Determine whether the functions f(x) = 2 cos x + 3 sin x and g(x) = 3 cos x − 2 sin x are linearly dependent or linearly independent on the real line Solution: We
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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+3sinx3cosx2sinx2sinx+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 realline).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
y1=e5xandy2=e3xare solutions. Furthermore, by Theorem 5 in §3.1, the general
solution to the DE isy=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. 1MAT 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 ar2+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 c1f+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, c2=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, c1=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 linearcombination 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) =
e3xare linearly independent on the real line.Solution:We computeW(f,g,h):
W(f,g,h) =
f g h f 0g0h0 f00g00h00
e xe2xe3x e x2e2x3e3x e x4e2x9e3x =ex2e2x3e3x4e2x9e3x
e2xe3x4e2x9e3x
+e2xe3x2e2x3e3x
=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. 2MAT 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:
h0(x) =2x3lnx+x21x
= (12lnx)x3 h00(x) = (3)(12lnx)x4+2x
x3= (6lnx5)x4Plugging these into the Wronskian, we have
W(f,g,h) =
f g h f 0g0h0 f00g00h00
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
1x2012x3x3
1 6x45x4
With these simplifications, we expand along the first row, which conveniently contains a0 entry:
W(f,g,h) =ex2x3x3
6x45x4
x21x3 15x4 +0 =ex10x76x7x2(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. 3MAT 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, y2=e2x, andy3=e3x.Solution:We lety=c1ex+c2e2x+c3e3x. Theny0=c1ex+2c2e2x+3c3e3xandy00=
c1ex+4c2e2x+9c3e3x, so evaluating these functions atx=0 and matching them to the
initial conditions, we obtain the linear system c1+c2+c3=0
c1+2c2+3c3=0
c1+4c2+9c3=3
We solve this linear system by row reduction of an augmented matrix to echelon form: 241 1 10
1 2 30
1 4 93
3 5 241 1 10
0 1 20
0 3 83
3 5R2 R2R1,R3 R3R1
241 1 10
0 1 20
0 0 23
3 5R3 R33R2
241 1 0
320 1 03
0 0 132
3 5 R3 12R3,R2 R22R3,R1 R1R3
241 0 032
0 1 03
0 0 132
3 5R1 R1R2
Thenc1=c3=32
andc2=3, soy=32 ex3e2x+32e3xis 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, y2=excosx, andy3=exsinx.Solution:We lety=c1ex+c2excosx+c3exsinx. Then
y0=c1ex+c2ex(cosxsinx) +c3ex(sinx+cosx)
y00=c1ex2c2exsinx+2c3excosx
Evaluating these functions atx=0 and matching them to the initial conditions, we obtain the linear system c1+c2=1
c1+c2+c3=0
c1+2c3=0
4MAT 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 y0=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. Whydo these observations not contradict part (b) of Theorem 3?Solution:We first check that these are solutions:
x