Help with first order, Bernoulli ODE

[scorpion990]

Help with first order, "Bernoulli" ODE

We just covered:
-First order linear ordinary differential equations
-Bernoulli Equations
-Simple substitutions.

This problem was assigned. Its supposedly a Bernoulli equation with respect to y, but I can't figure it out...

http://img520.imageshack.us/img520/12/23331767fh5.png

When I solve for dx/dy, I get dx/dy = x^3 -y/x, which is not a Bernoulli equation because of the factor of 1/x, and not x. Help?

 

[John Creighto]

Well, you can check your solution by plugging it back into the differential equation. Did you use the same method to solve it as suggested in wikipedia?

http://en.wikipedia.org/wiki/Bernoulli_differential_equation

 

[scorpion990]

I didn't get an answer at all. My problem is that I could not convert it into a form which I can use Bernoulli's substitutions on it. I tried finding an explicit for both y(x) and x(y).

 

[John Creighto]

I doesn't look like Bernoulli's equation but I wonder if you can use similar techniques. Is there a change of variables you can apply to put it into a form that you know how to solve?

 

[scorpion990]

A basic u = x^4-y substitution did not yield a linear (or a Bernoulli) differential equation =( I'm stumped. Does anybody mind steering me in the right direction?

 

[trambolin]

Can you try forming the equation for the variable x?

 

[Rainbow Child]

This problem was assigned. Its supposedly a Bernoulli equation with respect to y, but I can't figure it out...

The ODE $(x^4-y(x))\,y'(x)=x$ is not a Bernoulli equation and furthermore is not a simple one
You can transformed it into a Riccati one by the transformation

$$x=\frac{\sqrt{t(u)}}{2^{1/6}},\,y(x)=\frac{u}{2^{2/3}},\,y'(x)=\frac{2\,\sqrt{t(u)}}{t'(u)}$$

which makes the ODE

$$t'(u)-t(u)^2=-u$$

Now letting
$$t(u)=-\frac{w'(u)}{w(u)}$$
we arrive to

$$w''(u)-u\,w(u)=0$$

which is the definition of the Airy function.

 

[scorpion990]

Ya..As it turns out, our teacher gave us the wrong problem...
Thanks though! I really appreciate it!

(Its kind of interesting how tiny changes in the terms creates such a huge difference in difficulty)