MHB Finding a value for a y(0) = a in a IVP

  • Thread starter Thread starter nathancurtis111
  • Start date Start date
  • Tags Tags
    Ivp Value
nathancurtis111
Messages
10
Reaction score
0
Find the value of a ∈ R for which the initial value problem:

(S) {(xy' +y)/(1+x2y2)=1
{y(0)= a

has a solution and, for this value of a, find explicitly the maximal solution of (S).

I'm assuming I first find the implicit solution of the original diff eq (which I'm not too sure how to do given the equation) and then find a number a which will not make the eq undefined? I'm not too sure how to go about this one.
 
Last edited:
Physics news on Phys.org
You have some derivatives in an expression there, but it's not a differential equation, because there's no equals sign. Is there more to the expression?
 
My apologies, it is equal to 1, will edit original post to include that.
 
Try the substitution $u=xy$. Then $u'=y+xy'$. Can you continue?
 
I have the equation ##F^x=m\frac {d}{dt}(\gamma v^x)##, where ##\gamma## is the Lorentz factor, and ##x## is a superscript, not an exponent. In my textbook the solution is given as ##\frac {F^x}{m}t=\frac {v^x}{\sqrt {1-v^{x^2}/c^2}}##. What bothers me is, when I separate the variables I get ##\frac {F^x}{m}dt=d(\gamma v^x)##. Can I simply consider ##d(\gamma v^x)## the variable of integration without any further considerations? Can I simply make the substitution ##\gamma v^x = u## and then...

Similar threads

  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
6
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 8 ·
Replies
8
Views
5K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K