Solve Dy/dx=x+y^2 | Integrating Factor & Rearrange

[cerium]

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

 

[lineintegral1]

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.

 

[cerium]

Im really confused I thought I rearranged and moved the y^2 to the other side, I am I missing a really easy point, how could I change it to make it linear

 

[D H]

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.

 

[ali salah mah]

solve z''+x*z=0?

 

[Dickfore]

if you take
<br /> y = -\frac{z&#039;}{z}<br />
then:
<br /> y&#039; = -\frac{z&#039;&#039;}{z} + \frac{(z&#039;)^2}{z^2}<br />
and your equation becomes of 2nd order, but a linear one:
<br /> z&#039;&#039; + x \, z = 0<br />
The solution for z is in terms of Airy functions.

 

[ali salah mah]

solve z''+x*z=0 to find z?

 

[ali salah mah]

solve z''+x*z=0?

solve z''+x*z=0 to find z?

if you take
<br /> y = -\frac{z&#039;}{z}<br />
then:
<br /> y&#039; = -\frac{z&#039;&#039;}{z} + \frac{(z&#039;)^2}{z^2}<br />
and your equation becomes of 2nd order, but a linear one:
<br /> z&#039;&#039; + x \, z = 0<br />
The solution for z is in terms of Airy functions.

what is the solution for [z′′+x*z=0]

 

[ali salah mah]

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

 

[Dickfore]

See this:

http://www.wolframalpha.com/input/?i=DSolve[y%27%27[x]%2Bx*y[x]%3D%3D0%2Cy[x]%2Cx]#