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Solve a differential equation for the initial conditions,

  • Thread starter mesa
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  • #1
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Homework Statement



Solve y' = (y^3)(t^2) for the initial condition y(0)=0 and state in which interval in 't' this solution exists.

The Attempt at a Solution



First I divided both sides by y^3 and then subtracted t^2 from each as well,

-t^2 + y'(y^-3) = 0

then solved,

(-t^3)/3 + (y^-2)/-2 = C

and then uh oh for y = 0 so chugging ahead and leaving 'C' alone for now,

y^-2 = (C + (t^3)/3)/-2

so,

y = [-2/(C+(t^3)/3)]^(1/2)

Therefore (C + (t^3)/3) ≤ 0

I am sure (as usual) I am missing something simple :P
 

Answers and Replies

  • #2
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and then uh oh for y = 0 so chugging ahead and leaving 'C' alone for now,

y^-2 = (C + (t^3)/3)/-2
I cannot follow this. According to the question, y is zero when t is zero. What do you get for C?
 
  • #3
pasmith
Homework Helper
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Homework Statement



Solve y' = (y^3)(t^2) for the initial condition y(0)=0 and state in which interval in 't' this solution exists.

The Attempt at a Solution



First I divided both sides by y^3 and then subtracted t^2 from each as well,

-t^2 + y'(y^-3) = 0

then solved,

(-t^3)/3 + (y^-2)/-2 = C

and then uh oh for y = 0 so chugging ahead and leaving 'C' alone for now,

y^-2 = (C + (t^3)/3)/-2

so,

y = [-2/(C+(t^3)/3)]^(1/2)

Therefore (C + (t^3)/3) ≤ 0
For [itex]y(0) \neq 0[/itex] it is the case that the solution only exists when [itex]3C + t^3 < 0[/itex]. But for [itex]y(0) = 0[/itex] that method doesn't work at all. And in fact [itex]y(t) = 0[/itex] is a solution of [itex]y' = y^3t^2[/itex] which satisfies [itex]y(0) = 0[/itex].
 
  • #4
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18
I cannot follow this. According to the question, y is zero when t is zero. What do you get for C?
Sorry, let me explain where I am coming from. We have,

(-t^3)/3 + (y^-2)/-2 = C

With the (1/-2y^2) term I get infinity for 'C'. That's my sticking point, so obviously I',ve goofed this up and there is another way to solve for 'C'?
 
  • #5
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18
For [itex]y(0) \neq 0[/itex] it is the case that the solution only exists when [itex]3C + t^3 < 0[/itex]. But for [itex]y(0) = 0[/itex] that method doesn't work at all. And in fact [itex]y(t) = 0[/itex] is a solution of [itex]y' = y^3t^2[/itex] which satisfies [itex]y(0) = 0[/itex].
I'm sorry but I am not quite following what you are saying. Should I not solve for the general form?
 
  • #6
pasmith
Homework Helper
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I'm sorry but I am not quite following what you are saying. Should I not solve for the general form?
You've done that, and shown that there is no choice of C for which y(0) = 0. At that point you need to find another approach.

In any event, if you have the initial condition [itex]y(0) = D[/itex], and it is obvious on inspection that [itex]y' = 0[/itex] when [itex]y = D[/itex], then the solution is [itex]y(t) = D[/itex] and you can save yourself some work.
 
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  • #7
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You've done that, and shown that there is no choice of C for which y(0) = 0. At that point you need to find another approach.

In any event, if you have the initial condition [itex]y(0) = D[/itex], and it is obvious on inspection that [itex]y' = 0[/itex] when [itex]y = D[/itex], then the solution is [itex]y(t) = D[/itex] and you can save yourself some work.
It's the not working in the general form that is bothering me but I think I have this one figured out.
Here is what I sent to my Professor,

Our y(0)=0 is saying our general form:
y = [-2/(t^3/3+C)]^(1/2) which is also = to 0 when t = 0

So we need a negative value for our ((t^3/3+C) to keep our square root happy.

we also have our other form,
(-t^3)/3 + y^(-2)/-2 = C and t=0 so,

0+ 1/0 = C, except I should have kept in mind that our division by 0 is still a negative if we take the limit as we approach 0 for our 'y' since we have that '-2' coefficient in the denominator so really we have,

lim y-->0 1/-2y^2 = 1/-0 = C which is = to -infinity

And then our square root in the general form is satisfied and the Universe doesn't blow up, thank goodness!
So we can now state the interval 't' is all real numbers. That seems right but...
I feel satisfied with this although I also like the explanation I just saw from the Professor.
 
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  • #8
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Edit:

I'm sorry but I am not quite following what you are saying. Should I not solve for the general form?
What (s)he was trying to state is that y = 0 is an equilibrium solution of the differential equation. What would this imply about the equation?
 
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  • #9
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What (s)he was trying to state is that y = 0 is an equilibrium solution of the differential equation. What would this imply about the equation?
I understand what they were getting at now, ignoring that we didn't have a solution for the general form in order to pursue the differential equation is where I was getting stuck. Both are solved at this point so I am satisfied unless someone else has something to add?
 

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