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Solve the differential equation. $$ y^{\prime}=x-y $$

Short Answer

Expert verified
Solution: \( y = x - 1 + Ce^{-x} \).

Step by step solution

01

Identify the type of differential equation

The given differential equation is \( y' = x - y \). This is a first-order linear differential equation.
02

Rearrange the equation to the standard form

Rearrange the equation into the standard form \( y'+py=q \), where \( p = -1 \) and \( q = x \). The equation becomes \( y' + y = x \).
03

Find the integrating factor

The integrating factor \( \mu(x) \) is given by \( e^{\int p \, dx} \). Here \( p = 1 \), so \( \mu(x) = e^{\int 1 \, dx} = e^x \).
04

Multiply through by the integrating factor

Multiply the whole differential equation by the integrating factor \( e^x \): \( e^x y' + e^x y = e^x x \).
05

Recognize the left-hand side as a derivative

Notice that \( e^x y' + e^x y = \frac{d}{dx}(e^x y) \). This means we can rewrite the equation as \( \frac{d}{dx}(e^x y) = e^x x \).
06

Integrate both sides with respect to \( x \)

Integrate both sides: \( \int \frac{d}{dx}(e^x y) \, dx = \int e^x x \, dx \). This gives \( e^x y = \int e^x x \, dx \).
07

Solve the right-side integral

Use integration by parts for \( \int e^x x \, dx \). Let \( u = x \) and \( dv = e^x \, dx \). Then \( du = dx \) and \( v = e^x \). The integral becomes \( e^x x - \int e^x \, dx = e^x x - e^x + C \).
08

Divide by the integrating factor

Solve for \( y \) by dividing through by \( e^x \): \( y = x - 1 + Ce^{-x} \).
09

State the general solution

The general solution to the differential equation \( y' = x - y \) is \( y = x - 1 + Ce^{-x} \).

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Key Concepts

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

First-Order Linear Differential Equation
A first-order linear differential equation is a relationship between a function and its first derivative. It takes the form \(y' + p(x)y = q(x)\), where \(p(x)\) and \(q(x)\) are functions of \(x\).
In these equations, the highest order derivative is the first derivative, hence the term 'first-order'. The equation we have, \(y' = x - y\), fits this form once rearranged to \(y' + y = x\).
This type of differential equation is called linear because the function \(y\) and its derivative appear to the power of one, and purely in a linear form. They are important in mathematical modeling of systems and phenomena because they can describe a wide range of physical processes.
Integrating Factor
An integrating factor is a crucial method for solving linear first-order differential equations. It is typically denoted as \(\mu(x)\) and is used to simplify the equation.
To find the integrating factor, we compute \(\mu(x) = e^{\int p(x) \, dx}\), where \(p(x)\) is the coefficient of \(y\) in the standard form.
For the equation \(y' + y = x\), we identify \(p(x) = 1\). Thus, the integrating factor is \(e^x\).
By multiplying the entire differential equation by this factor, the left-hand side becomes an exact derivative, allowing straightforward integration. This simplifies the approach to finding a solution.
Integration by Parts
Integration by parts is a technique to integrate products of functions. It is based on the formula \[\int u \, dv = uv - \int v \, du\]. This method is particularly handy when dealing with an integral like \(\int e^x x \, dx\).
In the process for solving such integrals, we choose parts: \(u = x\) and \(dv = e^x \, dx\). Then, differentiate and integrate respectively to find \(du = dx\) and \(v = e^x\).
Substituting these into the formula, we split the original integral into simpler parts and compute: \(e^x x - \int e^x \, dx\). Solving this leads to \(e^x x - e^x + C\), which is key in solving the differential equation.
General Solution
The general solution of a differential equation is an expression that encompasses all possible solutions. For first-order linear equations, the general solution includes a constant \(C\), representing any particular solution dependent on initial conditions.
In our problem, after solving the differential equation \(y' = x - y\) using integrating factors and integration by parts, we arrive at \(y = x - 1 + Ce^{-x}\).
Here, \(x - 1\) is the particular solution to the non-homogeneous part, and \(Ce^{-x}\) is the solution to the homogeneous part (when \(x = 0\)).
The constant \(C\) allows for adjustment to specific initial conditions, ensuring the solution fits given criteria in a real-world context.

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Most popular questions from this chapter

A certain small country has dollar 10 billion in paper currency in circulation, and each day dollar 50 million comes into the country's banks. The government decides to introduce new currency by having the banks replace old bills with new ones whenever old currency comes into the banks. Let \(x=x(t)\) denote the amount of new currency in circulation at time \(t,\) with \(x(0)=0\) (a) Formulate a mathematical model in the form of an initial-value problem that represents the "flow" of the new currency into circulation. (b) Solve the initial-value problem found in part (a). (c) How long will it take for the new bills to account for \(90 \%\) of the currency in circulation?

In an elementary chemical reaction, single molecules of two reactants \(A\) and \(B\) form a molecule of the product \(C\) : \(A+B \rightarrow C\). The law of mass action states the rate of reaction is proportional to the product of the concentrations of A and \(B\) : \(\frac{d[\mathrm{C}]}{d t}=k[\mathrm{A}][\mathrm{B}]\) (See Example 3.7 .4 .) Thus, if the initial concentrations are \([\mathrm{A}]=a\) moles \(/ \mathrm{L}\) and \([\mathrm{B}]=b\) moles \(/ \mathrm{L}\) and we write \(x=[\mathrm{C}]\) then we have \(\frac{d x}{d t}=k(a-x)(b-x)\) (a) Assuming that \(a \neq b\), find \(x\) as a function of \(t .\) Use the fact that the initial concentration of \(\mathrm{C}\) is \(0 .\) (b) Find \(x(t)\) assuming that \(a=b .\) How does this expression for \(x(t)\) simplify if it is known that \([\mathrm{C}]=\frac{1}{2} a\) after 20 seconds?

Find an equation of the curve that passes through the point \((0,2)\) and whose slope at \((x, y)\) is \(x / y .\)

Sketch a direction field for the differential equation. Then use it to sketch three solution curves. $$ y^{\prime}=x-y+1 $$

In a seasonal-growth model, a periodic function of time is introduced to account for seasonal variations in the rate of growth. Such variations could, for example, be caused by seasonal changes in the availability of food. (a) Find the solution of the seasonal-growth model $$ \frac{d P}{d t}=k P \cos (r t-\phi) \quad P(0)=P_{0} $$ where \(k, r,\) and \(\phi\) are positive constants. (b) By graphing the solution for several values of \(k, r,\) and \(\phi,\) explain how the values of \(k, r,\) and \(\phi\) affect the solution. What can you say about \(\lim _{t \rightarrow \infty} P(t) ?\)

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