Question: > 
> For the differential equation $\displaystyle a\frac{d^2x}{dt^2}+5\frac{dx}{dt}+5x=0,$ where $a>0$
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> Represent supercritical damping, Then range of $a$ as
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What i try :: Here we have campare with $\displaystyle A\frac{dx^2}{dt^2}+b\frac{dx}{dt}+C=0$

Then $A=a,B=5,C=5$

So here $D=B^2-4AC=25-5a$

Can anyone please tell me here $D>0$ or $D<0$ for supercritical damping.

Please help me , Thanks

Answer: 
There is a theorem that says that an integer $d$ is the "greatest common divisor" of two integers $a,b$ if and only if:

* $d$ is a divisor of both $a$ and $b$, i.e. $a=ud, b=vd$ for some integers $u, v$, and
* $d$ is a "linear combination" of $a$ and $b$, i.e. $d=xa+yb$ for some integers $x,y$.

So, knowing that $d=\gcd(a,b)$, to prove $md=gcd(ma, mb)$, you need to prove that $md$ is a common divisor of $ma, mb$ and that $md$ is a linear combination of $ma$ and $mb$.

* The first part is trivial: $a=ud$ implies $ma=u(md)$ and $b=vd$ implies $mb=v(md)$.
* The second part is what the proof you are referring to does: as $d=xa+yb$, obviously $md=x(ma)+y(mb)$, notably for *the same* $x, y$.
The key idea that is missing is that the gcd is the *least* positive number that can be written as a linear combination of the two. So $d=ax+by$ for some integers $x,y$ and $d$ is the smallest number that can be written that way.

Then you also know that gcd$(ma,mb)$ can be written as $ma(i)+mb(j)$ for some $i,j$ and is the smallest linear combination. (You make this point in the answer, but you do want to use different constants in the two linear combinations since you can't assume the same numbers work in both cases.)

Then the move you indicate is that gcd$(ma,mb)=ma(i)+mb(j)=m(ai+bj)$. So any linear combination of $ma,mb$ can be written as $m$ times a linear combination of $a,b$.

Now, what is the smallest linear combination of $a,b$? Because if you can make *that* as small as possible, you also find the smallest linear combination of $ma,mb$.