# Homework Help: System of Differential Equations

1. Sep 29, 2010

1. The problem statement, all variables and given/known data

$$\frac{dx}{dt} +\frac{dy}{dt} + y - x = e^{2t}$$
$$\frac{d^2x}{dt^2} +\frac{dy}{dt} = 3e^{2t}$$

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

$$\frac{d^2x}{dt^2}-\frac{dx}{dt} -y +x = 2e^{2t}$$

But I still have x AND y in here.

Any hints on this?

2. Sep 29, 2010

### Staff: Mentor

It seems to me that both x and y will have to involve e2t.

3. Sep 29, 2010

Hi Mark44 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. Sep 29, 2010

### Staff: Mentor

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. Sep 29, 2010

### jackmell

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:

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

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:

\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}

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

$$D(D-1)x-D^2(D+1)x=De^{2x}-(D+1)3e^{2x}$$

$$(D^2-D-D^3-D^2)x=2e^{2x}-(6e^2x}+3e^{2x})$$

$$(D^3+D)x=e^{2t}$$

Solve that with the set of $(a_1,a_2,a_3)$ arbitrary constants. Do the same for y with $(b_1,b_2)$, 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. Sep 30, 2010

Hi again Mark44 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.'

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. Sep 30, 2010

### fzero

This equation can be directly integrated to

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

8. Sep 30, 2010

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.