Differential Equations: System of equations

  • Thread starter SpiffyEh
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  • #1
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Homework Statement


I have the following 2 equations dx/dt = 3x+y and dy/dt = sqrt(y)
I need to find the general solution to the system.

Homework Equations



dx/dt = 3x+y and dy/dt = sqrt(y)

The Attempt at a Solution



i did the y equation first using separation of variables and got
2sqrt(y) = t+c
so... y = (2t+A)^2 where A = 2c

then i plugged this into the 2nd equation for y so i have dx/dt = 3x + (2t+2t)^2
i'm not sure how to solve this. I tried using method of undetermined coefficients but i keep getting stuck. Can someone please show me step by step how to solve this? Thank you
 

Answers and Replies

  • #2
tiny-tim
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Hi SpiffyEh! :smile:

(have a square-root: √ :wink:)
2sqrt(y) = t+c
so... y = (2t+A)^2 where A = 2c

No, (t/2 + A). :wink:
then i plugged this into the 2nd equation for y so i have dx/dt = 3x + (2t+2t)^2
i'm not sure how to solve this. I tried using method of undetermined coefficients but i keep getting stuck.

(You need both a complementary solution and a particular solution … I assume you've got the former?)

Substituting u = t/2 + A may make it easier.

If it doesn't, show us how far you get, and where you're stuck, and then we'll know how to help! :smile:
 
  • #3
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oops sorry completely missed the t/2...

well when i start the method of undetermined coefficients i can ge the yh but the yp is where i have issues. Since it has to be something like the right hand side i can't figure out what to set it to. Is this the wrong method to use in this case?
 
  • #4
tiny-tim
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Hi SpiffyEh! :wink:

Whenever I see a polynomial on the RHS, I always try a polynomial as the particular solution. :smile:
 
  • #5
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i understand that, i just don't see how it works out because of the c being in two different parts of the equation. I don't know if im just being stupid and thinking of it wrong or not.
 
  • #6
tiny-tim
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(just got up :zzz: …)
i understand that, i just don't see how it works out because of the c being in two different parts of the equation. I don't know if im just being stupid and thinking of it wrong or not.

I'm confused … I thought you were trying to solve dx/dt = 3x + (t/2 + A)2 ? :confused:
 
  • #7
D H
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The constant of integration from integrating the first equation has nothing to do with the constant of integration from integrating the second equation. You will have two constants of integration here. Think of them as c1 and c2, or A and B if you want.
 

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