2.7 HW 7

  2.7.1 Problems listing
  2.7.2 Problem 3 section 5.5
  2.7.3 Problem 9 section 5.5
  2.7.4 Problem 11 section 5.5
  2.7.5 Problem 23 section 5.5
  2.7.6 Problem 32 section 5.5
  2.7.7 Problem 7 section 6.1
  2.7.8 Problem 17 section 6.1
  2.7.9 Problem 21 section 6.1
  2.7.10 Problem 25 section 6.1
  2.7.11 Problem 29 section 6.1
  2.7.12 Additional problem 1
  2.7.13 Additional problem 2
  2.7.14 Additional optional problem 3
  2.7.15 key solution for HW 7

2.7.1 Problems listing

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2.7.2 Problem 3 section 5.5

In Problems 1 through 20, find a particular solution yp of the given equation.(A)yy6y=2sin3x

Solution

The first step is to find the homogeneous solution yh in order to determine the basis solutions to check for any duplication with basis solutions for the particular solution.yy6y=0 The characteristic equation is r2r6=0(r3)(r+2)=0

Hence the roots are r1=3,r2=2. Therefore the basis solutions are  (1){e3x,e2x} Which implies yh=c1e3x+c2e2x Now that we found the basis solution, we turn our attention to finding yp. The RHS is sin3x. Looking at this function and all possible derivatives gives (2){sin3x,cos3x} Notice that we ignore any leading coefficients when doing this. Now we compare the above to the basis of the homogeneous solution found in (1) to check if there are duplication in basis or not. There is no duplication. Therefore we assume that particular solution yp is a linear combination of the functions in (2). This implies thatyp=Asin3x+Bcos3xyp=3Acos3x3Bsin3xyp=9Asin3x9Bcos3x

Substituting the above back in original ODE (A) givesypyp6yp=2sin3x(9Asin3x9Bcos3x)(3Acos3x3Bsin3x)6(Asin3x+Bcos3x)=2sin3xsin(3x)(9A+3B6A)+cos(3x)(9B3A6B)=2sin3xsin(3x)(15A+3B)+cos(3x)(15B3A)=2sin3x

Comparing coefficients gives(3)15A+3B=2(4)15B3A=0

Multiplying first equation by 5 and adding result to second equation gives(75A+15B)+(15B3A)=1078A=10A=1078=539

From (3)15(539)+3B=22513+3B=2B=225133=139

Hence the particular solution is yp=Asin3x+Bcos3x=539sin3x+139cos3x=139(cos3x5sin3x)

Therefore the general solution is y=yh+yp=c1e3x+c2e2x+139(cos3x5sin3x)

2.7.3 Problem 9 section 5.5

In Problems 1 through 20, find a particular solution yp of the given equation.(A)y+2y3y=1+xex Solution

The first step is to find the homogeneous solution yh in order to determine the basis solutions to check for any duplication with basis solutions for the particular solution.y+2y3y=0 The characteristic equation is r2+2r3=0(r+3)(r1)=0

Hence the roots are r1=3,r2=1. Therefore the basis solutions are  (1){e3x,ex} Which implies yh=c1e3x+c2ex Now that we found the basis solution, we turn our attention to finding yp. The RHS is 1+xex. Hence it basis functions are{1,xex} taking derivatives of each basis gives(2){1,(xex,ex)} Where we used () to group all basis generated from same one.

Now we compare the above to the basis of the homogeneous solution found in (1) to check if there are duplication in basis or not. We see duplication since ex is basis in both (1) and (2). Therefore we multiply the group which generated ex by x. The the above now becomes(2A){1,(x2ex,xex)} We compare again (1) against (2A) and now we see no duplication. Therefore we assume that particular solution yp is a linear combination of the functions in (2A). This implies thatyp=A+Bx2ex+Cxexyp=2Bxex+Bx2ex+Cex+Cxexyp=2Bex+2Bxex+2Bxex+Bx2ex+Cex+Cex+Cxex=xex(2B+2B+C)+x2ex(B)+ex(2B+2C)

Substituting the above back in original ODE (A) givesyp+2yp3yp=1+xexxex(2B+2B+C)+x2ex(B)+ex(2B+2C)+2(2Bxex+Bx2ex+Cex+Cxex)3(A+Bx2ex+Cxex)=1+xexxex(2B+2B+C+4B+2C3C)+ex(2B+2C+2C)+x2ex(B+2B3B)3A=1+xexxex(8B)+ex(2B+4C)3A=1+xex

