System of Differential Equations

In summary, the conversation discussed various approaches to solving a system of differential equations involving the operators D and D^2. One approach involved "guessing" solutions for x and y, while another involved using the annihilator method to eliminate one variable and solve for the other. Ultimately, the conversation concluded with a simpler integration method for solving the equations.
  • #1
Saladsamurai
3,020
7

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?
 
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  • #2
It seems to me that both x and y will have to involve e2t.
 
  • #3
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
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
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
Mark44 said:
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.'

jackmell said:
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
Saladsamurai said:
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
fzero said:
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
fzero said:
This equation can be directly integrated to

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

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

Related to System of Differential Equations

What is a system of differential equations?

A system of differential equations is a group of equations that describe the relationship between multiple variables and their rates of change over time. These equations are used to model dynamic systems in various fields such as physics, engineering, and biology.

What is the difference between a system of differential equations and a single differential equation?

The main difference between a system of differential equations and a single differential equation is that a system involves multiple equations that are interconnected, while a single equation only involves one equation. A system of differential equations is more complex and can provide a more accurate representation of real-world phenomena.

How is a system of differential equations solved?

Solving a system of differential equations involves finding a set of equations that satisfy all the conditions of the system. This can be done analytically (using mathematical methods) or numerically (using computational methods). The method used depends on the complexity of the system and the availability of initial conditions.

What are the applications of systems of differential equations?

Systems of differential equations have a wide range of applications in various fields, including physics, engineering, economics, and biology. They are used to model and predict the behavior of complex systems such as population dynamics, chemical reactions, and electrical circuits.

What are the limitations of systems of differential equations?

While systems of differential equations can accurately model many real-world phenomena, they have some limitations. These include the difficulty of finding exact solutions for complex systems, the need for initial conditions, and the possibility of instability or chaotic behavior in some systems.

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