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MAT 303 Spring 2013 Calculus IV with Applications

Homework #6 Solutions

Problems

Section 3.1: 26, 34, 40, 46

Section 3.2: 2, 8, 10, 14, 18, 24, 30 3.1.26. Determine whether the functionsf(x) =2cosx+3sinxandg(x) =3cosx

2sinxare linearly dependent or linearly independent on the real line.Solution:We compute the Wronskian of these functions:

W(f,g) =f g

f 0g0 =2cosx+3sinx3cosx2sinx

2sinx+3cosx3sinx2cosx

= (2cosx+3sinx)(3sinx2cosx)(3cosx2sinx)(2sinx+3cosx) = (12cosxsinx9sin2x4cos2x) + (12cosxsinx9cos2x4sin2x) =13(sin2x+cos2x) =13 Since the Wronskian is the constant function 13, which is not the 0 function, these func- tions are linearly independent on the real line (and in fact on any subinterval of the real

line).3.1.34. Find the general solution of the DEy00+2y015y=0.Solution:Guessing the solutiony=erx, we obtain the characteristic equationr22r+

15=0, which factors as(r5)(r+3) =0. Therefore,r=5 andr=3 are roots, so

y

1=e5xandy2=e3xare solutions. Furthermore, by Theorem 5 in §3.1, the general

solution to the DE is

y=c1e5x+c2e3x.3.1.40. Find the general solution of the DE 9y0012y0+4y=0.Solution:As above, the characteristic equation for the DE is 9r212r+4=0, which

factors as(3r2)2=0. Therefore, this equation has a double root atr=2/3. By Theorem 6 in §3.1, the general solution is then y=c1xe2x/3+c2e2x/3. 1

MAT 303 Spring 2013 Calculus IV with Applications

3.1.46. Find a homogeneous second-order DEay00+by0+cy=0 with general solution

y=c1e10x+c2e100x.Solution:This constant coefficient DE must havey1=e10xandy2=e100xas solutions, so we expectr10 andr100 to be factors of its characteristic polynomial. Then we may take ar

2+br+c=a(r10)(r100) =a(r2110r+1000),

so takinga=1, we have the corresponding DEy00110y0+1000y=0.3.2.2. Show directly that the functionsf(x) =5,g(x) =23x2, andh(x) =10+15x2

are linearly dependent on the real line.Solution:We find a nontrivial linear combinationc1f+c2g+c3hof these functions iden-

tically equal to 0. Since all 3 functions are polynomials inx, the function is 0 exactly when the coefficients on all the powers ofxare 0. Since c

1f+c2g+c3h=c1(5) +c2(23x2) +c3(10+15x2)

= (5c1+2c2+10c3) + (3c2+15c3)x2, we require that 5c1+2c2+10c3=0 and3c2+15c3=0. From the second equation, c

2=5c3. Substituting this into the first,

5c1+2c2+10c3=5c1+2(5c3) +10c3=5c1+20c3=0.

Thenc1=4c3, and there are no more constraints on theci. Choosing to setc3=1, c

1=4 andc2=5. We check that this nontrivial linear combination of functions is 0:

(4)(5) + (5)(23x2) + (1)(10+15x2) =20+1015x2+10+15x2=0. Rearranging this equation, we can express any single one of these functions as a linear

combination of the other two: for example, 10+15x2=4(5)5(103x2).3.2.8. Use the Wronskian to prove that the functionsf(x) =ex,g(x) =e2x, andh(x) =

e

3xare linearly independent on the real line.Solution:We computeW(f,g,h):

W(f,g,h) =

f g h f 0g0h0 f

00g00h00

e xe2xe3x e x2e2x3e3x e x4e2x9e3x =ex2e2x3e3x

4e2x9e3x

e2xe3x

4e2x9e3x

+e2xe3x

2e2x3e3x

=exe2xe3x((2943)(1914) + (1312)) =e6x(65+1) =2e6x. SinceW(x) =2e6x, which is not zero on the real line (and in fact nowhere 0), these three functions are linearly independent. 2

MAT 303 Spring 2013 Calculus IV with Applications

3.2.10. Use the Wronskian to prove that the functionsf(x) =ex,g(x) =x2, and

h(x) =x2lnxare linearly independent on the intervalx>0.Solution:We computeW(f,g,h). First, we compute derivatives ofh:

h

0(x) =2x3lnx+x21x

= (12lnx)x3 h

00(x) = (3)(12lnx)x4+2x

x3= (6lnx5)x4

Plugging these into the Wronskian, we have

W(f,g,h) =

f g h f 0g0h0 f

00g00h00

e xx2x2lnx e x2x3(12lnx)x3 e x6x4(6lnx5)x4 Rather than expand this directly, we make use of some additional properties of the deter- minant. One of these is that the determinant is unchanged if a multiple of one column is added to or subtracted from a different column. We subtract lnxtimes the second column from the third to cancel the lnxterms there:

