
Integration
Integration is the process of finding the antiderivative or the integral of a function. It's a key step when solving differential equations, especially in determining unknown functions that make equations exact.
In the context of exact differential equations:
  • Once we find the necessary partial derivatives, we use integration to determine the missing functions.
  • For example, from the condition \(2y \sin t = \frac{\partial N}{\partial t}\), we integrate \(2y \sin t\) with respect to \(t\) to find \(N(t, y)\).
Performing integration here gives \(N(t, y) = -2y \cos t + f(y)\), which is essential in making the differential equation exact.
Function of Arbitrary Constants
Functions of arbitrary constants, often recognized as functions of arbitrary variables in this context, represent the undetermined part of an integral in a solution to a differential equation. This concept is most visible post-integration.
When integrating partial derivatives:
  • We include arbitrary functions in the integral, as these account for other possible solutions that lead to exactness.
  • In our solution, \(f(y)\) is such an arbitrary function relying solely on \(y\), indicating the generality of potential solutions.
This flexibility is crucial, as it implies a family of solutions that could make the original equation exact, reflecting the solution's comprehensiveness.

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Most popular questions from this chapter

A drag chute must be designed to reduce the speed of a 3000-lb dragster from 220 mph to \(50 \mathrm{mph}\) in \(4 \mathrm{sec}\). Assume that the drag force is proportional to the velocity. (a) What value of the drag coefficient \(k\) is needed to accomplish this? (b) How far will the dragster travel in the 4-sec interval?

The motion of a body of mass \(m\), gravitationally attracted to Earth in the presence of a resisting drag force proportional to the square of its velocity, is given by $$ m \frac{d v}{d t}=-\frac{G m M_{e}}{r^{2}}+\kappa v^{2} $$ [recall equation (13)]. In this equation, \(r\) is the radial distance of the body from the center of Earth, \(G\) is the universal gravitational constant, \(M_{e}\) is the mass of Earth, and \(v=d r / d t\). Note that the drag force is positive, since it acts in the positive \(r\) direction. (a) Assume that the body is released from rest at an altitude \(h\) above the surface of Earth. Recast the differential equation so that distance \(r\) is the independent variable. State an appropriate initial condition for the new problem. (b) Show that the impact velocity can be expressed as $$ v_{\text {impact }}=-\left[2 G M_{e} \int_{0}^{h} \frac{e^{-2(\kappa / m) s}}{\left(R_{e}+s\right)^{2}} d s\right]^{1 / 2}, $$ where \(R_{e}\) represents the radius of Earth. (The minus sign reflects the fact that \(v=d r / d t<0 .)\)

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?

Radiocarbon Dating Carbon-14 is a radioactive isotope of carbon produced in the upper atmosphere by radiation from the sun. Plants absorb carbon dioxide from the air, and living organisms, in turn, eat the plants. The ratio of normal carbon (carbon-12) to carbon- 14 in the air and in living things at any given time is nearly constant. When a living creature dies, however, the carbon- 14 begins to decrease as a result of radioactive decay. By comparing the ameunts of carton-14 and carbon12 present, the amount of carbon- 14 that has decayed can therefore be ascertained. Let \(Q(t)\) denote the amount of carbon- 14 present at time \(t\) after death. If we assume its behavior is modeled by the differential equation \(Q^{\prime}(c)=-k Q(f)\), then \(Q(l)=Q(0) e^{-k t}\). Knowing the half-life of carbon- 14 , we can determine the constant \(k\). Given a specimen to be dated, we can measure its radioactive content and deduce \(Q(t)\). Knowing the amount of carbon- 12 present enables us to determine \(Q(0)\). Therefore, we can use the solution of the differential equation \(Q(t)=Q(0) e^{-k r}\) to deduce the age, \(f\), of the radicactive sample. (a) The half-life of carbon- 14 is nominally 5730 years. Suppose remains have been found in which it is estimated that \(30 \%\) of the original amount of carbon-14 is present. Fstimate the age of the remains. (b) The half-life of carbon- 14 is not known precisely. Let tus assume that its half-life is \(5730 \pm 30\) years. Determine how this half-life uncertainty affects the age estimate you computed in (a); that is, what is the corresponding uncertainty in the age of the remains? (c) It is claimed that radiocarbon dating cannot be used to date objects older than about 60,000 years. To appreciate this practical limitation, compute the ratio \(Q(60,000) / Q(0)\), assuming a half-life of 5730 years.

A tank, containing 1000 gal of liquid, has a brine solution entering at a constant rate of \(2 \mathrm{gal} / \mathrm{min}\). The well-stirred solution leaves the tank at the same rate. The concentration within the tank is monitored and is found to be the function of time specified. In each exercise, determine (a) the amount of salt initially present within the tank. (b) the inflow concentration \(c_{i}(t)\), where \(c_{i}(t)\) denotes the concentration of salt in the brine solution flowing into the tank. $$c(t)=\frac{1}{20}\left(1-e^{-t / 500}\right) \mathrm{lb} / \mathrm{gal}$$
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