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Mathematics LibreTexts

17.2: Nonhomogeneous Linear Equations

  • Gilbert Strang & Edwin “Jed” Herman
  • OpenStax

( \newcommand{\kernel}{\mathrm{null}\,}\)

Learning Objectives
  • Write the general solution to a nonhomogeneous differential equation.
  • Solve a nonhomogeneous differential equation by the method of undetermined coefficients.
  • Solve a nonhomogeneous differential equation by the method of variation of parameters.

In this section, we examine how to solve nonhomogeneous differential equations. The terminology and methods are different from those we used for homogeneous equations, so let’s start by defining some new terms.

General Solution to a Nonhomogeneous Linear Equation

Consider the nonhomogeneous linear differential equation

a2(x)y+a1(x)y+a0(x)y=r(x).

The associated homogeneous equation

a2(x)y+a1(x)y+a0(x)y=0

is called the complementary equation. We will see that solving the complementary equation is an important step in solving a nonhomogeneous differential equation.

Definition: Particular Solution

A solution yp(x) of a differential equation that contains no arbitrary constants is called a particular solution to the equation.

GENERAL Solution TO A NONHOMOGENEOUS EQUATION

Let yp(x) be any particular solution to the nonhomogeneous linear differential equation

a2(x)y+a1(x)y+a0(x)y=r(x).

Also, let c1y1(x)+c2y2(x) denote the general solution to the complementary equation. Then, the general solution to the nonhomogeneous equation is given by

y(x)=c1y1(x)+c2y2(x)+yp(x).

Proof

To prove y(x) is the general solution, we must first show that it solves the differential equation and, second, that any solution to the differential equation can be written in that form. Substituting y(x) into the differential equation, we have

a2(x)y+a1(x)y+a0(x)y=a2(x)(c1y1+c2y2+yp)+a1(x)(c1y1+c2y2+yp)+a0(x)(c1y1+c2y2+yp)=[a2(x)(c1y1+c2y2)+a1(x)(c1y1+c2y2)+a0(x)(c1y1+c2y2)]+a2(x)yp+a1(x)yp+a0(x)yp=0+r(x)=r(x).

So y(x) is a solution.

Now, let z(x) be any solution to a2(x)y+a1(x)y+a0(x)y=r(x). Then

a2(x)(zyp)+a1(x)(zyp)+a0(x)(zyp)=(a2(x)z+a1(x)z+a0(x)z)(a2(x)yp+a1(x)yp+a0(x)yp)=r(x)r(x)=0,

so z(x)yp(x) is a solution to the complementary equation. But, c1y1(x)+c2y2(x) is the general solution to the complementary equation, so there are constants c1 and c2 such that

z(x)yp(x)=c1y1(x)+c2y2(x).

Hence, we see that

z(x)=c1y1(x)+c2y2(x)+yp(x).

Example 17.2.1: Verifying the General Solution

Given that yp(x)=x is a particular solution to the differential equation y+y=x, write the general solution and check by verifying that the solution satisfies the equation.

Solution

The complementary equation is y+y=0, which has the general solution c1cosx+c2sinx. So, the general solution to the nonhomogeneous equation is

y(x)=c1cosx+c2sinx+x.

To verify that this is a solution, substitute it into the differential equation. We have

y(x)=c1sinx+c2cosx+1

and

y(x)=c1cosxc2sinx.

Then

y(x)+y(x)=c1cosxc2sinx+c1cosx+c2sinx+x=x.

So, y(x) is a solution to y+y=x.

Exercise 17.2.1

Given that yp(x)=2 is a particular solution to y3y4y=8, write the general solution and verify that the general solution satisfies the equation.

Hint

Find the general solution to the complementary equation.

Answer

y(x)=c1ex+c2e4x2

In the preceding section, we learned how to solve homogeneous equations with constant coefficients. Therefore, for nonhomogeneous equations of the form ay+by+cy=r(x), we already know how to solve the complementary equation, and the problem boils down to finding a particular solution for the nonhomogeneous equation. We now examine two techniques for this: the method of undetermined coefficients and the method of variation of parameters.

Undetermined Coefficients

The method of undetermined coefficients involves making educated guesses about the form of the particular solution based on the form of r(x). When we take derivatives of polynomials, exponential functions, sines, and cosines, we get polynomials, exponential functions, sines, and cosines. So when r(x) has one of these forms, it is possible that the solution to the nonhomogeneous differential equation might take that same form. Let’s look at some examples to see how this works.

