/*! 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 7 Solve the differential equations... [FREE SOLUTION] | 91Ó°ÊÓ

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Solve the differential equations. \(2 y^{\prime}=e^{x / 2}+y\)

Short Answer

Expert verified
The solution is \(y = e^{\frac{1}{2}x} \left( \frac{x}{2} + C \right)\).

Step by step solution

01

Rewrite the Differential Equation

Begin by rearranging terms in the given differential equation \(2 y^{\prime} = e^{x / 2} + y\). Move the term involving \(y\) to the left side to isolate \(y^{\prime}\) as follows: \(2 y^{\prime} - y = e^{x / 2}\). Then rewrite it in standard linear first-order form: \(y^{\prime} - \frac{1}{2}y = \frac{1}{2} e^{x / 2}\).
02

Identify Integrating Factor

For a linear differential equation of the form \(y^{\prime} + P(x) y = Q(x)\), the integrating factor \(\mu(x)\) is given by \(e^{\int P(x) \, dx}\). Here, \(P(x) = -\frac{1}{2}\), so the integrating factor is: \[ \mu(x) = e^{-\int \frac{1}{2} \, dx} = e^{-\frac{1}{2}x} \].
03

Multiply the Differential Equation by the Integrating Factor

Multiply the entire differential equation by \(\mu(x) = e^{-\frac{1}{2}x}\). You get:\[e^{-\frac{1}{2}x} y^{\prime} - \frac{1}{2} e^{-\frac{1}{2}x} y = \frac{1}{2} e^{0} \].Notice that the left-hand side is the derivative of a product: \((e^{-\frac{1}{2}x} y)^{\prime}\).
04

Integrate Both Sides

Observe that \((e^{-\frac{1}{2}x} y)^{\prime} = \frac{1}{2}\). Integrate both sides with respect to \(x\):\(\int (e^{-\frac{1}{2}x} y)^{\prime} \, dx = \int \frac{1}{2} \, dx\), leading to:\[e^{-\frac{1}{2}x} y = \frac{x}{2} + C\], where \(C\) is the integration constant.
05

Solve for y

To find \(y\), multiply through by \(e^{\frac{1}{2}x}\) to get:\[y = e^{\frac{1}{2}x} \left( \frac{x}{2} + C \right)\].This represents the general solution for the differential equation.

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Key Concepts

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

Integrating Factor
An integrating factor is a crucial tool for solving first-order linear differential equations. It is used to transform a non-exact differential equation into an exact one, enabling easier integration. The goal is to simplify the equation into a form where the derivative of a product appears, allowing for straightforward integration.
To find the integrating factor, consider a general first-order linear differential equation:
  • \( y^{\prime} + P(x) y = Q(x) \)
Here, \( P(x) \) is a function of \( x \). The integrating factor \( \mu(x) \) is defined as:\[\mu(x) = e^{\int P(x) \, dx}\]By multiplying the entire differential equation by \( \mu(x) \), the equation becomes exact, making it easier to find a solution. In our problem, \( P(x) = -\frac{1}{2} \), so the integrating factor is:\[\mu(x) = e^{-\frac{1}{2}x}\]Using this integrating factor, the differential equation becomes more manageable and transforms into a derivative of a product.
General Solution
Finding the general solution to a differential equation involves determining a function or a set of functions that satisfy the equation. After applying the integrating factor, the equation is now in a form:\[(e^{-\frac{1}{2}x} y)^{\prime} = \frac{1}{2}\]The left side of the equation appears as a derivative of a product, which is a key indicator that integration can be performed straightforwardly. Integrating both sides gives:\[e^{-\frac{1}{2}x} y = \frac{x}{2} + C\]where \( C \) is an integration constant. This process transforms the differential equation into the general solution form. To express \( y \) explicitly, simply solve for \( y \) by multiplying both sides by \( e^{\frac{1}{2}x} \):
  • \( y = e^{\frac{1}{2}x} \left( \frac{x}{2} + C \right) \)
This is the general solution and represents all possible solutions to the differential equation, encompassing different scenarios depending on \( C \).
Integration Constant
The integration constant, represented as \( C \) in the general solution, is a vital addition when integrating both sides of a differential equation. When integrating, the constant accounts for all possible initial conditions and scenarios that may be applied to the general solution.
Every differential equation represents a family of solutions, where each specific solution within that family corresponds to a particular initial condition or integration constant. For example, in the general solution:\[ y = e^{\frac{1}{2}x} \left( \frac{x}{2} + C \right)\]the integration constant \( C \) allows this equation to adapt to different initial values by providing a unique result for each value of \( C \).
Determining \( C \) often depends on additional information, such as initial conditions provided within a specific problem setting.

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Most popular questions from this chapter

Use a CAS to explore graphically each of the differential equations.Perform the following steps to help with your explorations. a. Plot a slope field for the differential equation in the given \(x y\) -window. b. Find the general solution of the differential equation using your CAS DE solver. c. Graph the solutions for the values of the arbitrary constant \(C=-2,-1,0,1,2\) superimposed on your slope field plot. d. Find and graph the solution that satisfies the specified initial condition over the interval \([0, b].\) e. Find the Euler numerical approximation to the solution of the initial value problem with 4 subintervals of the \(x\) -interval and plot the Euler approximation superimposed on the graph produced in part (d). f. Repeat part (e) for \(8,16,\) and 32 subintervals. Plot these three Euler approximations superimposed on the graph from part (e). g. Find the error ( \(y\) (exact) \(-y\) (Euler)) at the specified point \(x=b\) for each of your four Euler approximations. Discuss the improvement in the percentage error.$$y^{\prime}=-x / y, \quad y(0)=2 ; \quad-3 \leq x \leq 3, \quad-3 \leq y \leq 3 ; b=2.$$

Show that the second-order differential equation \(y^{\prime \prime}=F\left(x, y, y^{\prime}\right)\) can be reduced to a system of two first-order differential equations $$\begin{array}{l}\frac{d y}{d x}=z \\\\\frac{d z}{d x}=F(x, y, z)\end{array}$$ Can something similar be done to the \(n\) th-order differential equation \(y^{(n)}=F\left(x, y, y^{\prime}, y^{\prime \prime}, \ldots, y^{(n-1)}\right) ?\)

An economic model Consider the following economic model. Let \(P\) be the price of a single item on the market. Let \(Q\) be the quantity of the item available on the market. Both \(P\) and \(Q\) are functions of time. If one considers price and quantity as two interacting species, the following model might be proposed: $$\begin{aligned}\frac{d P}{d t} &=a P\left(\frac{b}{Q}-P\right) \\\\\frac{d Q}{d t} &=c Q(f P-Q)\end{aligned}$$ where \(a, b, c,\) and \(f\) are positive constants. Justify and discuss the adequacy of the model. a. If \(a=1, b=20,000, c=1,\) and \(f=30,\) find the equilibrium points of this system. If possible, classify each equilibrium point with respect to its stability. If a point cannot be readily classified, give some explanation. b. Perform a graphical stability analysis to determine what will happen to the levels of \(P\) and \(Q\) as time increases. c. Give an economic interpretation of the curves that determine the equilibrium points.

Use Euler's method to calculate the first three approximations to the given initial value problem for the specified increment size. Calculate the exact solution and investigate the accuracy of your approximations. Round your results to four decimal places. $$y^{\prime}=x(1-y), \quad y(1)=0, \quad d x=0.2$$

Write an equivalent first-order differential equatio and initial condition for \(y\). $$y=1+\int_{0}^{x} y(t) d t.$$

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