Finding the general solution from a given particular solution.

**bsanghera** · October 7th 2009, 12:46 AM

for

One solution of this differential equation is

Using this information, what is the general solution of the DE?

y(t) = ?

**chisigma** · October 7th 2009, 01:45 AM

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

$\chi$ $\sigma$

**bsanghera** · October 7th 2009, 01:53 AM

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:

What is the straight-line solution of this DE?

**mr fantastic** · October 7th 2009, 02:04 AM

Originally Posted by bsanghera

sorri i accidently posted the wrong question lol i meant to post this one:
for:

What is the straight-line solution of this DE?

The answer is already in this thread. Look for it.

**bsanghera** · October 7th 2009, 02:30 AM

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!!!

**mr fantastic** · October 7th 2009, 02:44 AM

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 ....

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Straight-line solution?!?!?

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