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17. SECOND-ORDER DIFFERENTIAL EQUATIONS. SECOND-ORDER DIFFERENTIAL EQUATIONS. 17.2 Nonhomogeneous Linear Equations. In this section, we will learn how to solve: Second-order nonhomogeneous linear differential equations with constant coefficients. NONHOMOGENEOUS LNR. EQNS. Equation 1.

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SECOND-ORDER DIFFERENTIAL EQUATIONS

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Second order differential equations

17

SECOND-ORDER DIFFERENTIAL EQUATIONS


Second order differential equations

SECOND-ORDER DIFFERENTIAL EQUATIONS

17.2

Nonhomogeneous Linear Equations

  • In this section, we will learn how to solve:

  • Second-order nonhomogeneous linear

  • differential equations with constant coefficients.


Nonhomogeneous lnr eqns

NONHOMOGENEOUS LNR. EQNS.

Equation 1

  • Second-order nonhomogeneous linear differential equations with constant coefficients are equations of the form ay’’ + by’ + cy = G(x)

  • where:

    • a, b, and c are constants.

    • G is a continuous function.


Complementary equation

COMPLEMENTARY EQUATION

Equation 2

  • The related homogeneous equation ay’’ + by’ + cy = 0is called the complementary equation.

    • It plays an important role in the solution of the original nonhomogeneous equation 1.


Nonhomogeneous lnr eqns1

NONHOMOGENEOUS LNR. EQNS.

Theorem 3

  • The general solution of the nonhomogeneous differential equation 1 can be written as y(x) = yp(x) + yc(x)

  • where:

    • yp is a particular solution of Equation 1.

    • yc is the general solution of Equation 2.


Nonhomogeneous lnr eqns2

NONHOMOGENEOUS LNR. EQNS.

Proof

  • All we have to do is verify that, if y is any solution of Equation 1, then y – yp is a solution of the complementary Equation 2.

    • Indeed, a(y – yp)’’ + b(y – yp)’ + c(y – yp) = ay’’ – ayp’’ + by’ – byp’ + cy – cyp= (ay’’ + by’ + cy) – (ayp’’ + byp’ + cyp) = g(x) – g(x) = 0


Nonhomogeneous lnr eqns3

NONHOMOGENEOUS LNR. EQNS.

  • We know from Section 17.1 how to solve the complementary equation.

    • Recall that the solution is: yc = c1y1 + c2y2 where y1 and y2 are linearly independent solutions of Equation 2.


Nonhomogeneous lnr eqns4

NONHOMOGENEOUS LNR. EQNS.

  • Thus, Theorem 3 says that:

    • We know the general solution of the nonhomogeneous equation as soon as we know a particular solution yp.


Methods to find particular solution

METHODS TO FIND PARTICULAR SOLUTION

  • There are two methods for finding a particular solution:

    • The method of undetermined coefficients is straightforward, but works only for a restricted class of functions G.

    • The method of variation of parameters works for every function G, but is usually more difficult to apply in practice.


Undetermined coefficients

UNDETERMINED COEFFICIENTS

  • We first illustrate the method of undetermined coefficients for the equation

  • ay’’ + by’ + cy = G(x)

  • where G(x) is a polynomial.


Undetermined coefficients1

UNDETERMINED COEFFICIENTS

  • It is reasonable to guess that there is a particular solution yp that is a polynomial of the same degree as G:

    • If y is a polynomial, then ay’’ + by’ + cyis also a polynomial.


Undetermined coefficients2

UNDETERMINED COEFFICIENTS

  • Thus, we substitute yp(x) = a polynomial (of the same degree as G) into the differential equation and determine the coefficients.


Undetermined coefficients3

UNDETERMINED COEFFICIENTS

Example 1

  • Solve the equation y’’ + y’ – 2y = x2

    • The auxiliary equation of y’’ + y’ – 2y = 0 is: r2 + r – 2 = (r – 1)(r + 2) = 0 with roots r = 1, –2.

