Second Order ODE second term tending to 0 - question reads something must be unsolve

Mar 2016
170
0
Australia
Hi,

I have a 2nd order ODE of general form:

y= Ae^(17x) + Be^(-x)

Initial conditions are:
y(0)=0; and
y'(0)=2

From this I got A = 1/9 and B= -1/9

Leading to:

1/9e^(17x) - 1/9e^(-x)

Even I can see the second term is tending to zero - as it was when I tested a few points.

In the question the instructions were:

"Find (if possible) the general solution to each of the following 2nd-order ODE by trying y=e^(landa x). If you can't give a solution, give a reason.
(a) y''+4y'+25y = 0
(b) y''-16y'-17y = 0; subject to y(0) = 0 and y'(0)=2"

Therefore I assume there is something unsolvable - do you think that would refer to not having initial conditions for part (a) or have something to do with the second half of the equation for (a) tending to zero?

For part (a) there were 2 complex roots but I thought that still meant it was solvable and at our level we are given the trig equation to substitute the values into - so I can't work out what might have gone wrong there.

I am just a bit confused as both seemed solvable yet the caveat in the question re not being able to find an solution leads me to think there is something more to this.

Kind regards
Beetle
 

romsek

MHF Helper
Nov 2013
6,738
3,034
California
Re: Second Order ODE second term tending to 0 - question reads something must be unso

it certainly doesn't tend to zero when $x < 0$

the solution you posted satisfies the differential equation and the initial conditions and is the correct answer.

I'm not sure I understand your confusion.
 
Mar 2016
170
0
Australia
Re: Second Order ODE second term tending to 0 - question reads something must be unso

Thanks, I am glad I am right I am just confused by my lecturer who indicated there could be some problem solving on the ODE's ie "Find a solution if possible...If you can't find a solution, give a reason". He wouldn't put that there for nothing so trying to figure out why it is there - that is the confusion.
 

romsek

MHF Helper
Nov 2013
6,738
3,034
California
Re: Second Order ODE second term tending to 0 - question reads something must be unso

Thanks, I am glad I am right I am just confused by my lecturer who indicated there could be some problem solving on the ODE's ie "Find a solution if possible...If you can't find a solution, give a reason". He wouldn't put that there for nothing so trying to figure out why it is there - that is the confusion.
Have you all been dealing with complex roots and exponentials?

Maybe he meant that the first problem had no real roots. This certainly doesn't mean the differential equation has no solution. It just means the solutions are strictly periodic.
 
Mar 2016
170
0
Australia
Re: Second Order ODE second term tending to 0 - question reads something must be unso

We learnt about complex numbers which I struggled with (as it is taught in year 11 & 12 and the last maths I did was 1st year in 1996 - most of which is now taught at school!). With the 2nd order ODE he made it really easy, told us to use e^(landa x) and then wrote out the general solutions for the different scenarios:
1. Two real (different) roots = y=A^(e landa1 x) + Be^(landa 2x)
2. Complex conjugate y=e^(alpha x) [Acos (Beta x) + Bsin (Beta x)
3. Repeated real root y=(Ax+b)e^(landa x).
That is what is covered in lectures. I am external so don't get to go to tutes which would help a lot.

By the way thank you very much for your help.

Kind regards
Beetle