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Chapter 2: Problem 23

Find the general solution. \(y^{\prime}+y=t\)

Short Answer

Expert verified
Question: Find the general solution to the first-order linear differential equation: \(y^{\prime} + y = t\). Answer: The general solution to the given differential equation is \(y(t) = t - 1 + Ce^{-t}\), where C is the constant of integration.

Step by step solution

01

Find the integrating factor

The given differential equation is of the form \( y^{\prime} + p(t)y = q(t) \), with \(p(t) = 1\) and \(q(t) = t\). The integrating factor, denoted by \(I(t)\), is defined as \(I(t) = e^{\int p(t) \, dt}\). In this case, the integrating factor is: \(I(t) = e^{\int 1 \, dt} = e^{t}\)
02

Multiply the equation by the integrating factor

Now, let's multiply both sides of the equation by the integrating factor \(e^{t}\) to obtain our new equation: \((y^{\prime} + y)e^{t} = te^{t}\) Simplifying further and using product rule, we get: \((ye^{t})' = te^{t}\)
03

Integrate both sides of the equation

Now, we can integrate both sides of the equation to cancel out the derivative on the left side: \(\int (ye^{t})' \, dt = \int te^{t}\, dt\) Performing integration by parts on the right side: Let \(u = t\) and \(dv = e^{t}dt\). Then, \(du = dt\) and \(v = e^{t}\). Using the integration by parts formula \(\int u dv = uv - \int v du\), we get: \(ye^{t} = te^{t} - \int e^{t}\, dt\) Which then gives: \(ye^{t} = te^{t} - e^{t} + C\)
04

Solve for the general solution (y)

Finally, we can solve for the general solution by dividing both sides by \(e^{t}\): \(y = \frac{te^{t} - e^{t} + C}{e^{t}}\) Simplifying the expression, we have the general solution of the given differential equation as: \(y(t) = t - 1 + Ce^{-t}\)

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Key Concepts

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

Integrating Factor
The integrating factor is a crucial technique when dealing with linear first-order differential equations. A differential equation is termed linear when it aligns with the format:
  • \(y' + p(t)y = q(t)\)
In our problem, this translates to \(p(t) = 1\) and \(q(t) = t\).
Our mission is to simplify the integration process by transforming the equation into an easily integrable form.

The integrating factor \(I(t)\) is calculated as:
  • \(I(t) = e^{\int p(t) \, dt}\)
Since \(p(t) = 1\), this becomes:
  • \(I(t) = e^{\int 1 \, dt} = e^{t}\)
With this factor, the differential equation is adjusted:
  • \((y' + y)e^{t} = te^{t}\)
This manipulation creates an equation where the left side is the derivative of a product, \((ye^{t})'\), enabling easier integration.
General Solution
The objective is to find the general solution of our differential equation, providing a formula for all possible solutions.
After obtaining the integrated form using the integrating factor, we need to solve for \(y\) which is our dependent variable.

Following the integrating process, we get:
  • \(ye^{t} = te^{t} - e^{t} + C\)
To isolate \(y\), divide the entire equation by \(e^{t}\):
  • \(y = \frac{te^{t} - e^{t} + C}{e^{t}}\)
Simplifying this gives:
  • \(y(t) = t - 1 + Ce^{-t}\)
Here, \(C\) is an integration constant that encapsulates the infinite number of potential solutions to the differential equation. The term \(C e^{-t}\) accounts for the arbitrary nature of constants when solving differentials.
Integration by Parts
Integration by parts is an essential technique, especially when facing integrals involving a product of functions. Like in our exercise, we often meet situations requiring integration that may not be directly evident.

In our equation, we had to integrate \(te^{t}\).
Using integration by parts, we apply the formula:
  • \(\int u \, dv = uv - \int v \, du\)
Classically, select:
  • \(u = t\)
  • \(dv = e^{t}dt\)
This choice yields:
  • \(du = dt\)
  • \(v = e^{t}\)
Applying the integration by parts formula results in:
  • \(ye^{t} = te^{t} - e^{t} + C\)
This application transforms the problem into simpler parts that are easier to integrate, ultimately aiding in deriving the solution for \(y\). By recognizing when and how to use integration by parts, you can handle more complex integral expressions efficiently.

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Most popular questions from this chapter

(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. $$ (2 y-\sin y) y^{\prime}+t=\sin t, \quad y(0)=0 $$

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 \(v=d r / d t<0 .)\)

A tank initially holds 500 gal of a brine solution having a concentration of \(0.1 \mathrm{lb}\) of salt per gallon. At some instant, fresh water begins to enter the tank at a rate of 10 \(\mathrm{gal} / \mathrm{min}\) and the well- stirred mixture leaves at the same rate. How long will it take before the concentration of salt is reduced to \(0.01 \mathrm{lb} / \mathrm{gal}\) ?

Consider a population modeled by the initial value problem $$ \frac{d P}{d t}=(1-P) P+M, \quad P(0)=P_{0} $$ where the migration rate \(M\) is constant. [The model (8) is derived from equation (6) by setting the constants \(r\) and \(P_{*}\) to unity. We did this so that we can focus on the effect \(M\) has on the solutions.] For the given values of \(M\) and \(P(0)\), (a) Determine all the equilibrium populations (the nonnegative equilibrium solutions) of differential equation (8). As in Example 1, sketch a diagram showing those regions in the first quadrant of the \(t P\)-plane where the population is increasing \(\left[P^{\prime}(t)>0\right]\) and those regions where the population is decreasing \(\left[P^{\prime}(t)<0\right]\). (b) Describe the qualitative behavior of the solution as time increases. Use the information obtained in (a) as well as the insights provided by the figures in Exercises 11-13 (these figures provide specific but representative examples of the possibilities). $$ M=2, \quad P(0)=0 $$

An object undergoes one-dimensional motion along the \(x\)-axis subject to the given decelerating forces. At time \(t=0\), the object's position is \(x=0\) and its velocity is \(v=v_{0}\). In each case, the decelerating force is a function of the object's position \(x(t)\) or its velocity \(v(t)\) or both. Transform the problem into one having distance \(x\) as the independent variable. Determine the position \(x_{f}\) at which the object comes to rest. (If the object does not come to rest, \(x_{f}=\infty\).) $$ m \frac{d v}{d t}=-k x v^{2} $$
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