Solving the IVP: $y'(x) = \frac{1}{1-(xy)^2}, y(-1)=1$ I am only starting to learn diff.equations and have the following initial value problem:
$y'(x) = \frac{1}{1-(xy)^2}, y(-1)=1$. So, since $y'(x)$ is undefined at $(-1,1)$ do we say that the solution of the initial value problem doesn't exist or do we say that it "jumps" to infinity (or negative infinity?) as $y'(x)$ at the given point is infinite? Is $y(-1)=1$ still a solution of the problem? Is this point the only solution? I am lost.
EDITED: Given the response below, if we solve such a problem with Euler's method, will it not provide any solution? I assume the Existence and Uniqueness theorem will also not be applicable. How can we ensure that the solution exists?
 A: A diferential equation $F(x,y(x),y'(x))=0$ is determined by a function $F:\Omega\to\mathbb R$ defined on an open set $\Omega$ contained in $\mathbb R^3$. It simply does not make any sense to consider an initial condition for the equation which is not in the domain of $F$.
In your situation, it is not that there is no solution: it does not even make sense to look for one.
Now, if you consider instead the equation $(1-(xy(x))^2)y'(x)-1=0$, which is different to yours because it has a different domain, then it makes sense to look for solutions with your initial condition, but obviously there are none.
A: Just because
$y'(-1)$ is infinite
does not mean
that the solution
is not defined.
For example,
$\sqrt{x}$ has derivative
$\frac{1}{2\sqrt{x}}
$
so its slope is infinite at the origin.
A: $\dfrac{dy}{dx}=\dfrac{1}{1-(xy)^2}$ with $y(-1)=1$
$\dfrac{dx}{dy}=1-y^2x^2$ with $x(1)=-1$
Let $x=\dfrac{1}{u}$ ,
Then $\dfrac{dx}{dy}=-\dfrac{1}{u^2}\dfrac{du}{dy}$
$\therefore-\dfrac{1}{u^2}\dfrac{du}{dy}=1-\dfrac{y^2}{u^2}$ with $u(1)=-1$
$\dfrac{du}{dy}=y^2-u^2$ with $u(1)=-1$
Let $u=\dfrac{1}{v}\dfrac{dv}{dy}$ ,
Then $\dfrac{du}{dy}=\dfrac{1}{v}\dfrac{d^2v}{dy^2}-\dfrac{1}{v^2}\left(\dfrac{dv}{dy}\right)^2$
$\therefore\dfrac{1}{v}\dfrac{d^2v}{dy^2}-\dfrac{1}{v^2}\left(\dfrac{dv}{dy}\right)^2=y^2-\dfrac{1}{v^2}\left(\dfrac{dv}{dy}\right)^2$
$\dfrac{d^2v}{dy^2}-y^2v=0$
One of the form of the general solution is $v=C_1D_{-\frac{1}{2}}(\sqrt2y)+C_2D_{-\frac{1}{2}}(i\sqrt2y)$ (according to http://www.wolframalpha.com/input/?i=v%22-y%5E2v%3D0 and http://zh.scribd.com/doc/14623310/Safari-on-the-country-of-the-Special-Functions-Safari-au-pays-des-fonctions-speciales)
How does this ODE cannot solve?
