Initial Value Problem Question - Differential Equations

  • #1

Homework Statement


Solve the initial value problem where y(0) = 2.


Homework Equations


dy/dt = 3 - 5(y^(1/2))


The Attempt at a Solution


I tried the separable equation method but when it came time to take the integral of
1/[3 - 5(y^(1/2)], every solution I got became too complex to solve for y.
I'm thinking maybe you can just somehow put this into the linear form and use Leibniz's method by multiplying the equation by u(t) but I don't see how I can do this.
All I need is some info on how to make the integral simpler (so I can eventually solve for y) or how I can change this into linear form.
 

Answers and Replies

  • #2
1,753
1
[tex]\frac{dy}{dt}=3-5y^{\frac 1 2}[/tex]

Correct? If so, have you learned these 2 methods: Integrating factor or Variation of parameters?
 
  • #3
That is correct (the y^1/2 is a square root but I can't really type that).
I've learned the integrating factor method, I believe, but not variation of parameters.

Try explaining what you're thinking and I'll see if I understand or recognize what I have learned :)
 
  • #4
1,753
1
Well you can't easily integrate this problem, so you will need to use the integrator factor.
 
  • #5
Will you demonstrate how to do this? or get me started? Maybe it's obvious but I can't seem to figure it out.
 
  • #6
I know how to do the integrating factor method, but not with this problem. How would i get rid of the square root so as to have this in its linear form? (dy/dt + p(t)y = g(t))
 
  • #7
Defennder
Homework Helper
2,591
5
Separation of variables works here. The expression is also easy to integrate with an appropriate substitution.
 
  • #8
Any hints as to what that substitution might be? I've tried letting u = 3-5(y^(1/2)) and
u = y^(1/2) and I've even tried multiplying top and bottom by 3 + 5(y^(1/2)) in order to remove the square root from the bottom. Nothing has worked for me so far...I must be missing something really obvious.
 
  • #9
gabbagabbahey
Homework Helper
Gold Member
5,002
7
Show us your work for the u=y^1/2 substitution; you should be able to integrate it without much trouble using this sub.
 
  • #10
Defennder
Homework Helper
2,591
5
Any hints as to what that substitution might be? I've tried letting u = 3-5(y^(1/2)) and
u = y^(1/2) and I've even tried multiplying top and bottom by 3 + 5(y^(1/2)) in order to remove the square root from the bottom. Nothing has worked for me so far...I must be missing something really obvious.
Why doesn't this work?
 
  • #11
Ok. Maybe I just made a simple mathematical error the first time around. Let me give it another try and I'll get back to you.
 
  • #12
DOH! haha. Ok I see it now. For some reason it didn't work last night (I must have been tired). I was able to fully integrate.
Thanks so much guys for your help! (and your patience :) )
 
  • #13
1,753
1
Separation of variables works here. The expression is also easy to integrate with an appropriate substitution.
Oh crap it is ... lol, just divide the right and get dt on the right ... sorry!!!
 
  • #14
Um...ACTUALLY I have one more question. First off, I accidentally made a typo...it's actually 5 - 3(y^(1/2)) NOT 3 - 5(y^(1/2)). Second, after I took the integral of 1/(5-3(y^(1/2))) I got (-10/9)ln l 5 - 3(y^(1/2)) l + (2/9)(5-3(y^(1/2))) = t + C. How would I solve for y in this case? Any hints would be appreciated :)
 
  • #15
Defennder
Homework Helper
2,591
5
Do you really have to solve in terms of y? Remember that the only thing you're required to do is to solve an initial value problem. You're given the initial condition. So just plug that in to find C. It's not always possible to express a function f(y,t) = g(t) in terms of y only on the left hand side. This has something to do with the implicit function theorem:
http://en.wikipedia.org/wiki/Implicit_function_theorem
 
  • #16
I don't know if I HAVE to solve in terms of y, but all the examples/problems that we've done in class have been in that form so I really don't know.
So I take it there is no real way of finding it it in terms of y on the left hand side?
 
  • #17
Actually, I take that back. There is a second part to the problem. I have to use Euler's method to approximate at several values of t. I take it I need to first solve for y in this IVP problem so I can use it in Euler's method. Any ideas on how I'm supposed to do this?
 

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