Comparing coefficients3A=12B+4C=08B=1

Hence B=18 and from second equation 4C=28, or C=116 and A=13. Therefore the particular solution is yp=A+Bx2ex+Cxex=13+18x2ex116xex=13+116(2x2x)ex

Therefore the general solution is y=yh+yp=c1e3x+c2ex13+116(2x2x)ex

2.7.4 Problem 11 section 5.5

In Problems 1 through 20, find a particular solution yp of the given equation.(A)y(3)+4y=3x1 Solution

The first step is to find the homogeneous solution yh in order to determine the basis solutions to check for any duplication with basis solutions for the particular solution.y(3)+4y=0 The characteristic equation is r3+4r=0r(r2+4)=0

Hence the roots are r1=0,r2=±2i. Therefore the basis solutions are  (1){1,cos(2x),sin(2x)} Which implies yh=c1+c2cos(2x)+c3sin(2x) Now that we found the basis solution, we turn our attention to finding yp. The RHS is 3x1. Hence it basis functions are(2){1,x} Taking derivatives does not add any new basis. Now we compare the above to the basis of the homogeneous solution found in (1) to check if there are duplication in basis or not. We see duplication the constant is in both (1) and (2). Therefore we multiply the group by x. We took the whole basis as one group, since the constant 1 above is generated by taking derivative of x, so it is really in the same group. The above now becomes, after multiplying everything by x(2A){x,x2} We compare again (1) against (2A) and now we see no duplication. Therefore we assume that particular solution yp is a linear combination of the functions in (2A). This implies thatyp=Ax+Bx2yp=A+2Bxyp=2Byp(3)=0

Substituting the above back in original ODE (A) givesyp(3)+yp=3x10+4(A+2Bx)=3x14A+8Bx=3x1

Comparing coefficients4A=18B=3

Hence A=14,B=38.  Therefore the particular solution is yp=Ax+Bx2=14x+38x2=18(3x22x)

Therefore the general solution is y=yh+yp=c1+c2cos(2x)+c3sin(2x)14x+38x2

2.7.5 Problem 23 section 5.5

In Problems 21 through 30, set up the appropriate form of a particular solution yp, but do not determine the values of the coefficients.(A)y+4y=3xcos(2x) Solution

The first step is to find the homogeneous solution yh in order to determine the basis solutions to check for any duplication with basis solutions for the particular solution.y+4y=0 The characteristic equation is r2+4=0 Hence the roots are r=±2i. Therefore the basis solutions are  (1){cos(2x),sin(2x)} Which implies yh=c1cos(2x)+c2sin(2x) Now that we found the basis solution, we turn our attention to finding yp. The RHS is 3xcos(2x). Hence it basis functions are(2){xcos(2x)} Taking all possible derivatives of the above gives(2A){xcos(2x),cos(2x),xsin(2x),sin(2x)} Where in the above all signs and coefficients were ignored.

Now we compare the above to the basis of the homogeneous solution found in (1) to check if there are duplication in basis or not. We see duplication as cos(2x),sin(2x) are in both. Therefore we multiply the group by x. We took the whole basis as one group since everything above was generated from (2). The above now becomes, after multiplying each term by x(2B){x2cos(2x),xcos(2x),x2sin(2x),xsin(2x)} Now we compare (2B) again with (1) and see no duplication. Hence yp=Ax2cos(2x)+Bxcos(2x)+Cx2sin(2x)+Dxsin(2x)

2.7.6 Problem 32 section 5.5

Solve the initial value problems in Problems 31 through 40.(A)y+3y+2y=exy(0)=0y(0)=3

Solution

The first step is to find the homogeneous solution yh in order to determine the basis solutions to check for any duplication with basis solutions for the particular solution.y+3y+2=0 The characteristic equation is r2+3r+2=0(r+2)(r+1)=0

Hence the roots are r1=2,r2=1. Therefore the basis solutions are  (1){e2x,ex} Which implies yh=c1e2x+c2ex Now that we found the basis solution, we turn our attention to finding yp. The RHS is ex. Hence it basis functions are(2){ex} Taking all derivatives does not any terms. We also see no duplication between (2) and (1). Hence letyp=Aexyp=Aexyp=Aex

Substituting these into (A) givesyp+3yp+2yp=exAex+3Aex+2Aex=exex(A+3A+2A)=ex

Hence 6A=1A=16

Thereforeyp=16ex Therefore the complete solution isy=yh+yp(3)=c1e2x+c2ex+16ex

We are now ready to apply the initial conditions.  y(0)=0,y(0)=3. Applying first IC to (3) gives(4)0=c1+c2+16 Taking derivative of (3) givesy=2c1e2xc2ex+16ex Applying second IC to the above gives(5)3=2c1c2+16 We now need to solve (4,5) for c1,c2. Adding (4,5) gives3=(c1+c2+16)+(2c1c2+16)3=13c1c1=133=83

