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My question is theoretical, but it really bothers me.

The way I see differential equation's solutions is that someone observed that, for instance, $Ce^{Ax}$ solves first order equations, etc. The solutions that we know are valid, but how can we be sure that there are no other solutions that we are neglecting?

Furthermore, could the existence of such solutions be a problem in the future, since we have incomplete answers for all the problems that involve differential equations?

Blue
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Iliasp
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1 Answers1

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For linear ODE we can tell number of solutions:

The linear ODE for $y(x)$ of $n$th order has $n$ linearly independent solutions and as many undetermined constants. An ODE is linear if $y,y',y'',...,y^{(n)}$ occur linearly. So the ODE $y'+P(x)y=Q(x)$ has only one solution and one undetermined constant. $y''+y=0$ has many pairs of linearly independent solutions: $(\sin x, \cos x), (e^{ix},e^{-ix}),[\sin(x-a), \cos(x-a)]$. In case of many pairs one member of one pair can be written as a linear combination of other two pairs.

Number of constants in a solution decide the oder of the ODE. In your case $y=C e^{Ax}$, will satisfy a 3nd order ODE. To know it you have to eliminate the constants using differentiation. $$y=C e^{Ax} \implies ye^{-Ax}=C \implies y'e^{-Ax}-Aye^{-Ax}=0 \implies y'/y=A \implies yy''-y'^2=0$$

This second order non-linear ODE, according to me we cannot comment about its number of solutions. The solution of the non-linear ODE $y''=y'^2$ is $y=A-\log(x+B)$: two constants but single solution.

I hope that it helps.

Z Ahmed
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