# Differentials

• February 20th 2010, 12:43 AM
IBD
Differentials
Hi guys! I ve got this very difficult exercise for my uni. Could someone please help? Thank you in advance.

http://img2.immage.de/200270494rszexercise2.jpg
• February 20th 2010, 01:02 AM
Prove It
Quote:

Originally Posted by IBD
Hi guys! I ve got this very difficult exercise for my uni. Could someone please help? Thank you in advance.

http://img2.immage.de/200270494rszexercise2.jpg

What's the exercise?
• February 20th 2010, 01:11 AM
IBD
Thanks for looking into my problem Prove It. I have inserted a picture of the exercise but it seems that you could not see it. I have now uploaded it as an attachment. Thanks again
• February 20th 2010, 01:39 AM
Prove It
Why not substitute D and S into the equation for $p''$?

This will give you a second order constand coefficient ODE.
• February 20th 2010, 09:16 PM
IBD
Thanks Prove it for your response. I will substitute into p'' however i don't get p' this way. could you please let me know how i will obtain the differential equation that the first question asks for?
• February 20th 2010, 09:31 PM
Prove It
Quote:

Originally Posted by IBD
Thanks Prove it for your response. I will substitute into p'' however i don't get p' this way. could you please let me know how i will obtain the differential equation that the first question asks for?

You should end up with

$p''(t) = a[d_0 + d_1p(t) - (s_0 + s_1p(t))]$

$= ad_0 + ad_1p(t) - as_0 - as_1p(t)$

$= (ad_1 - as_1)p(t) + ad_0 - as_0$.

So $p''(t) + (as_1 - ad_1)p(t) = ad_0 - as_0$.

Solve the homogeneous characteristic equation:

$m^2 + as_1 - ad_1 = 0$

$m^2 = ad_1 - as_1$

$m^2 = a(d_1 - s_1)$.

Now since $d_1 < 0$ and $s_1 > 0$, this means $d_1 - s_1 < 0$.

Also, since $a > 0$, this means $a(d_1 - s_1) < 0$.

So you have $m^2$ equaling a negative number. What does this tell you about $m$? Can you solve the DE now?
• February 20th 2010, 11:30 PM
IBD
My dearest Prove It can you please send me a pm with some contact information (email)? I cannot send you a pm because I am a new member and don't have 15 posts. I need to discuss something in private. Thanks