Question

I solved this system of equations, however it's an even numbered problem so I can't get any feedback. Can you solve it for me and compare?

The image has both my solution and the problem from the book. Thanks!

http://i.imgur.com/xidGJu4.jpg

Answer 1

heh is that Zill?

Answer 2

looks like it! I have the fifth edition and this is problem #5 on page 409, a section on variation of parameters

Answer 3

Yes sir! :) Pretty good book. I have the fifth edition, it's 604 for me (problem 4). I solved it using undetermined coefficients (it was in the UD subsection).

Answer 4

I have only A First Course in Differential Equations -- you probably have one of the more comprehensive ones. oops it is in undetermined coefficients -- good catch. It's on the same page that the variation of parameters chapter begins on hence the section title header is misleading

Answer 5

Ah gotcha. My book is Engineering Mathematics, it's used in 3 courses in the engineering program so it probably is the more comprehensive textbooks out there.

Answer 6

If yours is #5, does that mean you have the solution to it in the back?

Answer 7

no need! I can solve this by hand

Answer 8

alright, let's take a look... first consider the homogeneous case:$$\mathbf{X}'=\begin{pmatrix}4&\tfrac13\\9&6\end{pmatrix}\mathbf{X}$$First step first -- find our eigenvalues so let's inspect our characteristic polynomial:$$\det(\mathbf{X}-\lambda I)=\begin{vmatrix}4-\lambda&\tfrac13\\9&6-\lambda\end{vmatrix}=(4-\lambda)(6-\lambda)-\frac13(9)=\lambda^2-10\lambda+21$$

Answer 9

wrong one :p

Answer 10

oh shoot I kept looking at #5

Answer 11

this is actually #4 in my book as well though it's written in non-matrix form:$$\frac{dx}{dt}=x-4y+4t+9e^{6t}\\\frac{dy}{dt}=4x+y-t+e^{6t}$$

Answer 12

So as a matrix differential equation we first consider our homogeneous part:$$\mathbf{X}'=\begin{pmatrix}1&-4\\4&1\end{pmatrix}\mathbf{X}$$Now we determine our eigenvalues by determining the roots of our characteristic polynomial:$$\det(\mathbf{X}-\lambda I)=\begin{vmatrix}1-\lambda&-4\\4&1-\lambda\end{vmatrix}=(1-\lambda)^2-(-4)4=\lambda^2-2\lambda+17$$Since \(17\) is prime there is no way to factorize this over the integers so use the quadratic formula:$$\lambda=\frac{2\pm\sqrt{4-4(17)}}2=1\pm\sqrt{-16}=1\pm4i$$... so we've found two complex eigenvalues.

Answer 13

okay so my homogeneous solution should be correct unless I made a silly mistake in writing out the actual solution

Answer 14

Consider then an eigenvector for the first eigenvalue \(\lambda_1=1+4i\) denoted by
\(\mathbf{K}=\begin{pmatrix}k_1\\k_2\end{pmatrix}\). We know that \((\mathbf{A}-\lambda_1\mathbf{I})\mathbf{K}=0\), so we may equate components:$$(1-\lambda_1)k_1-4k_2=0\\4k_1+(1-\lambda_1)k_2=0$$

Answer 15

$$-ik_1-k_2=0\\k_1-ik_2=0\\k_2=-ik_1,k_1=ik_2$$letting \(k_1=1\) we find \(k_2=-i\) so we've determined \(\begin{pmatrix}1\\-i\end{pmatrix}\) is the corresponding eigenvector for \(\lambda_1=1+4i\)

Answer 16

For \(\lambda_2=1-4i\) we see:$$(1-\lambda_2)k_1-4k_2=0\\4k_1+(1-\lambda_2)k_2=0$$substitute \(\lambda_2=1-4i\):$$4ik_1-4k_2=0\Longleftrightarrow ik_1-k_2=0\\4k_1+4ik_2=0\Longleftrightarrow k_1+ik_2=0\\k_2=ik_1,k_1=-ik_2$$again, let \(k_1=1\) and we observe \(k_2=i\) so we've determined \(\begin{pmatrix}1\\i\end{pmatrix}\) is the corresponding eigenvector for \(\lambda_2\)

Answer 17

... thus we have the general solution \(\mathbf{X}=c_1\begin{pmatrix}1\\-i\end{pmatrix}e^{(1+4i)t}+c_2\begin{pmatrix}1\\1\end{pmatrix}e^{(1-4i)t}\)

Answer 18

oops that scond matrix should be \(\begin{pmatrix}1\\i\end{pmatrix}\). note we can also express this in alternate ways by noting \(e^{(1+4i)t}=e^te^{4it}\) and \(e^{4it}=\cos4t+i\sin4t\), etc. so we conclude: $$e^{(1-4i)t}=e^te^{-4i}=e^t(\cos(-4t)+i\sin(-4t))=e^t(\cos4t-i\sin4y)\\e^{(1+4i)t}=e^t(\cos4t+i\sin 4t)$$

Answer 19

for our nonhomogeneous case, consider we have:$$\mathbf{F}(t)=\begin{pmatrix}4\\-1\end{pmatrix}t+\begin{pmatrix}9\\1\end{pmatrix}e^{6t}$$therefore we assume a particular solution of the form:$$\mathbf{X}_c(t)=\begin{pmatrix}a_1\\b_2\end{pmatrix}t+\begin{pmatrix}a_2\\b_2\end{pmatrix}e^{6t}$$

Answer 20

don't we also have to consider A3 and B3 (power of t^0)?

Answer 21

coefficient of t^0 I mean

Answer 22

yes exactly I corrected that in the reply I was typing

Answer 23

but now I just accidentally deleted it

Answer 24

$$\mathbf{X}_p(t)=\begin{pmatrix}a_1\\b_1\end{pmatrix}t+\begin{pmatrix}a_2\\b_2\end{pmatrix}e^{6t}+\begin{pmatrix}a_3\\b_3\end{pmatrix}$$

Answer 25

therefore we find$$\mathbf{X}_p'(t)=\begin{pmatrix}a_1\\b_1\end{pmatrix}+6\begin{pmatrix}a_2\\b_2\end{pmatrix}e^{6t}$$

Answer 26

now we substitute into our differential equation:$$\mathbf{X}'=\begin{pmatrix}1&-4\\4&1\end{pmatrix}\mathbf{X}+\begin{pmatrix}4t+9e^{6t}\\-t+\ \ \ e^{6t}\end{pmatrix}\\\begin{pmatrix}a_1+6a_2e^{6t}\\b_1+6b_2e^{6t}\end{pmatrix}=\begin{pmatrix}a_1t+a_2e^{6t}+a_3-4(b_1t+b_2e^{6t}+b_3)\\4(a_1t+a_2e^{6t}+a_3)+b_1t+b_2e^{6t}+b_3\end{pmatrix}\begin{pmatrix}4t+9e^{6t}\\-t+e^{6t}\end{pmatrix}$$

Answer 27

plus between those two matrices

Answer 28

$$\begin{pmatrix}a_1+6a_2e^{6t}\\b_1+6b_2e^{6t}\end{pmatrix}=\begin{pmatrix}(a_1-4b_1+4)t+(a_2-4b_2+9)e^{6t}-4b_3\\(4a_1+b_1-1)t+(4a_2+b_2+1)e^{6t}+b_3\end{pmatrix}$$equate the coefficients of each term for each pair of rows:$$a_1=-4b_3\\a_1-4b_1+4=0\\6a_2=a_2-4b_2+9\\b_1=b_3\\6b_2=4a_2+b_3+1\\4a_1+b_1-1=0$$