# What does it mean mathematically to set some of the integration constants in the general solution to a linear differential equation, equal to zero?

I'm trying to calculate the position of a particle in a quadrapole magnet depending on the entry position $x_0$ and the combined (constant) physical parameters $k$. Given an equation

$$x(t) =\frac{(\frac{x''(t)}{k})''}{k},$$

solving via assuming that $x(t) = e^{\lambda t}$ et,c...

I arrive at the general solution

$$x(t) = c_1\cos(\sqrt{k}\cdot t)+c_2\sin(\sqrt{k}\cdot t)+c_3e^{-\sqrt{k}\cdot t}+c_4e^{\sqrt{k}\cdot t}$$

with $c_1,c_2,c_3,c_4$ arbitrary constants. What would it mean mathematically if I were to set say $c_1,c_2,c_3 = 0$, assuming I don't have other constraints (in my example I would have additional $x(0) = x_0$, but as far as I can see that doesn't forbid it).

Given that they are arbitrary, I can't see a problem with it. Of course, if you have additional starting conditions, you have to set the constants accordingly, but in my example $c_4 = x_0$ seems to do the job and leaves me with a much simpler solution. So why would I ever NOT eliminate every unnecessary term?

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So, the answer is: either you're lucky, and the solution you want is indeed (0, 0, 0, $x_0$) (but you should check this!), or you don't have enough constraints to pin down the solution you want exactly, and you can't calculate the position of your particle yet, because you don't know enough about how it started.
 so if I just have the contraint given in the answer,either any combination which satisfies $x_0=c_1+c_3+c_4$ will describe the location or I can't pin down the exact location? In this specific example to keep it short I omitted the second equation $y(t)=\frac{x''(t)}{k}$, which also has a constraint which comes down to $y_0 = k(-c_1+c_3+c_4)$. If these are the only constraints I have, I have 2 equations for 3 variable...constants( you know ) with $x_0,y_0$ being parameters. So I will HAVE to just set one right?The other to will have a fixed relationship. – ananon Sep 15 '12 at 16:02 A reason for setting to zero some constant of integration may be that you want your solution to have particular properties (for example, being differentiable), so that some "ugly" pieces must be discarded. – Andrea Orta Sep 15 '12 at 16:07 Yes - don't forget that, for example, "I don't want this particle to fly off to infinity as t increases" is a constraint too (which would set $c_4 = 0$). – Billy Sep 15 '12 at 16:13 Using your solution, compute $x(0)$, $x'(0)$, $x''(0)$, $x'''(0)$ as a linear combination of the $c_i$. You'll find that not less than knowing $x(0)$, $x'(0)$, $x''(0)$, $x'''(0)$ determines the solution. – Hagen von Eitzen Sep 15 '12 at 16:53