# Thread: Finding solution to a DFQ

1. ## Finding solution to a seperable ODE

Hi!

I have this DFQ;

Im struggling on how to find the solution on this one. I think im starting wrong somehow.

Could it be something like;

$\displaystyle y = (e^-t + C)^2$

How do I continue solving $\displaystyle y(0) = 1$

Any help is appriciated!

2. 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:

$\displaystyle \begin{gathered} 2 = C - 1 \Rightarrow C = 3 \hfill \\ \sqrt y = \frac{{3 - e^{ - t} }} {2} \hfill \\ \end{gathered}$

3. 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;

$\displaystyle y = e^{-cost} + C$

4. 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;

$\displaystyle y = Ce^{k}$

???

5. Originally Posted by jokke22
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;

$\displaystyle y = e^{-cost} + C$
Correct!

Originally Posted by jokke22
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;

$\displaystyle y = Ce^{k}$

???
Correct!

Now, if you have initial conditions, apply them to find C, and then write out the exact solution(s).

--Chris

6. 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;
$\displaystyle y(x+2pi) = y(x)$

?

7. If I try to initialize;
$\displaystyle y = e^{-cost} + C$
$\displaystyle y(0) = 1$

$\displaystyle y = e^{-cost} + C = 1$

$\displaystyle = 0,37 + C = 1$

$\displaystyle C = 0.63$

Would this be correct?

8. Originally Posted by jokke22
If I try to initialize;
$\displaystyle y = e^{-cost} + C$
$\displaystyle y(0) = 1$

$\displaystyle y = e^{-cost} + C = 1$

$\displaystyle = 0,37 + C = 1$

$\displaystyle C = 0.63$

Would this be correct?

Yes, that would be correct.

However, I'm not quite sure how to show its periodic...

--Chris

9. 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...