Example 17.2.2: Undetermined Coefficients When r(x) Is a Polynomial

Find the general solution to y+4y+3y=3x.

Solution

The complementary equation is y+4y+3y=0, with general solution c1ex+c2e3x. Since r(x)=3x, the particular solution might have the form yp(x)=Ax+B. If this is the case, then we have yp(x)=A and yp(x)=0. For yp to be a solution to the differential equation, we must find values for A and B such that

y+4y+3y=3x0+4(A)+3(Ax+B)=3x3Ax+(4A+3B)=3x.

Setting coefficients of like terms equal, we have

3A=34A+3B=0.

Then, A=1 and B=43, so yp(x)=x43 and the general solution is

y(x)=c1ex+c2e3x+x43.

In Example 17.2.2, notice that even though r(x) did not include a constant term, it was necessary for us to include the constant term in our guess. If we had assumed a solution of the form yp=Ax (with no constant term), we would not have been able to find a solution. (Verify this!) If the function r(x) is a polynomial, our guess for the particular solution should be a polynomial of the same degree, and it must include all lower-order terms, regardless of whether they are present in r(x).

Example 17.2.3: Undetermined Coefficients When r(x) Is an Exponential

Find the general solution to yy2y=2e3x.

Solution

The complementary equation is yy2y=0, with the general solution c1ex+c2e2x. Since r(x)=2e3x, the particular solution might have the form yp(x)=Ae3x. Then, we have yp(x)=3Ae3x and yp(x)=9Ae3x. For yp to be a solution to the differential equation, we must find a value for A such that

yy2y=2e3x9Ae3x3Ae3x2Ae3x=2e3x4Ae3x=2e3x.

So, 4A=2 and A=1/2. Then, yp(x)=(12)e3x, and the general solution is

y(x)=c1ex+c2e2x+12e3x.

Exercise 17.2.3

Find the general solution to y4y+4y=7sintcost.

Hint

Use yp(t)=Asint+Bcost as a guess for the particular solution.

Answer

y(t)=c1e2t+c2te2t+sint+cost

In the previous checkpoint, r(x) included both sine and cosine terms. However, even if r(x) included a sine term only or a cosine term only, both terms must be present in the guess. The method of undetermined coefficients also works with products of polynomials, exponentials, sines, and cosines. Some of the key forms of r(x) and the associated guesses for yp(x) are summarized in Table 17.2.1.

Table 17.2.1: Key Forms for the Method of Undetermined Coefficients
r(x) Initial guess for yp(x)
k (a constant) A (a constant)
ax+b Ax+B (Note: The guess must include both terms even if b=0.)
ax2+bx+c Ax2+Bx+C (Note: The guess must include all three terms even if b or c are zero.)
Higher-order polynomials Polynomial of the same order as r(x)
aeλx Aeλx
acosβx+bsinβx Acosβx+Bsinβx (Note: The guess must include both terms even if either a=0 or b=0.)
aeαxcosβx+beαxsinβx Aeαxcosβx+Beαxsinβx
(ax2+bx+c)eλx (Ax2+Bx+C)eλx
(a2x2+a1x+a0)cosβx+(b2x2+b1x+b0)sinβx (A2x2+A1x+A0)cosβx+(B2x2+B1x+B0)sinβx
(a2x2+a1x+a0)eαxcosβx+(b2x2+b1x+b0)eαxsinβx (A2x2+A1x+A0)eαxcosβx+(B2x2+B1x+B0)eαxsinβx

Keep in mind that there is a key pitfall to this method. Consider the differential equation y+5y+6y=3e2x. Based on the form of r(x), we guess a particular solution of the form yp(x)=Ae2x. But when we substitute this expression into the differential equation to find a value for A,we run into a problem. We have

yp(x)=2Ae2x

and

yp=4Ae2x,

so we want

y+5y+6y=3e2x4Ae2x+5(2Ae2x)+6Ae2x=3e2x4Ae2x10Ae2x+6Ae2x=3e2x0=3e2x,

which is not possible.