W(f,g,h) =

e xx20 e x2x3x3 e x6x45x4 Using another property of the determinant, we factor the scalarexout of the first column, so that it multiplies the determinant of the remaining matrix:

W(f,g,h) =ex

1x20

12x3x3

1 6x45x4

With these simplifications, we expand along the first row, which conveniently contains a

0 entry:

W(f,g,h) =ex2x3x3

6x45x4

x21x3 15x4 +0 =ex

10x76x7x2(5x4x3)

=exx7(x2+5x+4) =ex(x+1)(x+4)x 7. This function is defined and continuous for allx>0. Furthermore, none of the factors in its numerator is 0 forx>0, so it is in fact nowhere 0 on this interval. Since their Wronskian is not identically 0, these functions are linearly independent on this interval. 3

MAT 303 Spring 2013 Calculus IV with Applications

3.2.14. Find a particular solution to the DEy(3)6y00+11y06y=0 matching the

initial conditionsy(0) =0,y0(0) =0,y00(0) =3 that is a linear combination ofy1=ex, y

2=e2x, andy3=e3x.Solution:We lety=c1ex+c2e2x+c3e3x. Theny0=c1ex+2c2e2x+3c3e3xandy00=

c

1ex+4c2e2x+9c3e3x, so evaluating these functions atx=0 and matching them to the

initial conditions, we obtain the linear system c

1+c2+c3=0

c

1+2c2+3c3=0

c

1+4c2+9c3=3

We solve this linear system by row reduction of an augmented matrix to echelon form: 2

41 1 10

1 2 30

1 4 93

3 5 2

41 1 10

0 1 20

0 3 83

3 5

R2 R2R1,R3 R3R1

2

41 1 10

0 1 20

0 0 23

3 5

R3 R33R2

2

41 1 0

32

0 1 03

0 0 132

3 5 R3 12

R3,R2 R22R3,R1 R1R3

2

41 0 032

0 1 03

0 0 132

3 5

R1 R1R2

Thenc1=c3=32

andc2=3, soy=32 ex3e2x+32

e3xis the solution to the IVP.3.2.18. Find a particular solution to the DEy(3)3y00+4y02y=0 matching the initial

conditionsy(0) =1,y0(0) =0,y00(0) =0 that is a linear combination ofy1=ex, y

2=excosx, andy3=exsinx.Solution:We lety=c1ex+c2excosx+c3exsinx. Then

y

0=c1ex+c2ex(cosxsinx) +c3ex(sinx+cosx)

y

00=c1ex2c2exsinx+2c3excosx

Evaluating these functions atx=0 and matching them to the initial conditions, we obtain the linear system c

1+c2=1

c

1+c2+c3=0

c

1+2c3=0

4

MAT 303 Spring 2013 Calculus IV with Applications

By the third equation,c1=2c3. Substituting this into the second,c2c3=0, soc2=c3. Finally, in the first equation,2c3+c3=1, soc3=1,c2=1, andc1=2(1) =2.

Theny=2exexcosxexsinx=ex(2cosxsinx)is the solution to the IVP.3.2.24. The nonhomogeneous DEy002y0+2y=2xhas the particular solutionyp=x+

1 and the complementary solutionyc=c1excosx+c2exsinx. Find a solution satisfying

the initial conditionsy(0) =4,y0(0) =8.Solution:The general solution to this DE is of the form y=yp+yc=x+1+c1excosx+c2exsinx. Then y

0=1+c1ex(cosxsinx) +c2ex(sinx+cosx).

Evaluating atx=0 and applying the initial conditions,y(0) =1+c1=4 andy0(0) =

1+c1+c2=8. Thenc1=3 andc2=4, so the solution to the IVP is

y=x+1+ex(3cosx+4sinx).3.2.30. Verify thaty1=xandy2=x2are linearly independent solutions on the entire

real line of the equationx2y002xy0+2y=0, but thatW(x,x2)vanishes atx=0. Why

do these observations not contradict part (b) of Theorem 3?Solution:We first check that these are solutions:

x

2y0012xy01+2y1=x2(0)2x(1) +2(x) =x(2+2) =0

x

2y0022xy02+2y2=x2(2)2x(2x) +2(x2) =x2(24+2) =0

We then compute their Wronskian:

W(x,x2) =x x2

1 2x =x(2x)1(x2) =x2. This function is not identically 0, so the two functionsxandx2are linearly independent on the real line, but it is 0 at preciselyx=0. We note that Theorem 3 applies only to normalized homogeneous linear DEs. Normaliz- ing this DE, we obtain y 002x y0+2x 2y=0, the coefficient functions of which are continuous forx6=0. Thus, any interval on which thetheoremappliesdoesnotincludex=0, theonlypointatwhichW(x) =0, soW(x,x2) is nonzero on every such interval. This is consistent with the linear independence of the solutionsxandx2. 5quotesdbs_dbs17.pdfusesText_23

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