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

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.

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?

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!

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?

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?