Looking closely, we see that, in this case, the general solution to the complementary equation is c1e2x+c2e3x. The exponential function in r(x) is actually a solution to the complementary equation, so, as we just saw, all the terms on the left side of the equation cancel out. We can still use the method of undetermined coefficients in this case, but we have to alter our guess by multiplying it by x. Using the new guess, yp(x)=Axe2x, we have

yp(x)=A(e2x2xe2x

and

yp(x)=4Ae2x+4Axe2x.

Substitution gives

y+5y+6y=3e2x(4Ae2x+4Axe2x)+5(Ae2x2Axe2x)+6Axe2x=3e2x4Ae2x+4Axe2x+5Ae2x10Axe2x+6Axe2x=3e2xAe2x=3e2x.

So, A=3 and yp(x)=3xe2x. This gives us the following general solution

y(x)=c1e2x+c2e3x+3xe2x.

Note that if xe2x were also a solution to the complementary equation, we would have to multiply by x again, and we would try yp(x)=Ax2e2x.

PROBLEM-SOLVING STRATEGY: METHOD OF UNDETERMINED COEFFICIENTS
  1. Solve the complementary equation and write down the general solution.
  2. Based on the form of r(x), make an initial guess for yp(x).
  3. Check whether any term in the guess foryp(x) is a solution to the complementary equation. If so, multiply the guess by x. Repeat this step until there are no terms in yp(x) that solve the complementary equation.
  4. Substitute yp(x) into the differential equation and equate like terms to find values for the unknown coefficients in yp(x).
  5. Add the general solution to the complementary equation and the particular solution you just found to obtain the general solution to the nonhomogeneous equation.
Example 17.2.3: Solving Nonhomogeneous Equations

Find the general solutions to the following differential equations.

  1. y9y=6cos3x
  2. x+2x+x=4et
  3. y2y+5y=10x23x3
  4. y3y=12t
Solution
  1. The complementary equation is y9y=0, which has the general solution c1e3x+c2e3x(step 1). Based on the form of r(x)=6cos3x, our initial guess for the particular solution is yp(x)=Acos3x+Bsin3x (step 2). None of the terms in yp(x) solve the complementary equation, so this is a valid guess (step 3).
    Now we want to find values for A and B, so substitute yp into the differential equation. We have

    yp(x)=3Asin3x+3Bcos3x and yp(x)=9Acos3x9Bsin3x,

    so we want to find values of A and B such that

    y9y=6cos3x9Acos3x9Bsin3x9(Acos3x+Bsin3x)=6cos3x18Acos3x18Bsin3x=6cos3x.

    Therefore,

    18A=618B=0.

    This gives A=13 and B=0, so yp(x)=(13)cos3x (step 4).
    Putting everything together, we have the general solution

    y(x)=c1e3x+c2e3x+13cos3x.

  2. The complementary equation is x+2x+x=0, which has the general solution c1et+c2tet (step 1). Based on the form r(t)=4et, our initial guess for the particular solution is xp(t)=Aet (step 2). However, we see that this guess solves the complementary equation, so we must multiply by t, which gives a new guess: xp(t)=Atet (step 3). Checking this new guess, we see that it, too, solves the complementary equation, so we must multiply by t again, which gives xp(t)=At2et (step 3 again). Now, checking this guess, we see that xp(t) does not solve the complementary equation, so this is a valid guess (step 3 yet again).
    We now want to find a value for A, so we substitute xp into the differential equation. We have

    xp(t)=At2et, soxp(t)=2AtetAt2et

    and xp(t)=2Aet2Atet(2AtetAt2et)=2Aet4Atet+At2et.
    Substituting into the differential equation, we want to find a value of A so that

    x+2x+x=4et2Aet4Atet+At2et+2(2AtetAt2et)+At2et=4et2Aet=4et.

    This gives A=2, so xp(t)=2t2et (step 4). Putting everything together, we have the general solution

    x(t)=c1et+c2tet+2t2et.

  3. The complementary equation is y2y+5y=0, which has the general solution c1excos2x+c2exsin2x (step 1). Based on the form r(x)=10x23x3, our initial guess for the particular solution is yp(x)=Ax2+Bx+C (step 2). None of the terms in yp(x) solve the complementary equation, so this is a valid guess (step 3). We now want to find values for A, B, and C, so we substitute yp into the differential equation. We have yp(x)=2Ax+B and yp(x)=2A, so we want to find values of A, B, and C such that

    y2y+5y=10x23x32A2(2Ax+B)+5(Ax2+Bx+C)=10x23x35Ax2+(5B4A)x+(5C2B+2A)=10x23x3.