From (4)0=c1+c2+160=83+c2+16c2=52

Therefore the complete solution (3) becomesy(x)=c1e2x+c2ex+16ex=83e2x+52ex+16ex=16(16e2x+15ex+ex)

2.7.7 Problem 7 section 6.1

In Problems 1 through 26, find the (real) eigenvalues and associated eigenvectors of the given matrix A. Find a basis for each eigenspace of dimension 2 or larger.A=[10864] Solution

We first need to find the eigenvalues. These are found by solving |AλI|=0. Hence|10λ864λ|=0(10λ)(4λ)+48=0λ26λ+8=0(λ4)(λ2)=0

Hence λ1=4,λ2=2. For each eigenvalue we find its associated eigenvectors.

λ1=4

We nee to solve Av=λv. This becomes (AλI)v=0. Therefore [1048644][v1v2]=[00](1)[6868][v1v2]=[00]

Augmented matrix[680680] R2R2R1[680000] Therefore (1) becomes(1A)[6800][v1v2]=[00] v2 is free variable. Let v2=1. Then from first row 6v18=0 or v1=86. Hence v1=[861]=[86]=[43] λ1=2

We nee to solve Av=λv. This becomes (AλI)v=0. Therefore [1028642][v1v2]=[00](1)[8866][v1v2]=[00]

Augmented matrix[880660] R2R268R1[880000] Therefore (1) becomes(1A)[8800][v1v2]=[00] v2 is free variable. Let v2=1. Then from first row 8v18=0 or v1=1. Hence v2=[11] This table gives summary of the result



eigenvalue λ associated eigenvector v




4 [43]


2 [11]


2.7.8 Problem 17 section 6.1

In Problems 1 through 26, find the (real) eigenvalues and associated eigenvectors of the given matrix A. Find a basis for each eigenspace of dimension 2 or larger.A=[352020021] Solution

We first need to find the eigenvalues. These are found by solving |AλI|=0. Hence|3λ5202λ0021λ|=0 Expanding along the first columns. (3λ)|2λ021λ|=0(3λ)(2λ)(1λ)=0

Hence roots (eigenvalues) are λ1=3,λ2=2,λ3=1. For each eigenvalue we find its associated eigenvectors.

λ1=3

We nee to solve Av=λv. This becomes (AλI)v=0. Therefore [335202300213][v1v2v3]=[000](1)[052010022][v1v2v3]=[000]

R3R2+2R2[052010002][v1v2v3]=[000] Free variable is v1. Let v1=1. Last row gives v3=0. Second row gives v2=0. Hence the eigenvector is v1=[100] λ1=2

We nee to solve Av=λv. This becomes (AλI)v=0. Therefore [325202200212][v1v2v3]=[000](1)[152000021][v1v2v3]=[000]

Swap R2,R3 (for clarify only)[152021000][v1v2v3]=[000] Free variable is v3. Let v3=1. From second row 2v2v3=0. Hence v2=12. First row gives v1+5v22v3=0. Hence v1=5(12)+2=12.  Hence the eigenvector is v2=[12121]=[112] λ1=1

We nee to solve Av=λv. This becomes (AλI)v=0. Therefore [315202100211][v1v2v3]=[000](1)[252010020][v1v2v3]=[000]

R3R32R2[252010000][v1v2v3]=[000] Free variable is v3. Let v3=1. From second row v2=0. First row gives 2v1=2v3. Hence v1=1.  Hence the eigenvector is v3=[101] This table gives summary of the result



eigenvalue λ associated eigenvector v




3 [100]


2 [112]


1 [101]


2.7.9 Problem 21 section 6.1

In Problems 1 through 26, find the (real) eigenvalues and associated eigenvectors of the given matrix A. Find a basis for each eigenspace of dimension 2 or larger.A=[431211002] Solution

We first need to find the eigenvalues. These are found by solving |AλI|=0. Hence|4λ3121λ1002λ|=0 Expanding along the last row(1)3+3(2λ)|4λ321λ|=0(2λ)((4λ)(1λ)+6)=0(2λ)(λ23λ+2)=0(2λ)(λ2)(λ1)=0

Hence the eigenvalues are λ1=2 of algebraic multiplicity 2 and λ2=1. For each eigenvalue we find its associated eigenvectors.

