ODE Interval of Validity
The problem is stated as such;
Solve the I.V.P $$y'=\frac{1+3x^2}{3y^2-6y}; y(0)=1 $$and determine the interval in which the solution is valid.
The hint given is: To find the interval of definition, look for points where the integral curve has a vertical tangent line.
I was able to solve the IVP with the following solution;
$$y^3-3y^2-x-x^3+2=0$$
But with the hint, I am under the assumption that the curve has vertical tangent lines when the denominator of the derivative is equal to 0, which would be when $y=0$ and $y=2$. But the solution is give as $|x|<1$. What am I doing wrong?
$\endgroup$ 61 Answer
$\begingroup$$y=0$ or $y=2$ are where problems occur (considering that the numerator never vanishes). But the question is asking about where this occurs in terms of $x$. Since you know the relevant $y$ values, you can plug them in and search for the closest solutions to the initial point $x=0$. This gives the equations
$$2-x-x^3=0,2-x-x^3=4$$
which you can solve (though the easiest way to do this by hand is just by guess-and-check). Then $(-1,1)$ is the largest open interval containing $0$ and not containing either $-1$ or $1$.
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