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Solve Dy/dx=x+y^2

  • Thread starter cerium
  • Start date
  • #1
15
0

Homework Statement



do I use the integrating factor for this question and if I do when i rearrange y^2 to the other side into the form of p(x)y does x become -1

Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2


The ODE is not linear, so the linear integrating factor you are most likely talking about does not apply (in the same way as you have learned). Are there other methods you have learned? Are there ways of converting an equation like this into a linear equation? Check and see what you can find. If you need more help, let us know.
 
  • #3
15
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Im really confused I thought I rearranged and moved the y^2 to the other side, Im I missing a really easy point, how could I change it to make it linear
 
  • #4
D H
Staff Emeritus
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You posted this in the homework section, so I would presume it is for some class. Which one? This certainly isn't the typical calc III type of problem. An ODE class, perhaps, but since this is a nonlinear ODE the techniques used to solve linear ODEs won't work here.

In fact, this is a Riccati equation.
 
  • #5


solve z''+x*z=0?
 
Last edited:
  • #6
2,967
5


if you take
[tex]
y = -\frac{z'}{z}
[/tex]
then:
[tex]
y' = -\frac{z''}{z} + \frac{(z')^2}{z^2}
[/tex]
and your equation becomes of 2nd order, but a linear one:
[tex]
z'' + x \, z = 0
[/tex]
The solution for z is in terms of Airy functions.
 
  • #7


solve z''+x*z=0 to find z?
 
  • #8


solve z''+x*z=0?
solve z''+x*z=0 to find z?
if you take
[tex]
y = -\frac{z'}{z}
[/tex]
then:
[tex]
y' = -\frac{z''}{z} + \frac{(z')^2}{z^2}
[/tex]
and your equation becomes of 2nd order, but a linear one:
[tex]
z'' + x \, z = 0
[/tex]
The solution for z is in terms of Airy functions.
what is the solution for [z′′+x*z=0]
 
  • #9


what the solution for z in terms Airy functions for z′′+x*z=0 ?
 

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