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It seems to be common practice when solving ODE's to keep a count of what conditions you have used. I was under the impression that once a condition has been used it cannot be used again.

However, I have been considering the following Ordinary Differential Equation:

$ xf''(x) - xf'(x) - f(x) = 0 \quad$ with $\quad f(0)=0 $ , $\quad f'(0) = 3$


Applying Laplace Transforms using each initial condition, then get the following separable ODE:

$ (s-s^2)F' -2sF = 0$.

This ODE has general solution $F = k(s-1)^{-2}$.

If we apply the inverse Laplace Transform we obtain the following general solution:

$f(x) = kxe^x$.

If we use our second initial condition $f'(0) = 3$, we get $k = 3$, so

$f(x) = 3xe^x$.


Obviously this upsets the rough rule of thumb of 'once you use the condition once, it won't be useful again'. Obviously this is a poor rule of thumb, but I'm just wondering if there's a deeper reason for this, and what a more correct but similar rule of thumb would look like. The answer is probably obvious enough, but any help would be appreciated.

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This initial value problem is tricky because the ODE has a singularity at $x=0$ (the coefficient of the highest order term $f''$ vanishes there), which is exactly where the initial conditions are given. This is the underlying reason for the unusual pattern in computations.

In more concrete terms, you haven't really used the condition $f'(0)=3$ when obtaining the Laplace transform of the equation. Just repeat the computation with $f'(0)=a$ where $a$ is any number; the result is the same. Indeed, $a$ appears only in $\mathcal L[f''] = s^2F(s)-a$, but this gets differentiated in $s$ due to the factor of $x$ next to $f''$.

So the rule of thumb wasn't really contradicted. And it's a good one: it says that when solving an $n$th order ODE, we need $n$ linear conditions to reduce the dimension of the space of candidates for $f$ from $n$ to $0$. Once a linear condition has been imposed, it reduces the space by one dimension, and after that it's not useful: within the remaining subspace it already holds. In the above example, it's more like you used $0\cdot f'(0)= 0\cdot 3$, which doesn't reduce the space of eligible solutions.

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