## Question about differential equations!

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evinda
Posts: 32
Joined: Fri May 03, 2013 10:16 pm UTC

### Question about differential equations!

Hi!! I am looking at some exercises and got stuck at this one:

If the graph of the solution u of the differential equation [math]y''-4y'+29y=0[/math] intersects the graph of the solution v of the differential equation [math]y''+4y'+13y=0[/math] at the point (0,0),find u,v so that: (see attachment)

I found that u=c1*e^(2x)*sin(5x)+c2*e^(2x)*cos(5x) and v=d1*e^(-2x)*sin(3x)+d2*e^(-2x)*cos(3x)

and I thought that u(0)=v(0),because the graphs intersect at the point (0,0) and I got that c2=d2=0.

Is it right so far??And how can I continue?
Attachments dif.gif (898 Bytes) Viewed 2324 times

flownt
Posts: 70
Joined: Sat Feb 09, 2013 5:24 pm UTC

### Re: Question about differential equations!

You probably made a slight mistake in posing the problem, because as it stands, limx->oo v=0 and limx->oo u=oo which means that automatically limx->oo v4/u=0

You might've swapped the -4y' and 4y' or u and v

evinda
Posts: 32
Joined: Fri May 03, 2013 10:16 pm UTC

### Re: Question about differential equations!

flownt wrote:You probably made a slight mistake in posing the problem, because as it stands, limx->oo v=0 and limx->oo u=oo which means that automatically limx->oo v4/u=0

You might've swapped the -4y' and 4y' or u and v

No,I think that u and v are right..I have checked them again...Maybe,I don't have to find u and v,by solving the differential equations that are given,but with another way?
Because,the exercise asks me to find u and v,so that we have the result from the attachment above tooyoo
Posts: 100
Joined: Sat Jan 22, 2011 5:39 pm UTC

### Re: Question about differential equations!

evinda wrote:Is it right so far??And how can I continue?

It's what I got. However, something must be wrong, or you need to do something pretty damn fancy. I asked mathematica to compute the limit and didn't get anything sensible.

flownt
Posts: 70
Joined: Sat Feb 09, 2013 5:24 pm UTC

### Re: Question about differential equations!

I agree that as posed you're heading in the right direction, but independent of the constants, your limit already is 0 instead of 5/6, so the problem as it stands is ill posed.

This does assume real (and not complex) x, but I think that is a safe assumption. Taking the limit x to 0 seems sane, are you sure that is not in fact the question?

Demki
Posts: 199
Joined: Fri Nov 30, 2012 9:29 pm UTC

### Re: Question about differential equations!

flownt wrote:You probably made a slight mistake in posing the problem, because as it stands, limx->oo v=0 and limx->oo u=oo which means that automatically limx->oo v4/u=0

You might've swapped the -4y' and 4y' or u and v

Actually, limx->oo u(x) is undefined because of the sin(x) part, the value of u fluctuates between positive and negative numbers, and sometimes 0. The limit is not defined, even if you move it to the denominator.
(The limit is not positive infinity because for each M>0 there is no N>0 such that for each x>N, u(x)>M. The same applies for negative infinity, just change the '>'s to '<'s.)

You must've copied/read the question wrong, either at the differential equations or at the limit requirement.

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