1. Solving dy/dx = (1+3x^2)/(3y^2-6y)

$dy/dx=(1+3x^2)/(3y^2-6y)$ with $y(0)=1$

It was easy enough getting it to the solution:

$y^3-3y^2-x-x^3+2=0$.

Now it wants me to find the solution where it's valid. The answer in the back of the book is that $|x|<1$ and the hint was find where the integral curve has a vertical tangent. I'm not really sure how they arrived to this solution though.

Thanks for any help

2. You got that, I presume, by multiplying both sides by $3y^2- 6y$ and integrating. Of course, that is valid only as long as the denominator of the derivative is not 0- that is as long as $3y^2- 6y= 3y(y- 2)\ne 0$. At points where it is 0, that is where y= 0 or y= 2, the denominator is 0 so the derivative does not exist (in other words, the graph has a vertical tangent). Since y lies between 0 and 2, this solution must be valid only for y between 0 and 2.

Now, if you put y= 0 in your solution, you get $-x- x^3+ 2= 0$ and it is not too difficult to see that x= 1 is a solution to that. $-x- x^2+ 2= -(x- 1)(x^2+ x+ 2)$ and $x^2+ x+ 2= 0$ has no real roots so x= 1 is the only real solution.

If you put y= 2 in your solution, you get $8- 12- x- x^2+ 2= -2- x- x^2$ and it is not difficult to see that x= -1 is the only real solution to that equation. Since x=0 is between -1 and 1, we see that this solution is valid only for x between -1 and 1- that is, |x|< 1.

3. thanks for the explanation

My brain is so fried today from all the math. Here's to you: