Question

Long one (copied from Boyce & Diprima, 10th ed.):
Consider the system
x1′ =−2x1 +x2, x2′ =x1 −2x2.
(a) Solve the first equation for x2 and substitute into the second equation, thereby obtaining a second order equation for x1. Solve this equation for x1 and then determine x2 also.

(I got this part up until the last sentence, I have x2'=-x1-x1' and x1=-x2'-x1'. Not sure what method to solve to get a function that works.. should I start guessing?)

(b)Find the solution of the given system that also satisfies the initial conditions x1(0)=2, x2(0) = 3.

Answer 1

I just noticed that in part (a) it wants me to obtain a 2nd order de for x1.
x2=x1'+2x1
subbing in 2nd eq gives: x2'=-2x1'-3x1
Does it want me to derive my equation for x2? Can I do that without knowing what x1 and x2 are functions of?

Sorry if the post is confusing, please let me know if I've done anything wrong.

Answer 2

So, the system that you're given is
\[\large\left\{\begin{aligned}x_1^{\prime} &= -2x_1+x_2\\ x_2^{\prime} &= x_1-2x_2\end{aligned}\right.\]
To write this as a single second order equation, solve for \(\large x_2\) in the first equation to get \(\large x_2 = x_1^{\prime}+2x_1\). Now since \(\large x_1\) and \(\large x_2\) are differentiable functions, we now differentiate this equation to see that \(\large x_2^{\prime}=x_1^{\prime\prime}+2x_1^{\prime}\), We can now substitute our values for \(\large x_2\) and \(\large x_2^{\prime}\) into the second equation to see that
\[\large x_2^{\prime}=x_1-2x_2 \implies (x_1^{\prime\prime}+2x_1^{\prime}) = x_1 - 2(x_1^{\prime}+2x_1) \implies x_1^{\prime\prime}+4x_1^{\prime}+3x_1 = 0.\]
Can you take things from here and solve for \(\large x_1\)?

I hope this makes sense! :-)

Answer 3

I noticed that the last equation cut off. It should read \(\large x_1^{\prime\prime} + 4x_1^{\prime} + 3x_1 = 0\).

Answer 4

Ok, so I've got\[x _{1}'' + 2x _{1}'=x _{1}-2x _{1}'-4x _{1}\]

which means x1 is
\[-\frac{ (x _{1}'' + 4x _{1}') }{ 3 }\]

So that's one second order de. I am assuming they want me to get an actual solution, I am going to try doing the characteristic eq to find roots.

Answer 5

ALRIGHT, I solved the characteristic eq and got:
\[x _{1}= c _{1}e ^{-t} + c _{2}e ^{-3}\]
(Anything wrong?)
Then I did the same procedure for x_{2}
\[x _{1}=x _{2}'+2x _{2}\]
\[x _{1}= x _{2}''+2x _{2}\]
x2''+2x2'+5x2=0
Roots for x2 are -1 (+/-) 2i

Answer 6

So your \(\large x_1 = c_1e^{-t}+c_2e^{-3t}\) looks good, but now that you have this, recall that we had \(\large x_2=x_1^{\prime}+2x_1\). What you can do now is substitute the solution you found for \(\large x_1\) into this equation to see that
\[\large \begin{aligned}x_2 &= x_1^{\prime}+2x_1\\ & = (c_1e^{-t}+c_2e^{-3t})^{\prime} + 2(c_1e^{-t}+c_2e^{-3t})\\ &=-c_1e^{-t}-3c_2e^{-3t} + 2c_1e^{-t}+2c_2e^{-3t} \\ &= c_1e^{-t}-c_2e^{-3t}\end{aligned}\]
Thus, \(\large x_2=c_1e^{-t}-c_2e^{-3t}\). Now plug in your initial conditions \(\large x_1(0)=2\) and \(\large x_2(0)=3 \) into \(\large x_1\) and \(\large x_2\) to get
\[\large \left\{\begin{aligned} c_1+c_2 &=2\\ c_1 - c_2 &= 3 \end{aligned}\right.\]and solve for the constants \(\large c_1\) and \(\large c_2\). Then plug those into \(\large x_1\) and \(\large x_2\) to get the particular solution for the system.

Does this make sense? :-)

Answer 7

Yes, thank you! It had crossed my mind to do that, should have tried it!
I got c2=-1/2 and c1=5/2 so everything works.