    • So, the solution of the complementary equation is: yc = c1ex + c2e–2x


Undetermined coefficients4

UNDETERMINED COEFFICIENTS

Example 1

  • Since G(x) = x2 is a polynomial of degree 2, we seek a particular solution of the form

  • yp(x) = Ax2 + Bx + C

  • Then,

    • yp’ = 2Ax + B

    • yp’’ = 2A


Undetermined coefficients5

UNDETERMINED COEFFICIENTS

Example 1

  • So, substituting into the given differential equation, we have:

  • (2A) + (2Ax +B) – 2(Ax2 + Bx +C) = x2

  • or

  • –2Ax2 + (2A – 2B)x + (2A +B – 2C) = x2


Undetermined coefficients6

UNDETERMINED COEFFICIENTS

Example 1

  • Polynomials are equal when their coefficients are equal.

  • Thus, –2A = 1 2A – 2B = 0 2A + B – 2C = 0

    • The solution of this system of equations is: A = –½ B = –½ C = –¾


Undetermined coefficients7

UNDETERMINED COEFFICIENTS

Example 1

  • A particular solution, therefore, is: yp(x) = –½x2 –½x – ¾

  • By Theorem 3, the general solution is: y = yc + yp = c1ex + c2e-2x – ½x2 – ½x – ¾


Undetermined coefficients8

UNDETERMINED COEFFICIENTS

  • Suppose G(x) (right side of Equation 1) is of the form Cekx, where C and k are constants.

  • Then, we take as a trial solution a function of the same form, yp(x) = Aekx.

    • This is because the derivatives of ekxare constant multiples of ekx.


Undetermined coefficients9

UNDETERMINED COEFFICIENTS

  • The figure shows four solutions of the differential equation in Example 1 in terms of:

    • The particular solution yp

    • The functions f(x) = ex and g(x) = e–2x


Undetermined coefficients10

UNDETERMINED COEFFICIENTS

Example 2

  • Solve y’’ + 4y = e3x

    • The auxiliary equation is:

      r2 + 4 = 0

      with roots ±2i.

    • So, the solution of the complementary equation is: yc(x) = c1 cos 2x + c2 sin 2x


Undetermined coefficients11

UNDETERMINED COEFFICIENTS

Example 2

  • For a particular solution, we try: yp(x) = Ae3x

  • Then,

    • yp’ = 3Ae3x

    • yp’’ = 9Ae3x


Undetermined coefficients12

UNDETERMINED COEFFICIENTS

Example 2

  • Substituting into the differential equation, we have:

  • 9Ae3x + 4(Ae3x) = e3x

    • So,13Ae3x = e3xand A = 1/13


Undetermined coefficients13

UNDETERMINED COEFFICIENTS

Example 2

  • Thus, a particular solution is:

  • yp(x) = 1/13 e3x

  • The general solution is:

  • y(x) = c1 cos 2x + c2 sin 2x + 1/13 e3x


Undetermined coefficients14

UNDETERMINED COEFFICIENTS

  • Suppose G(x) is either C cos kx or C sin kx.

    • Then, because of the rules for differentiating the sine and cosine functions, we take as a trial particular solution a function of the form yp(x) = A cos kx + B sin kx


Undetermined coefficients15

UNDETERMINED COEFFICIENTS

  • The figure shows solutions of the differential equation in Example 2 in terms of yp and the functions f(x) = cos 2x and g(x) = sin 2x.


Undetermined coefficients16

UNDETERMINED COEFFICIENTS

  • Notice that:

    • All solutions approach ∞as x→ ∞.

    • All solutions (except yp) resemble sine functions when x is negative.


