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Chapter 8: Problem 6

Find the general solution of the following equations. $$y^{\prime}(x)=-y+2$$

Short Answer

Expert verified
Question: Find the general solution of the given first-order linear differential equation: \(y'(x) + y = 2\). Answer: The general solution of the given differential equation is \(y(x) = 2 + \frac{C}{e^{x}}\), where \(C\) is the constant of integration.

Step by step solution

01

Identify the type of differential equation

The given equation is a first-order linear differential equation as it can be written in the form \(y'(x) + P(x)y = Q(x)\).
02

Write the equation in standard form

The given equation can be written in standard form as: $$y'(x) + y = 2$$ Here, \(P(x) = 1\) and \(Q(x) = 2\).
03

Find the integrating factor

The integrating factor is given by \(IF(x) = e^{\int P(x) dx}\). In this case, \(P(x)=1\). So, the integrating factor will be: $$IF(x) = e^{\int 1 dx} = e^{x}$$
04

Multiply the equation by the integrating factor

We multiply the entire equation by the integrating factor \(e^x\): $$e^{x}y'(x) + e^{x}y = 2e^{x}$$
05

Integrate both sides of the equation with respect to x

Now integrate both sides of the equation with respect to \(x\): $$\int (e^{x}y'(x) + e^{x}y) dx = \int 2e^{x} dx$$ The left side of the equation can be recognized as the derivative of the product \(y(x)e^x\): $$\frac{d}{dx}(y(x)e^{x}) = 2e^{x}$$ Now integrate both sides: $$y(x)e^{x} = 2\int e^{x} dx$$ Integration of the right side gives: $$y(x)e^{x} = 2e^{x} + C$$ Here, \(C\) is the constant of integration.
06

Solve for y

Finally, we solve for \(y\) by dividing both sides by \(e^x\): $$y(x) = 2 + \frac{C}{e^{x}}$$ This is the general solution of the given differential equation.

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Most popular questions from this chapter

A special class of first-order linear equations have the form \(a(t) y^{\prime}(t)+a^{\prime}(t) y(t)=f(t),\) where \(a\) and \(f\) are given functions of \(t.\) Notice that the left side of this equation can be written as the derivative of a product, so the equation has the form $$a(t) y^{\prime}(t)+a^{\prime}(t) y(t)=\frac{d}{d t}(a(t) y(t))=f(t)$$ Therefore, the equation can be solved by integrating both sides with respect to \(t.\) Use this idea to solve the following initial value problems. $$t^{3} y^{\prime}(t)+3 t^{2} y=\frac{1+t}{t}, y(1)=6$$

Suppose Euler's method is applied to the initial value problem \(y^{\prime}(t)=a y, y(0)=1,\) which has the exact solution \(y(t)=e^{a t} .\) For this exercise, let \(h\) denote the time step (rather than \(\Delta t\) ). The grid points are then given by \(t_{k}=k h .\) We let \(u_{k}\) be the Euler approximation to the exact solution \(y\left(t_{k}\right),\) for \(k=0,1,2, \ldots\) a. Show that Euler's method applied to this problem can be written \(u_{0}=1, u_{k+1}=(1+a h) u_{k},\) for \(k=0,1,2, \ldots\) b. Show by substitution that \(u_{k}=(1+a h)^{k}\) is a solution of the equations in part (a), for \(k=0,1,2, \dots\) c. \(\lim _{h \rightarrow 0}(1+a h)^{1 / h}=e^{a} .\) Use this fact to show that as the time step goes to zero \(\left(h \rightarrow 0, \text { with } t_{k}=k h \text { fixed }\right),\) the approximations given by Euler's method approach the exact solution of the initial value problem; that is, \(\lim _{h \rightarrow 0} u_{k}=\lim _{h \rightarrow 0}(1+a h)^{k}=y\left(t_{k}\right)=e^{a t_{k}}\).

Suppose an object with an initial temperature of \(T_{0}>0\) is put in surroundings with an ambient temperature of \(A\) where \(A<\frac{T_{0}}{2}\). Let \(t_{1 / 2}\) be the time required for the object to cool to \(\frac{T_{0}}{2}\) a. Show that \(t_{1 / 2}=-\frac{1}{k} \ln \left[\frac{T_{0}-2 A}{2\left(T_{0}-A\right)}\right]\) b. Does \(t_{1 / 2}\) increase or decrease as \(k\) increases? Explain. c. Why is the condition \(A<\frac{T_{0}}{2}\) needed?

Let \(y(t)\) be the concentration of a substance in a chemical reaction (typical units are moles/liter). The change in the concentration, under appropriate conditions, is modeled by the equation \(\frac{d y}{d t}=-k y^{n},\) where \(k>0\) is a rate constant and the positive integer \(n\) is the order of the reaction. a. Show that for a first-order reaction \((n=1)\), the concentration obeys an exponential decay law. b. Solve the initial value problem for a second-order reaction \((n=2)\) assuming \(y(0)=y_{0}\) c. Graph the concentration for a first-order and second-order reaction with \(k=0.1\) and \(y_{0}=1\)

An endowment is an investment account in which the balance ideally remains constant and withdrawals are made on the interest earned by the account. Such an account may be modeled by the initial value problem \(B^{\prime}(t)=r B-m,\) for \(t \geq 0,\) with \(B(0)=B_{0} .\) The constant \(r>0\) reflects the annual interest rate, \(m>0\) is the annual rate of withdrawal, \(B_{0}\) is the initial balance in the account, and \(t\) is measured in years. a. Solve the initial value problem with \(r=0.05, m=\$ 1000 /\) year, and \(B_{0}=\$ 15,000 .\) Does the balance in the account increase or decrease? b. If \(r=0.05\) and \(B_{0}=\$ 50,000,\) what is the annual withdrawal rate \(m\) that ensures a constant balance in the account? What is the constant balance?
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