# Thread: Finding the general solution from a given particular solution.

1. ## Finding the general solution from a given particular solution.

for One solution of this differential equation is Using this information, what is the general solution of the DE?

y(t) = ?

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

3. ## Straight-line solution?!?!?

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
sorri i accidently posted the wrong question lol i meant to post this one:
for:
What is the straight-line solution of this DE?

4. 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?
$\displaystyle y = e^{-t} + t - 1$ is not linear! However, $\displaystyle y = t - 1$ is ....