Undetermined coefficients17

UNDETERMINED COEFFICIENTS

Example 3

  • Solve y’’ + y’ – 2y = sin x

    • We try a particular solutionyp(x) = A cos x + B sin x

    • Then, yp’ = –A sin x +B cos xyp’’ = –A cos x – B sin x


Undetermined coefficients18

UNDETERMINED COEFFICIENTS

Example 3

  • So, substitution in the differential equation gives: (–A cos x – B sin x) + (–A sin x +B cos x) – 2(A cos x +B sin x) = sin x

  • or

  • (–3A +B) cos x + (–A – 3B) sin x = sin x


Undetermined coefficients19

UNDETERMINED COEFFICIENTS

Example 3

  • This is true if:

  • –3A + B = 0 and –A – 3B = 1

    • The solution of this system is: A = –1/10 B = –3/10

    • So, a particular solution is:

      yp(x) = –1/10 cos x – 3/10 sin x


Undetermined coefficients20

UNDETERMINED COEFFICIENTS

Example 3

  • In Example 1, we determined that the solution of the complementary equation is:

  • yc = c1ex + c2e–2x

    • So, the general solution of the given equation is: y(x) = c1ex + c2e–2x – 1/10 (cos x – 3 sin x)


Undetermined coefficients21

UNDETERMINED COEFFICIENTS

  • If G(x) is a product of functions of the preceding types, we take the trial solution to be a product of functions of the same type.

    • For instance, in solving the differential equation y’’ + 2y’ + 4y =x cos 3x we could try yp(x) = (Ax +B) cos 3x + (Cx +D) sin 3x


Undetermined coefficients22

UNDETERMINED COEFFICIENTS

  • If G(x) is a sum of functions of these types, we use the principle of superposition, which says that:

    • If yp1 and yp2 are solutions of ay’’ + by’ + cy =G1(x) ay’’ + by’ + cy =G2(x)respectively, then yp1 + yp2 is a solution of ay’’ + by’ + cy =G1(x) + G2(x)


Undetermined coefficients23

UNDETERMINED COEFFICIENTS

Example 4

  • Solve y’’ – 4y =xex + cos 2x

    • The auxiliary equation is: r2 – 4 = 0 with roots ±2.

    • So, the solution of the complementary equation is: yc(x) = c1e2x +c2e–2x


Undetermined coefficients24

UNDETERMINED COEFFICIENTS

Example 4

  • For the equation y’’ – 4y = xex, we try:

  • yp1(x) = (Ax +B)ex

  • Then,

    • y’p1= (Ax +A + B)ex

    • y’’p1= (Ax + 2A + B)ex


Undetermined coefficients25

UNDETERMINED COEFFICIENTS

Example 4

  • So, substitution in the equation gives:

  • (Ax + 2A +B)ex – 4(Ax +B)ex =xex

  • or

  • (–3Ax + 2A – 3B)ex = xex


Undetermined coefficients26

UNDETERMINED COEFFICIENTS

Example 4

  • Thus, –3A = 1 and 2A – 3B = 0

    • So, A = –⅓, B = –2/9, and yp1(x) = (–⅓x – 2/9)ex


Undetermined coefficients27

UNDETERMINED COEFFICIENTS

Example 4

  • For the equation y’’ – 4y = cos 2x, we try:yp2(x) = C cos 2x +D sin 2x

    • Substitution gives: –4C cos 2x – 4D sin 2x – 4(C cos 2x +D sin 2x) = cos 2x or – 8C cos 2x – 8D sin 2x = cos 2x


Undetermined coefficients28

UNDETERMINED COEFFICIENTS

Example 4

  • Thus, –8C = 1, –8D = 0, andyp2(x) = –1/8 cos 2x

    • By the superposition principle, the general solution is: y =yc +yp1 + yp2

      = c1e2x +c2e-2x– (1/3 x + 2/9)ex–1/8 cos 2x


Undetermined coefficients29

UNDETERMINED COEFFICIENTS

  • Here, we show the particular solution yp = yp1 + yp2 of the differential equation in Example 4.

    • The other solutions are given in terms of f(x) = e2xand g(x) = e–2x.


Undetermined coefficients30

UNDETERMINED COEFFICIENTS

  • Finally, we note that the recommended trial solution yp sometimes turns out to be a solution of the complementary equation.

    • So, it can’t be a solution of the nonhomogeneous equation.

    • In such cases, we multiply the recommended trial solution by x (or by x2 if necessary) so that no term in yp(x) is a solution of the complementary equation.


