Finding the general solution from a given particular solution.

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October 6th 2009, 11:46 PM
bsanghera

Finding the general solution from a given particular solution.

for http://webwork2.math.ucsb.edu/webwor...09fa7a1871.png One solution of this differential equation is http://webwork2.math.ucsb.edu/webwor...881b1bd471.png Using this information, what is the general solution of the DE?

y(t) = ?
October 7th 2009, 12:45 AM
chisigma

The general solution of the 'incomplete equation' $y^{'} + y =0$ is $y_{g}(t)= c\cdot e^{-t}$ . The particular solution of the 'complete equation' $y^{'} + y = t$ You know is $y_{p} (t) = e^{-t} + t -1$ , so that the general solution of the 'complete equation' is...

$y(t)=y_{g} (t) + y_{p} (t)= (1+c)\cdot e^{-t} + t -1$

Kind regards

$\chi$ $\sigma$
October 7th 2009, 12:53 AM
bsanghera

Straight-line solution?!?!?

Quote:

Originally Posted by chisigma

The general solution of the 'incomplete equation' $y^{'} + y =0$ is $y_{g}(t)= c\cdot e^{-t}$ . The particular solution of the 'complete equation' $y^{'} + y = t$ You know is $y_{p} (t) = e^{-t} + t -1$ , so that the general solution of the 'complete equation' is...

$y(t)=y_{g} (t) + y_{p} (t)= (1+c)\cdot e^{-t} + t -1$

Kind regards

$\chi$ $\sigma$

sorri i accidently posted the wrong question lol i meant to post this one:
for:
http://webwork2.math.ucsb.edu/webwor...09fa7a1871.png What is the straight-line solution of this DE?
October 7th 2009, 01:04 AM
mr fantastic

Quote:

Originally Posted by bsanghera

sorri i accidently posted the wrong question lol i meant to post this one:
for:
http://webwork2.math.ucsb.edu/webwor...09fa7a1871.png What is the straight-line solution of this DE?

The answer is already in this thread. Look for it.
October 7th 2009, 01:30 AM
bsanghera

E^(-t) +t -1 is what I think the answer is but I checked on the website where I am submitting it and it tells me my answer is incorrect that is why I asked again!!!
October 7th 2009, 01:44 AM
mr fantastic

Quote:

Originally Posted by bsanghera

E^(-t) +t -1 is what I think the answer is but I checked on the website where I am submitting it and it tells me my answer is incorrect that is why I asked again!!!

$y = e^{-t} + t - 1$ is not linear! However, $y = t - 1$ is ....