System of Differential Equations

  • #1
3,003
6

Homework Statement



[tex]\frac{dx}{dt} +\frac{dy}{dt} + y - x = e^{2t}[/tex]
[tex]\frac{d^2x}{dt^2} +\frac{dy}{dt} = 3e^{2t}[/tex]


I am a little stumped on this one. I tried writing it using D as the differential operator d/dt:

D(x+y) +x - y = e2t
D2x+Dy = 3e2t

but that doesn't do anything but change the variables to something else... if I subtract the 1st EQ from second I get

[tex]\frac{d^2x}{dt^2}-\frac{dx}{dt} -y +x = 2e^{2t}[/tex]

But I still have x AND y in here.

Any hints on this?
 

Answers and Replies

  • #2
35,238
7,058
It seems to me that both x and y will have to involve e2t.
 
  • #3
3,003
6
Hi Mark44 :smile: I am not sure if you are giving me a hint or just making a comment. I am sure that they both will, but I am trying to eliminate all 'x' or all 'y' from one equation so that I can solve for it. Just like with a system of regular equations (i.e. not DEs).

Just not sure how to approach it.
 
  • #4
35,238
7,058
I'm giving you a hint. I was able to find solutions for x and y by "guessing" that each was some multiple of e2t. The solutions I found worked, but it's possible that I didn't get the complete solution.

Alternatively, and more in line with what you're trying, you could solve for y' in the second equation, integrate to get y, and then substitute for y and y' in the first equation. Then you would have a single 2nd order, linear equation in one dependent variable.
 
  • #5
1,800
53
One approach is to treat the system in terms of the differential operators and then just use simple elimination keeping in mind the terms are operators. So writing it as:

[tex]
\begin{aligned}
(D-1)x&+(D+1)y&=e^{2t}\\
D^2 x&+Dy&=3e^{2t}
\end{aligned}
[/tex]

Now, to solve for y, I can operate on the first with the D^2 operator and on the second with the (D-1) operator:

[tex]
\begin{aligned}
D(D-1)x&+D(D+1)y&=De^{2x} \\
D^2(D+1)x&+D(D+1)y&=(D+1)3e^{2x}
\end{aligned}
[/tex]

Notice now I can subtract the second from the first and obtain:

[tex]D(D-1)x-D^2(D+1)x=De^{2x}-(D+1)3e^{2x}[/tex]

[tex](D^2-D-D^3-D^2)x=2e^{2x}-(6e^2x}+3e^{2x})[/tex]

[tex](D^3+D)x=e^{2t}[/tex]

Solve that with the set of [itex](a_1,a_2,a_3)[/itex] arbitrary constants. Do the same for y with [itex](b_1,b_2)[/itex], back-substitute into the DE, equate coefficients to determine the relationship between the a and b coefficients.

See "Differential Equations" by Rainville and Bedient
 
  • #6
3,003
6
I'm giving you a hint. I was able to find solutions for x and y by "guessing" that each was some multiple of e2t. The solutions I found worked, but it's possible that I didn't get the complete solution.

Alternatively, and more in line with what you're trying, you could solve for y' in the second equation, integrate to get y, and then substitute for y and y' in the first equation. Then you would have a single 2nd order, linear equation in one dependent variable.

Hi again Mark44 :smile: I still don't see it. The second equation is

x'' + y' = 3e2t or y' = 3e2t - x''

I don't see how I can "integrate to obtain y." y' is in terms of 't' and 'x.'

One approach is to treat the system in terms of the differential operators and then just use simple elimination keeping in mind the terms are operators. So writing it as:

[tex]
\begin{aligned}
(D-1)x&+(D+1)y&=e^{2t}\\
D^2 x&+Dy&=3e^{2t}
\end{aligned}
[/tex]

Now, to solve for y, I can operate on the first with the D^2 operator and on the second with the (D-1) operator:

[tex]
\begin{aligned}
D(D-1)x&+D(D+1)y&=De^{2x} \\
D^2(D+1)x&+D(D+1)y&=(D+1)3e^{2x}
\end{aligned}
[/tex]

Notice now I can subtract the second from the first and obtain:

[tex]D(D-1)x-D^2(D+1)x=De^{2x}-(D+1)3e^{2x}[/tex]

[tex](D^2-D-D^3-D^2)x=2e^{2x}-(6e^2x}+3e^{2x})[/tex]

[tex](D^3+D)x=e^{2t}[/tex]

Solve that with the set of [itex](a_1,a_2,a_3)[/itex] arbitrary constants. Do the same for y with [itex](b_1,b_2)[/itex], back-substitute into the DE, equate coefficients to determine the relationship between the a and b coefficients.

See "Differential Equations" by Rainville and Bedient

As I recall, some refer to this as the annihilator method right? This was the approach I was going for, but I just did not have the proper foresight.
 
  • #7
fzero
Science Advisor
Homework Helper
Gold Member
3,119
289
Hi again Mark44 :smile: I still don't see it. The second equation is

x'' + y' = 3e2t or y' = 3e2t - x''

I don't see how I can "integrate to obtain y." y' is in terms of 't' and 'x.'

This equation can be directly integrated to

y = (3/2) e2t - x' + c
 
  • #8
3,003
6
This equation can be directly integrated to

y = (3/2) e2t - x' + c

OH! I still have to wrap my head around the fact that these are differential equations. As such, I did not have to 'solve' the second equation completely for y interms of t. I see now that x' in the above can be directly substituted into the 1st equation leaving only x in terms of t.

Thanks guys for your patience. :smile:

Casey
 
  • #9
1,800
53
This equation can be directly integrated to

y = (3/2) e2t - x' + c

Nicer way. Thank you. I didn't know that either. :)
 

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