Differential equation of second order

  • Thread starter greg997
  • Start date
  • #1
77
0
I would very much appreciate if anyone can help me with this problem. I got stuck at the end.
I have to find
1.the complementary function, particular integral and the general solution.
2. find complete solution when t=0, q=0, dq/dt=0.

-equation 2(d^2q)/(dt^2)+200dq/dt+5000q = 250
-in D operator form it looks like (2D^2+200d+5000)y=250
- subsituting m for D i receve repeated root m= -50
- so the complemenary function (At+B)e^(-50t)
- right side is constant so i subsituted constant into equation and solved that k= 0.05
- so the general solution is q=(At+B)e^(-50t)+0.05 right??

Now when t=0. q=0 , dq/dt=0
1) 0 = (A(0)+B)e^0+0.05
B= -0.05

And now i have problem
2) dq/dt= -50(At+B)e^(-50t) what the solution should be. I should find A but it dissapears because of t=0

I have no idea what to do with it.

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
Cyosis
Homework Helper
1,495
0
dq/dt= -50(At+B)e^(-50t)

This is not true, don't forget the product rule.
 
  • #3
77
0
would that be like?:

First i multiplied A and B by that e.

(At+e^-50t)+ (Be^-50t)

so the chain rule gives me

[(At)'(e^-50t)+ (e^-50t)'(At)] + (Be^-50t)'

[A(e^-50t)-(50e^-50t(At)) ] -50Be^-50t t=0

0= A - 50-50(-0.05)
A= 47,5???

Is that right?
 
  • #4
jambaugh
Science Advisor
Insights Author
Gold Member
2,305
295
Your equation is inhomogeneous. You've found the homogeneous solution. Now you must find the particular solution. Since the inhomogeneous term is just a constant I believe your particular solution will be as well.
y_p=250/5000
y = y_h + y_p.

Always verify that your function is indeed a solution...even if you know you "did it right". The verification step should have reminded you that you forgot the particular part.

NOW you can substitute your initial conditions and resolve the values of your arbitrary constants.
 
  • #5
Cyosis
Homework Helper
1,495
0
He already calculated the particular solution, 0.05=250/5000.

[A(e^-50t)-(50e^-50t(At)) ] -50Be^-50t t=0

0= A - 50-50(-0.05)

The second equality is not correct. for t=0, the second term gives?
 
  • #6
77
0
Hmm, would that be

0= A-0-50(-0.05)
A= -2.5??

So the final solution is q= (-2.5-0.05t)e^-50t???
 
  • #7
Cyosis
Homework Helper
1,495
0
You're being extremely sloppy.

A=-2.5 is correct.

q= (-2.5-0.05t)e^-50t?

This is not correct. First off the particular solution disappeared and secondly you put the value for A in B and the value for B in A. Stop being sloppy!

Also when you find a solution you can check if it's correct or not all by yourself by just putting it into the differential equation and see if the equality holds. You should really teach yourself to check your answers like that.
 
  • #8
77
0
I am really sorry for being sloopy. I am nervous. Thank you for your help and comments.

I checked the solution and I think i should look like:

q= (-2.5t-0.05)e^-50t+ 0.05

and when t=0

q= -0.05 +0.05
q=0

:biggrin:
 
  • #9
Cyosis
Homework Helper
1,495
0
Your solution is now complete and correct!
 

Related Threads on Differential equation of second order

  • Last Post
Replies
1
Views
491
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
2
Views
740
  • Last Post
Replies
3
Views
993
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
735
  • Last Post
Replies
5
Views
926
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
1
Views
963
Top