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This number, is it always equal to the order of the differential equation?

in which case, one can reduce this number?

thanks in advance

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If you are given $$\frac{{{d^n}y}}{{d{x^n}}} = f(x,y,y',y'', \ldots ,{y^{(n - 1)}})$$ then you need the values of $$y({x_0}),y'({x_0}),y''({x_0}), \ldots ,{y^{(n - 1)}}({x_0})$$ and the function itself. So you are right -- the number of initial conditions corresponds to the order of the differential equation. But this differential equation can, of course, be reduced to $$\frac{{dy}}{{dx}} = {y_1},\frac{{d{y_1}}}{{dx}} = {y_2}, \ldots ,\frac{{d{y_{n - 2}}}}{{dx}} = {y_{n - 1}},\frac{{d{y_{n - 1}}}}{{dx}} = f(x,{y_1},{y_2}, \ldots ,{y_{n - 1}})$$ using substitutions $${y_1} = y',{y_2} = y'', \ldots ,{y_{n - 1}} = {y^{(n - 1)}}.$$

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even in the second formulation (system) we need n condition, where n is the order of the matrix (= order of the equation) – MAK Oct 20 '12 at 20:32
Yes, we need $n$ conditions. – glebovg Oct 20 '12 at 20:35
what about characteristic Cauchy problems for PDEs? Is it the same? – MAK Oct 20 '12 at 20:38
What is the characteristic Cauchy problem? Are your referring to the method of characteristics? – glebovg Oct 20 '12 at 20:41
no, I mean when the initial surface is characteritic – MAK Oct 20 '12 at 20:43

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