non-linear ODE problem

[Sarahisme]

hey all, just having a bit of trouble with this problem...

 

 

i have done a little bit on it, just thought i'd let you guys look at it first, because i havent been able to type up my solution yet.

 

Cheers guys & gals

 

Sarah

[Perturbation]

The show part is easy enough (just remeber to use the original ODE for getting rid of g'). Well I guessed the solution g(t)=exp(t). Using the transformation we have

 

[math]z'+(a+2bg)z=-b\Rightarrow z'+(1+2e^{-t}e^t)z=-e^{-t}[/math]

 

Which has the solution [imath]z=-\frac{1}{2}e^{-t}[/imath] and transforming back will give you the general solution for y(t).

 

If the show bit is what you're having trouble with, gimme a shout.

[Sarahisme]

The show part is easy enough (just remeber to use the original ODE for getting rid of g'). Well I guessed the solution g(t)=exp(t). Using the transformation we have

 

[math]z'+(a+2bg)z=-b\Rightarrow z'+(1+2e^{-t}e^t)z=-e^{-t}[/math]

 

Which has the solution [imath]z=-\frac{1}{2}e^{-t}[/imath] and transforming back will give you the general solution for y(t).

 

If the show bit is what you're having trouble with' date=' gimme a shout.[/quote']

 

lol, yeah actually the 'show' bit is casuing me a bit of trouble anyy suggestions?

[Sarahisme]

lol, sorry i got that 'show' part out now. i put a plus instead of a minus at one point and everything went crazy, but its all good now!

[Sarahisme]

The show part is easy enough (just remeber to use the original ODE for getting rid of g'). Well I guessed the solution g(t)=exp(t). Using the transformation we have

 

[math]z'+(a+2bg)z=-b\Rightarrow z'+(1+2e^{-t}e^t)z=-e^{-t}[/math]

 

Which has the solution [imath]z=-\frac{1}{2}e^{-t}[/imath] and transforming back will give you the general solution for y(t).

 

If the show bit is what you're having trouble with' date=' gimme a shout.[/quote']

 

should there be an arbitrary constant of integration in there somewhere?

[Sarahisme]

i guessed the solution [math] g(t) = Ae^{t} [/math]

 

and after transforming back i got the GENERAL solution to be somethign which has TWO arbitrary constants:

 

[math] y(t) = \frac{1}{\frac{-1}{2A}e^{-t} + Ce^{-t(1 + 2A)}} + Ae^{t} [/math]

 

where C is constant of integration. and A is any real number.

 

 

what do you think?