Originally Posted by

**ThePerfectHacker** We cannot say that. My point is that the existence/uniqueness theorem was only proven for linear differencial equations. When we have non-linear differencial equations we have no idea whether it works or not without solving it.

The theorem says that given,

$\displaystyle y^{(n)}+f_{n-1}(x)y^{(n-1)}+...+f_1(x)y'+f_0(x)y = g(x)$

With $\displaystyle y(x_0)=y_0 , y'(x_0)=y'_0 , ... , y^{(n)} (x_0)=y^{(n)} _0$.

Where $\displaystyle f_{n-1}(x),...,f_0(x),g(x)$ are continous functions on some open interval $\displaystyle (a,b)$ and where $\displaystyle x_0\in (a,b)$ then there exists exactly one solution to that differencial equation.

So given (after division),

$\displaystyle y^{(4)} + \frac{t+1}{t-1} y'' + \tan t \ y = 0 \mbox{ and }0=y(0)=y'(0)=y''(0)=y'''(0)=y''''(0)$.

We must work on the interval containing $\displaystyle 0$ because if you look at the theorem the initial point for the equation is contained in the interval. But that is not it. We also require that $\displaystyle \frac{t+1}{t-1}$ and $\displaystyle \tan t$ be continous. Now, $\displaystyle \tan t$ is not continous at certain points (graph it). But if it needs to contain zero so the only open interval (largest interval) where $\displaystyle \tan t$ is continous is on $\displaystyle (-\pi/2,\pi/2)$. But we also require that $\displaystyle (t+1)/(t-1)$ be continous, so we need to exclude the point $\displaystyle t=1$. This point $\displaystyle t=1$ is contained in $\displaystyle (-\pi/2, \pi/2)$ we we need to omit it. This gives us that on $\displaystyle (-pi/2,1)$ these coefficient functions are continous. So there is a unique solution.