Need help with non-linear ODE IVP

[thelannonmonk]

Homework Statement

given: (y-xy)y' = y^(2) + 1 y(0)=0

Homework Equations

So I subbed y^(2) = u

The Attempt at a Solution

w/ u=y^(2) i got...

(y-xy)y' = y^(2) + 1
(1-x)yy' = y^(2)
(1-x)(u'/2) = u + 1

and it is here that I am stuck! I [think] I have to put it into the form u' + g(x)*y = b(x) but I can't do it! It is driving me crazy, I have been working on this single problem for over an hour, I am about to lose my sanity, especially since it's ALGEBRA that's tripping me up.

Any help would be greatly appreciated.

I was thinking of trying u = (y^(2) + 1) but I feel like that's leading me to nowhere as well.

Thanks in advance

 

[mighty2000]

.

I understand your frustration and I am here to help. Let's take a step back and look at the problem again. We have the equation (y-xy)y' = y^(2) + 1 with the initial condition y(0)=0. You have correctly substituted u=y^(2) to simplify the equation, but let's continue from there.

(1-x)(u'/2) = u + 1

First, let's expand the left side:

u'/2 - xu'/2 = u + 1

Now, let's move all terms involving u' to one side and all other terms to the other side:

u'/2 - u = xu'/2 + 1

Next, let's group the terms with u' together and factor out u':

u'(1/2 - x) = u + 2

Now, we can divide both sides by (1/2 - x):

u' = (u + 2)/(1/2 - x)

Finally, we can substitute back in u=y^(2):

y' = (y^(2) + 2)/(1/2 - x)

We now have the equation in the form y' + g(x)y = b(x), where g(x) = 1/(1/2 - x) and b(x) = (y^(2) + 2)/(1/2 - x). We can now use the method of integrating factors to solve for y. I hope this helps! Let me know if you have any further questions.