First order ODE initial value problem

  • Thread starter Shade
  • Start date
  • #1
Shade
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Homework Statement



Given the below stated equations I need to find the exact polynomial given the initial condition.

y(0) = 1
y = 4*t*sqrt(y)

Homework Equations




The Attempt at a Solution



I simply disregard the initial value condition and get y = t^4

How can I find the fourth order polynomial with the given initial value?

( see also a former thread https://www.physicsforums.com/showthread.php?t=111094 )
 

Answers and Replies

  • #2
Shade
13
0
Is it not possible to integrate a first order differential equation and also consider the initial value y(0) = 1 ?
 
  • #3
gammamcc
150
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Isn't there a typo? Where is y' ?
 
  • #4
Plastic Photon
138
0
I assume the equation is written y'=4ty^(1/2) in which case it appears you have dy/dt=y' and this is a seperable equation. Solve for y and then solve for the initial value.
 
  • #5
HallsofIvy
Science Advisor
Homework Helper
43,021
970

Homework Statement



Given the below stated equations I need to find the exact polynomial given the initial condition.

y(0) = 1
y = 4*t*sqrt(y)

Homework Equations




The Attempt at a Solution



I simply disregard the initial value condition and get y = t^4

How can I find the fourth order polynomial with the given initial value?

( see also a former thread https://www.physicsforums.com/showthread.php?t=111094 )

I can see any "attempt at a solution". Saying "I simply disregard the initial value condition and get y = t^4" makes no sense! How did you "get y= t^4"? Don't you have to integrate somewhere and doesn't that introduce a "constant of integration?

The differential equation is [itex]\frac{dy}{dt}= 4t\sqrt{y}= 4ty^{\frac{1}{2}}[/itex]

That can be written [itex]y^{-\frac{1}{2}}dy= 4t dt[/itex]. Now integrate both sides.

It's interesting that there are two distinct solutions (actually, there are an infinite number of solutions). The "uniqueness" part of the "existance and uniquenss" theorem is not satisfied.
 

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