λ1=2

We nee to solve Av=λv. This becomes (AλI)v=0. Therefore [423121210022][v1v2v3]=[000](1)[231231000][v1v2v3]=[000]

R2R2R1[231000000][v1v2v3]=[000] Free variables are v3.v2. This means this is a complete eigenvalue. Since it has algebraic multiplicity of 2 and have a geometric multiplicity of 2 as well. This means we can find two linearly independent eigenvectors from it. Let v2=s,v3=t. First row gives 2v13s+t=0 or v1=32s12t. Hence the solution is[v1v2v3]=[32s12tst]=s[3210]+t[1201]

Therefore the basis (eigenvectors) are{[3210],[1201]}{[320],[102]} Now that we found the eigenvectors associated with λ1=2, we will do the same for second eigenvalue.

λ2=1

We nee to solve Av=λv. This becomes (AλI)v=0. Therefore [413121110021][v1v2v3]=[000](1)[331221001][v1v2v3]=[000]

R2R223R1[3310023001][v1v2v3]=[000] R2R3+32R2[3310023000][v1v2v3]=[000] Free variable is v2, leading variables are v1,v3. Let v2=1. From second row, v3=0.First row gives 3v1=3. Hence v1=1.  Hence the eigenvector is v2=[110] This table gives summary of the result



eigenvalue λ associated eigenvector v




2 (multiplicity 2) {[320],[102]}


1 [110]


2.7.10 Problem 25 section 6.1

In Problems 1 through 26, find the (real) eigenvalues and associated eigenvectors of the given matrix A. Find a basis for each eigenspace of dimension 2 or larger.A=[1010011000200002] Solution

We first need to find the eigenvalues. These are found by solving |AλI|=0. Hence|1λ01001λ10002λ00002λ|=0 Since this is an upper triangle matrix, then the determinant is the product of the diagonal. Hence the above reduces to (1λ)2(2λ)2=0 Therefore the eigenvalues are λ1=1 of algebraic multiplicity 2 and λ2=2 also of algebraic multiplicity 2. For each eigenvalue we find its associated eigenvectors.

λ1=1

We nee to solve Av=λv. This becomes (AλI)v=0. Therefore [11010011100021000021][v1v2v3v4]=[0000][0010001000100001][v1v2v3v4]=[0000]

R2R2R1[0010000000100001][v1v2v3v4]=[0000] Swapping R3,R2 (for clarify)[0010001000000001][v1v2v3v4]=[0000] R2R2R1[0010000000000001][v1v2v3v4]=[0000] Swapping R4,R2 (for clarify)[0010000100000000][v1v2v3v4]=[0000] Hence leading variables are v3,v4 free variables are v1,v2. Let v1=s,v2=t. Second row gives v4=0. First row gives v3=0. Therefore the solution is[v1v2v3v4]=[st00]=s[1000]+t[0100] Therefore the two eigenvectors associated with this eigenvalues are {[1000],[0100]} λ2=2

We nee to solve Av=λv. This becomes (AλI)v=0. Therefore [12010012100022000022][v1v2v3v4]=[0000][1010011000000000][v1v2v3v4]=[0000]

Hence leading variables are v1,v2 free variables are v3,v4 Let v3=s,v4=t. Second row gives v2+v3=0 or v2=s. First row gives v1+s=0 or v1=s. Hence the solution is[v1v2v3v4]=[ssst]=s[1110]+t[0001] Therefore the two eigenvectors associated with this eigenvalues are {[1110],[0001]} This table gives summary of the result



eigenvalue λ associated eigenvector v




1 (multiplicity 2) {[1000],[0100]}


2 (multiplicity 2) {[1110],[0001]}


2.7.11 Problem 29 section 6.1

Find the complex conjugate eigenvalues and corresponding eigenvectors of the matrices given in Problems 27 through 32A=[03120] Solution

We first need to find the eigenvalues. These are found by solving |AλI|=0. Hence|λ312λ|=0λ2+36=0

Hence λ=±6i. For each eigenvalue we find its associated eigenvectors.

