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I had a system of differential equations and I ended up in this.

$$4x+2y+4z=0$$ $$2x+y+2z=0$$ $$4x+2y+4z=0$$

I don't know what to do from here.

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$(x,y,z)=(x,-2x-2z,z)$ –  pedja Jan 26 '12 at 12:35
    
There's always taking the null space... –  J. M. Jan 26 '12 at 12:35
    
Would you know what to do if you ended up with just one equation, $2x+y+2z=0$? –  Gerry Myerson Jan 26 '12 at 12:37

1 Answer 1

Note that your system of equations is actually a single equation in various avatars. (i.e.) one is scalar multiple of the other. Put another way, these equations convey the same information, nothing new!

$$4x+2y+4z=0 \iff 2(2x+y+2z=0) \iff 4x+2y+4z=0$$ Since there are three unknowns, and only one equations, two parameters are required to fix a solution. We can have a parametric solution as follows:

Set $x=\lambda$ and $y=\mu$. Then, $$2\lambda+\mu+2z=0 \implies z=\dfrac{-2\lambda-\mu}{2}$$

Hence, the triplet $(\lambda,\mu,\dfrac{-2\lambda-\mu}{2})$ with $\lambda,\mu \in \mathbb R$ is a solution to your system.

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That's kind of confusing. Firstly you mention that there is one equation and then you mention that there are two. –  papas Jan 26 '12 at 12:42
    
That was a typo! I am sorry. I fixed it now! –  user21436 Jan 26 '12 at 12:43

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