Undetermined coefficients31

UNDETERMINED COEFFICIENTS

Example 5

  • Solve y’’ + y = sin x

    • The auxiliary equation is: r2 + 1 = 0 with roots ±i.

    • So, the solution of the complementary equation is: yc(x) = c1 cos x + c2 sin x


Undetermined coefficients32

UNDETERMINED COEFFICIENTS

Example 5

  • Ordinarily, we would use the trial solution

  • yp(x) = A cos x +B sin x

    • However, we observe that it is a solution of the complementary equation.

    • So, instead, we try: yp(x) = Ax cos x +Bx sin x


Undetermined coefficients33

UNDETERMINED COEFFICIENTS

Example 5

  • Then,

    • yp’(x) = A cos x –Ax sin x +B sin x +Bx cos x

    • yp’’(x) = –2A sin x –Ax cos x + 2B cos x –Bx sin x


Undetermined coefficients34

UNDETERMINED COEFFICIENTS

Example 5

  • Substitution in the differential equation gives:

  • yp’’ + yp = –2A sin x + 2B cos x = sin x

    • So, A = –½ , B = 0, and yp(x) = –½x cos x

    • The general solution is: y(x) = c1 cos x +c2 sin x – ½ x cos x


Undetermined coefficients35

UNDETERMINED COEFFICIENTS

  • The graphs of four solutions of the differential equation in Example 5 are shown here.


Undetermined coefficients36

UNDETERMINED COEFFICIENTS

  • We summarize the method of undetermined coefficients as follows.


Summary part 1

SUMMARY—PART 1

  • If G(x) = ekxP(x), where P is a polynomial of degree n, then try:

  • yp(x) = ekxQ(x)

  • where Q(x) is an nth-degree polynomial (whose coefficients are determined by substituting in the differential equation).


Summary part 2

SUMMARY—PART 2

  • If G(x) = ekxP(x)cos mx or G(x) = ekxP(x) sin mx

  • where P is an nth-degree polynomial, then try:

  • yp(x) = ekxQ(x) cos mx +ekxR(x) sin mx

  • where Q and R are nth-degree polynomials.


Summary modification

SUMMARY—MODIFICATION

  • If any term of yp is a solution of the complementary equation, multiply ypby x (or by x2 if necessary).


Undetermined coefficients37

UNDETERMINED COEFFICIENTS

Example 6

  • Determine the form of the trial solution for the differential equation

  • y’’ – 4y’ + 13y =e2xcos 3x


Undetermined coefficients38

UNDETERMINED COEFFICIENTS

Example 6

  • G(x) has the form of part 2 of the summary, where k = 2, m = 3, and P(x) = 1.

    • So, at first glance, the form of the trial solution would be: yp(x) = e2x(A cos 3x +B sin 3x)


Undetermined coefficients39

UNDETERMINED COEFFICIENTS

Example 6

  • However, the auxiliary equation is: r2 – 4r + 13 = 0 with roots r = 2 ± 3i.

    • So, the solution of the complementary equation is: yc(x) = e2x(c1 cos 3x +c2 sin 3x)


Undetermined coefficients40

UNDETERMINED COEFFICIENTS

Example 6

  • This means that we have to multiply the suggested trial solution by x.

    • So, instead, we use: yp(x) = xe2x(A cos 3x +B sin 3x)


Variation of parameters

VARIATION OF PARAMETERS

Equation 4

  • Suppose we have already solved the homogeneous equation ay’’ + by’ + cy = 0 and written the solution as:

  • y(x) = c1y1(x) + c2y2(x)

  • where y1 and y2 are linearly independent solutions.


Variation of parameters1

VARIATION OF PARAMETERS

  • Let’s replace the constants (or parameters) c1 and c2 in Equation 4 by arbitrary functions u1(x)and u2(x).


Variation of parameters2

VARIATION OF PARAMETERS

Equation 5

  • We look for a particular solution of the nonhomogeneous equation ay’’ + by’ + cy = G(x) of the form

  • yp(x) = u1(x) y1(x) + u2(x) y2(x)


Variation of parameters3

VARIATION OF PARAMETERS

  • This method is called variation of parameters because we have varied the parameters c1 and c2 to make them functions.