    Therefore,

    5A=105B4A=35C2B+2A=3.

    This gives A=2, B=1, and C=1, so yp(x)=2x2+x1 (step 4). Putting everything together, we have the general solution

    y(x)=c1excos2x+c2exsin2x+2x2+x1.

  4. The complementary equation is y3y=0, which has the general solution c1e3t+c2 (step 1). Based on the form r(t)=−12t,r(t)=−12t, our initial guess for the particular solution is yp(t)=At+B (step 2). However, we see that the constant term in this guess solves the complementary equation, so we must multiply by t, which gives a new guess: yp(t)=At2+Bt (step 3). Checking this new guess, we see that none of the terms in yp(t) solve the complementary equation, so this is a valid guess (step 3 again). We now want to find values for A and B, so we substitute yp into the differential equation. We have yp(t)=2At+B and yp(t)=2A, so we want to find values of AA and BB such that

    y3y=12t2A3(2At+B)=12t6At+(2A3B)=12t.

    Therefore,

    6A=122A3B=0.

    This gives A=2 and B=4/3, so yp(t)=2t2+(4/3)t (step 4). Putting everything together, we have the general solution

    y(t)=c1e3t+c2+2t2+43t.

Exercise 17.2.3

Find the general solution to the following differential equations.

  1. y5y+4y=3ex
  2. y+y6y=52cos2t
Hint

Use the problem-solving strategy.

Answer a

y(x)=c1e4x+c2exxex

Answer b

y(t)=c1e3t+c2e2t5cos2t+sin2t

Variation of Parameters

Sometimes, r(x) is not a combination of polynomials, exponentials, or sines and cosines. When this is the case, the method of undetermined coefficients does not work, and we have to use another approach to find a particular solution to the differential equation. We use an approach called the method of variation of parameters.

To simplify our calculations a little, we are going to divide the differential equation through by a, so we have a leading coefficient of 1. Then the differential equation has the form

y+py+qy=r(x),

where p and q are constants.

If the general solution to the complementary equation is given by c1y1(x)+c2y2(x), we are going to look for a particular solution of the form

yp(x)=u(x)y1(x)+v(x)y2(x).

In this case, we use the two linearly independent solutions to the complementary equation to form our particular solution. However, we are assuming the coefficients are functions of x, rather than constants. We want to find functions u(x) and v(x) such that yp(x) satisfies the differential equation. We have

yp=uy1+vy2yp=uy1+uy1+vy2+vy2yp=(uy1+vy2)+uy1+uy1+vy2+vy2.

Substituting into the differential equation, we obtain

yp+pyp+qyp=[(uy1+vy2)+uy1+uy1+vy2+vy2]+p[uy1+uy1+vy2+vy2]+q[uy1+vy2]=u[y1+py1+qy1]+v[y2+py2+qy2]+(uy1+vy2)+p(uy1+vy2)+(uy1+vy2).

Note that y1 and y2 are solutions to the complementary equation, so the first two terms are zero. Thus, we have

(uy1+vy2)+p(uy1+vy2)+(uy1+vy2)=r(x).

If we simplify this equation by imposing the additional condition uy1+vy2=0, the first two terms are zero, and this reduces to uy1+vy2=r(x). So, with this additional condition, we have a system of two equations in two unknowns:

uy1+vy2=0uy1+vy2=r(x).

Solving this system gives us u and v, which we can integrate to find u and v.

Then, yp(x)=u(x)y1(x)+v(x)y2(x) is a particular solution to the differential equation. Solving this system of equations is sometimes challenging, so let’s take this opportunity to review Cramer’s rule, which allows us to solve the system of equations using determinants.

RULE: CRAMER’S RULE

The system of equations

a1z1+b1z2=r1a2z1+b2z2=r2

has a unique solution if and only if the determinant of the coefficients is not zero. In this case, the solution is given by

z1=r1b1r2b2a1b1a2b2andz2=a1r1a2r2a1b1a2b2.

