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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. –  Ryan Budney Mar 29 '12 at 18:15
    
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. –  Ryan Budney Mar 29 '12 at 18:17
    
You can also view the existence and uniqueness theorem as an application of something called Gronwall's Inequality. 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. –  Ryan Budney Mar 29 '12 at 18:18

1 Answer 1

up vote 1 down vote accepted

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? –  Mathematics Mar 29 '12 at 15:29
    
I have completed my answer. –  Julián Aguirre Mar 29 '12 at 17:01

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