it's a separable ODE:
$\displaystyle
\begin{gathered}
\frac{{dy}}
{{dt}} = \sqrt y e^{ - t} \hfill \\
\frac{{dy}}
{{\sqrt y }} = e^{ - t} dt \hfill \\
\end{gathered} $
integrate both sides:
$\displaystyle
2\sqrt y = C - e^{ - t} $
apply initial conditions:
$\displaystyle \begin{gathered}
2 = C - 1 \Rightarrow C = 3 \hfill \\
\sqrt y = \frac{{3 - e^{ - t} }}
{2} \hfill \\
\end{gathered}
$
I see! A little wrong calculations by me as I can see now. Thanks for your answere!
Just to check if I've done this correctly... Could you take a look on this one;
$\displaystyle \frac{{dy}} {{dt}} = y*sint$
Solving this;
$\displaystyle y = e^{-cost} + C$
If anyone or you Peritus can comfirm/correct this one as well;
$\displaystyle \frac{{dy}} {{dt}} = yk$
Where;
$\displaystyle \frac{{1}} {{y}}dy = kdt$
Where;
$\displaystyle ln|y| = k + C$
Where;
$\displaystyle e^{ln|y|} = e^{k + C}$
Which is;
$\displaystyle y = Ce^{k}$
???
Thanks Chris!
One question, how can I show that this is periodic;
$\displaystyle y = e^{-cost} + C$
I know cos has a periodic value =
$\displaystyle cos(x)2kpi$
First initialise with a value e.g
$\displaystyle y(0) = 1 $
and then;
$\displaystyle y(x+2pi) = y(x)$
?
Im not quite sure myself... Hehe!
However, when I have your "attention" and appriciate all the help I can get,
If you were to show values for k in;
$\displaystyle y = Ce^{k}$
and explain what happens in the cases of
$\displaystyle k < 0$
$\displaystyle k > 0$
$\displaystyle k = 0$
when
$\displaystyle k \Rightarrow infinity$(cant remeber the code for that sign)
How can you approach that...