Variation of parameters4

VARIATION OF PARAMETERS

Equation 6

  • Differentiating Equation 5, we get:

  • yp’ = (u1’y1 + u2’y2) + (u1y1’ + u2y2’)


Variation of parameters5

VARIATION OF PARAMETERS

  • Since u1 and u2 are arbitrary functions, we can impose two conditions on them.

    • One condition is that yp is a solution of the differential equation.

    • We can choose the other condition so as to simplify our calculations.


Variation of parameters6

VARIATION OF PARAMETERS

Equation 7

  • In view of the expression in Equation 6, let’s impose the condition that:

  • u1’y1 + u2’y2 = 0

    • Then, yp’’ = u1’y1’ + u2’y2’ + u1y1’’ + u2y2’’


Variation of parameters7

VARIATION OF PARAMETERS

Equation 8

  • Substituting in the differential equation, we get: a(u1’y1’ + u2’y2’ + u1y1’’ + u2y2’’) + b(u1y1’ + u2y2’) + c(u1y1 + u2y2) = Goru1(ay1” + by1’ + cy1) + u2(ay2” + by2” + cy2) + a(u1’y1’ + u2’y2’) = G


Variation of parameters8

VARIATION OF PARAMETERS

  • However, y1 and y2 are solutions of the complementary equation.

  • So, ay1’’ + by1’ + cy1 = 0 and ay2’’ + by2’ + cy2 = 0


Variation of parameters9

VARIATION OF PARAMETERS

Equation 9

  • Thus, Equation 8 simplifies to: a(u1’y1’ + u2’y2’) = G


Variation of parameters10

VARIATION OF PARAMETERS

  • Equations 7 and 9 form a system of two equations in the unknown functions u1’ and u2’.

    • After solving this system, we may be able to integrate to find u1 and u2 .

    • Then, the particular solution is given by Equation 5.


Variation of parameters11

VARIATION OF PARAMETERS

Example 7

  • Solve the equation y’’ + y = tan x, 0 < x < π/2

    • The auxiliary equation is:r2 + 1 = 0 with roots ±i.

    • So, the solution of y’’ + y = 0 is: c1 sin x +c2 cos x


Variation of parameters12

VARIATION OF PARAMETERS

Example 7

  • Using variation of parameters, we seek a solution of the form

  • yp(x) = u1(x) sin x +u2(x) cos x

    • Then, yp’ = (u1’ sin x +u2’ cos x) + (u1 cos x –u2 sin x)


Variation of parameters13

VARIATION OF PARAMETERS

E. g. 7—Equation 10

  • Set

  • u1’ sin x +u2’ cos x = 0

    • Then,yp’’ = u1’ cos x –u2’ sin x –u1 sin x –u2 cos x


Variation of parameters14

VARIATION OF PARAMETERS

E. g. 7—Equation 11

  • For yp to be a solution, we must have:yp’’ + yp =u1’ cos x –u2’ sin x = tan x


Variation of parameters15

VARIATION OF PARAMETERS

Example 7

  • Solving Equations 10 and 11, we get:

  • u1’(sin2x + cos2x) = cos x tan x

  • u1’ = sin x u1(x) = –cos x

    • We seek a particular solution.

    • So, we don’t need a constant of integration here.


Variation of parameters16

VARIATION OF PARAMETERS

Example 7

  • Then, from Equation 10, we obtain:


Variation of parameters17

VARIATION OF PARAMETERS

Example 7

  • So,u2(x) = sin x –ln(sec x + tan x)

    • Note that: sec x + tan x > 0 for 0 < x < π/2


Variation of parameters18

VARIATION OF PARAMETERS

Example 7

  • Therefore,

  • yp(x) = –cos x sin x + [sin x –ln(sec x + tan x)] cos x

  • = –cos xln(sec x + tan x)

    • The general solution is: y(x) = c1 sin x +c2 cos x – cos x ln(sec x + tan x)


Variation of parameters19

VARIATION OF PARAMETERS

  • The figure shows four solutions of the differential equation in Example 7.


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