λ1=6i

We nee to solve Av=λv. This becomes (AλI)v=0. Therefore [6i3126i][v1v2]=[00] R2R2+126iR1[6i300][v1v2]=[00] Leading variable is v1, free variable is v2. Let v2=1. From first row 6iv13v2=0 or v1=36i= 12i=12i. Hence the eigenvector is v1=[12i1]=[i2] λ1=6i

We nee to solve Av=λv. This becomes (AλI)v=0. Therefore [+6i312+6i][v1v2]=[00] R2R2126iR1[6i300][v1v2]=[00] Leading variable is v1, free variable is v2. Let v2=1. From first row 6iv13v2=0 or v1=36i= 12i=12i. Hence the eigenvector is v1=[12i1]=[i2] This table gives summary of the result



eigenvalue λ associated eigenvector v




6i [i2]


6i [i2]


2.7.12 Additional problem 1

Find particular solution to (1)y(7)(x)2y(6)+9y(5)16y(4)+24y(3)32y+16y=e2x+xsinx+x2 Solution

From HW6, we found yh asy(x)=c1+(c2ex+c3xex)+(c4cos2x+c5sin2x)+x(c6cos2x+c7sin2x) Therefore, we see the basis functions for yh are(2){1,ex,xex,cos2x,sin2x,xcos2x,xsin2x} Looking at RHS of (1), we see the basis functions for yp are{e2x,x2,xsinx} Taking derivative e2x does not generate new basis. Taking derivative of x2 generates x,1. And taking derivative of xsinx generates sinx,xcosx,cosx. Hence the above becomes(3){e2x,(x2,x,1),(xsinx,sinx,xcosx,cosx)} There are 3 groups. Comparing (2,3) we see there is one duplication, which is the constant term. Hence we need to multiply that one group by x. The above becomes(3A){e2x,x(x2,x,1),(xsinx,sinx,xcosx)}={e2x,(x3,x2,x),(xsinx,sinx,xcosx,cosx)} Now we again compare (3A) and (2). Now there is no duplication. Therefore the particular solution isyp=A1e2x+A2(x3)+A3(x2)+A4(x)+A5(xsinx)+A6(sinx)+A7(xcosx)+A8cosx

2.7.13 Additional problem 2

Let A=[t1000t2000t3] where t1,t2,t3 are distinct real numbers. Find the eigenvalues of A and the corresponding eigenvectors.

Solution

We first need to find the eigenvalues. These are found by solving |AλI|=0. Hence|t1λ000t2λ000t3λ|=0 Since this is a diagonal matrix, then the determinant is the product of the diagonal. Hence the above reduces to (t1λ)(t2λ)(t3λ)=0 Hence the eigenvalues are λ1=t1,λ2=t2,λ3=t3. For each eigenvalue we find its associated eigenvectors.

λ1=t1

We nee to solve Av=λv. This becomes (AλI)v=0. Therefore [t1t1000t2t1000t3t1][v1v2v3]=[000][0000t2t1000t3t1][v1v2v3]=[000]

Leading variables are v2,v3 and free variables is v1. Let v1=s. Second and third rows give v2=0,v3=0, this is because t1,t2,t3 are distinct real numbers therefore t2t10,t3t10. Therefore the solution isv1=[s00]=s[100] For s=1 this gives the eigenvectorv1=[100] λ2=t2

We nee to solve Av=λv. This becomes (AλI)v=0. Therefore [t1t2000t2t2000t3t2][v1v2v3]=[000][t1t20000000t3t2][v1v2v3]=[000]

Leading variables are v1,v3 and free variable is v2. Let v2=s. First and third rows give v1=0,v3=0, this is because t1,t2,t3 are distinct real numbers therefore t1t20,t3t20. Therefore the solution isv2=[0s0]=s[010] For s=1 this gives the eigenvectorv2=[010] λ3=t3

We nee to solve Av=λv. This becomes (AλI)v=0. Therefore [t1t3000t2t3000t3t3][v1v2v3]=[000][t1t3000t2t30000][v1v2v3]=[000]

Leading variables are v1,v2 and free variable is v3. Let v3=s. First and third rows give v1=0,v2=0, this is because t1,t2,t3 are distinct real numbers therefore t1t30,t2t30. Therefore the solution isv2=[00s]=s[001] For s=1 this gives the eigenvectorv3=[001] This table gives summary of the result



eigenvalue λ associated eigenvector v




t1 [100]


t2 [010]


t2 [001]


2.7.14 Additional optional problem 3

Extend the result in problem 2 to the case of n×n matrices. That is, let A be a matrix with entries t1,t2,,tn on the main diagonal and 0s everywhere else, where the ti are distinct real numbers. Find the eigenvalues and corresponding eigenvectors.

Solution

This follows immediately from the last problem. Therefore each eigenvalue will be λ1=t1,λ2=t2,,λn=tn. And corresponding eigenvectors are (each eigenvector is n×1. v1=[10000],v2=[01000],,vn=[00001] These eigenvectors are the standard basis for Rn.

2.7.15 key solution for HW 7

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