Strange equation

**chiph588@** · January 4th 2011, 01:25 PM

It's easy to solve $f'(x) = xf(x)$ , but is there a way to solve $f'(x) = xf(x-1)?$

**chisigma** · January 4th 2011, 01:57 PM

Setting $\displaystyle f(s)= \mathcal{L} \{y(t)}\}$ the first order DE equation $y^{'}(t)= t\ y(t-1)$ became the following linear first order DE in the variable s...

$f^{'} (s) = (1 -s\ e^{s})\ f(s) + e^{s}\ y(0)$ (1)

... that can be solved with a 'standard' approach...

Kind regards

$\chi$ $\sigma$

**Random Variable** · January 4th 2011, 02:23 PM

Can it be treated as the product of the Heaviside function with another function?

EDIT: Nevermind. That doesn't make any sense.

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