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For an $n$th order differential equation, why are there always $n$ solutions? Why exactly $n$, not $n - 1, n+1$ or infinite many?


Addendum by LePressentiment :

This is motivated by P176 on Strang's Intro to Lin Alg, 4th Ed. An $n$th order differential equation, how many basis functions does it have? I'd guess $n$, because the diffl eqn have $n$ linearly independent solutions (predicated on Julián Aguirre's answer), all of which look to span the nullspace/solution space of the homogeneous ODE.

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    Well, technically it's not true that order $n$ differential equations have $n$ independent solutions. For example, the first-order DE $y'=\sqrt{|y|}$ has infinitely many independent solutions, even with the initial condition $y(0)=0$. If you put some reasonable restrictions on the DE then you get to the theorem Julian cites.2012-03-29
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    The answer to *why* it's true this comes down to the proof of the *existence and uniqueness theorem* for ordinary differential equations. The main condition is that the DE has to satisfy a Lipschitz inequality (which the square root function does not). The key part of the proof is an application of the mean value theorem.2012-03-29
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    You can also view the existence and uniqueness theorem as an application of something called *Gronwall's Inequality*. http://en.wikipedia.org/wiki/Gronwall%27s_inequality This tells you that "nearby solutions stay nearby" in some precise sense, so if two solutions have the same initial condition, they have to stay the same.2012-03-29

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In general, a differential equation has an infinite number of solutions. The family of solutions of an equation of order $n$ depends on $n$ constants. If the equation is an homogeneous linear equation of order $n$ then there exist $n$ linearly independent solutions $y_1,\dots,y_n$ such that the general solution is $$ y=C_1y_1+\dots+C_ny_n, $$ where $C_1,\dots C_n$ are constants.

Consider the $n$-th order linear homogeneous equation $$ y^{(n)}+a_{n-1}(x)y^{(n-1)}+\dots+a_1(x)y'+a_0(x)y=0, $$ where the $a_i(x)$ are continuous functions on an interval, that without loss of generality we may assume that contains $x=0$. The theory of existence and uniqueness proves that there are solutions $y_1,\dots,y_n$ such that $$ y_1(0)=1,y_1'(0)=0,y_1''(0)=0,\dots,y^{(n-1)}(0)=0\\ y_2(0)=0,y_2'(0)=1,y_0''(0)=0,\dots,y^{(n-1)}(0)=0\\ y_3(0)=0,y_3'(0)=0,y_3''(0)=1,\dots,y^{(n-1)}(0)=0\\ \dots\\ y_n(0)=0,y_n'(0)=0,y_n''(0)=0,\dots,y^{(n-1)}(0)=1 $$ These solutions are linearly independent, and form a basis of the space of solutions. You can check the details in almost any book on ODE's.

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    why the homogeneous linear equation of order n has n linearly independent solutions?2012-03-29
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    I have completed my answer.2012-03-29
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    Does "theory" here imply that things aren't yet proven?2014-11-16
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    No. Theory does not have that meaning in science.2014-11-16
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    This is late question, but what about singular solutions ? first order differential equation should has 1 solution, but some have more singular solutions. ?2015-09-01