Example 17.2.4: Using Cramer’s Rule

Use Cramer’s rule to solve the following system of equations.

x2z1+2xz2=0z13x2z2=2x

Solution

We have

a1(x)=x2a2(x)=1b1(x)=2xb2(x)=3x2r1(x)=0r2(x)=2x.

Then,a1b1a2b2=x22x13x2=3x42x

and

r1b1r2b2=02x2x3x2=04x2=4x2.

Thus,

z1=r1b1r2b2a1b1a2b2=4x23x42x=4x3x3+2.

In addition,

a1r1a2r2=x2012x=2x30=2x3.

Thus,

z2=a1r1a2r2a1b1a2b2=2x33x42x=2x23x3+2.

Exercise 17.2.4

Use Cramer’s rule to solve the following system of equations.

2xz13z2=0x2z1+4xz2=x+1

Hint

Use the process from the previous example.

Answer

z1=3x+311x2,z2=2x+211x

PROBLEM-SOLVING STRATEGY: METHOD OF VARIATION OF PARAMETERS
  1. Solve the complementary equation and write down the general solution c1y1(x)+c2y2(x).
  2. Use Cramer’s rule or another suitable technique to find functions u(x) and v(x) satisfying uy1+vy2=0uy1+vy2=r(x).
  3. Integrate u and v to find u(x) and v(x). Then, yp(x)=u(x)y1(x)+v(x)y2(x) is a particular solution to the equation.
  4. Add the general solution to the complementary equation and the particular solution found in step 3 to obtain the general solution to the nonhomogeneous equation.
Example 17.2.5: Using the Method of Variation of Parameters

Find the general solution to the following differential equations.

  1. y2y+y=ett2
  2. y+y=3sin2x
Solution
  1. The complementary equation is y2y+y=0 with associated general solution c1et+c2tet. Therefore, y1(t)=et and y2(t)=tet. Calculating the derivatives, we get y1(t)=et and y2(t)=et+tet (step 1). Then, we want to find functions u(t) and v(t) so that

    uet+vtet=0uet+v(et+tet)=ett2.

    Applying Cramer’s rule (Equation ???), we have

    u=0tetett2et+tetettetetet+tet=0tet(ett2)et(et+tet)ettet=e2tte2t=1t

    and

    v′= \dfrac{\begin{array}{|ll|}e^t 0 \\ e^t \frac{e^t}{t^2} \end{array} }{\begin{array}{|lc|}e^t te^t \\ e^t e^t+te^t \end{array} } =\dfrac{e^t(\frac{e^t}{t^2})}{e^{2t}}=\dfrac{1}{t^2}\quad(\text{step 2}). \nonumber

    Integrating, we get

    \begin{align*} u &=−\int \dfrac{1}{t}dt=− \ln|t| \\[4pt] v &=\int \dfrac{1}{t^2}dt=−\dfrac{1}{t} \tag{step 3} \end{align*}

    Then we have

    \begin{align*}y_p &=−e^t \ln|t|−\frac{1}{t}te^t \\[4pt] &=−e^t \ln |t|−e^t \tag{step 4}.\end{align*}

    The e^t term is a solution to the complementary equation, so we don’t need to carry that term into our general solution explicitly. The general solution is

    y(t)=c_1e^t+c_2te^t−e^t \ln |t| \tag{step 5}

  2. The complementary equation is y″+y=0 with associated general solution c_1 \cos x+c_2 \sin x. So, y_1(x)= \cos x and y_2(x)= \sin x (step 1). Then, we want to find functions u′(x) and v′(x) such that

    \begin{align*} u′ \cos x+v′ \sin x &=0 \\[4pt] −u′ \sin x+v′ \cos x &=3 \sin _2 x \end{align*}. \nonumber

    Applying Cramer’s rule, we have

    u′= \dfrac{\begin{array}{|cc|}0 \sin x \\ 3 \sin ^2 x \cos x \end{array}}{ \begin{array}{|cc|} \cos x \sin x \\ − \sin x \cos x \end{array}}=\dfrac{0−3 \sin^3 x}{ \cos ^2 x+ \sin ^2 x}=−3 \sin^3 x \nonumber

    and

    v′=\dfrac{\begin{array}{|cc|} \cos x 0 \\ - \sin x 3 \sin^2 x \end{array}}{ \begin{array}{|cc|} \cos x \sin x \\ − \sin x \cos x \end{array}}=\dfrac{ 3 \sin^2x \cos x}{ 1}=3 \sin^2 x \cos x( \text{step 2}). \nonumber

    Integrating first to find u, we get

    u=\int −3 \sin^3 x dx=−3 \bigg[ −\dfrac{1}{3} \sin ^2 x \cos x+\dfrac{2}{3} \int \sin x dx \bigg]= \sin^2 x \cos x+2 \cos x. \nonumber

    Now, we integrate to find v. Using substitution (with w= \sin x), we get

    v= \int 3 \sin ^2 x \cos x dx=\int 3w^2dw=w^3=sin^3x.\nonumber

    Then,

    \begin{align*}y_p &=(\sin^2 x \cos x+2 \cos x) \cos x+(\sin^3 x)\sin x \\[4pt] &=\sin_2 x \cos _2 x+2 \cos _2 x+ \sin _4x \\[4pt] &=2 \cos_2 x+ \sin_2 x(\cos^2 x+\sin ^2 x) & & (\text{step 4}). \\[4pt] &=2 \cos _2 x+\sin_2x \\[4pt] &= \cos _2 x+1 \end{align*}

    The general solution is

    y(x)=c_1 \cos x+c_2 \sin x+1+ \cos^2 x(\text{step 5}).\nonumber

Exercise \PageIndex{5}

Find the general solution to the following differential equations.

  1. y″+y= \sec x
  2. x″−2x′+x=\dfrac{e^t}{t}
Hint

Follow the problem-solving strategy.

Answer a

y(x)=c_1 \cos x+c_2 \sin x+ \cos x \ln| \cos x|+x \sin x

Answer b

x(t)=c_1e^t+c_2te^t+te^t \ln|t|

Key Concepts

  • To solve a nonhomogeneous linear second-order differential equation, first find the general solution to the complementary equation, then find a particular solution to the nonhomogeneous equation.
  • Let y_p(x) be any particular solution to the nonhomogeneous linear differential equation a_2(x)y''+a_1(x)y′+a_0(x)y=r(x), \nonumber and let c_1y_1(x)+c_2y_2(x) denote the general solution to the complementary equation. Then, the general solution to the nonhomogeneous equation is given by y(x)=c_1y_1(x)+c_2y_2(x)+y_p(x). \nonumber
  • When r(x) is a combination of polynomials, exponential functions, sines, and cosines, use the method of undetermined coefficients to find the particular solution. To use this method, assume a solution in the same form as r(x), multiplying by x as necessary until the assumed solution is linearly independent of the general solution to the complementary equation. Then, substitute the assumed solution into the differential equation to find values for the coefficients.
  • When r(x) is not a combination of polynomials, exponential functions, or sines and cosines, use the method of variation of parameters to find the particular solution. This method involves using Cramer’s rule or another suitable technique to find functions and v′(x) satisfying \begin{align*}u′y_1+v′y_2 &=0 \\[4pt] u′y_1′+v′y_2′ &=r(x). \end{align*} Then, y_p(x)=u(x)y_1(x)+v(x)y_2(x) is a particular solution to the differential equation.

Key Equations

  • Complementary equation
    a_2(x)y″+a_1(x)y′+a_0(x)y=0
  • General solution to a nonhomogeneous linear differential equation
    y(x)=c_1y_1(x)+c_2y_2(x)+y_p(x)

Glossary

complementary equation
for the nonhomogeneous linear differential equation a+2(x)y″+a_1(x)y′+a_0(x)y=r(x), the associated homogeneous equation, called the complementary equation, is a_2(x)y''+a_1(x)y′+a_0(x)y=0
method of undetermined coefficients
a method that involves making a guess about the form of the particular solution, then solving for the coefficients in the guess
method of variation of parameters
a method that involves looking for particular solutions in the form y_p(x)=u(x)y_1(x)+v(x)y_2(x), where y_1 and y_2 are linearly independent solutions to the complementary equations, and then solving a system of equations to find u(x) and v(x)
particular solution
a solution y_p(x) of a differential equation that contains no arbitrary constants

This page titled 17.2: Nonhomogeneous Linear Equations is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Gilbert Strang & Edwin “Jed” Herman (OpenStax) via source content that was edited to the style and standards of the LibreTexts platform.

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