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Elimination of arbitrary constants 2

bergausstein

Active member
Jul 30, 2013
191
eliminate the arbitrary constant,

1. $\displaystyle y=mx+\frac{h}{m}$ h is a parameter, m to be eliminated.


2. $cxy+c^2x+4=0$ eliminate c

I don't know where to start please help me get started. thanks!
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
What do you get in both problems when you differentiate with respect to $x$?
 

bergausstein

Active member
Jul 30, 2013
191
for the first problem I get

$m^2-my'+h=0$ is this correct?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775

bergausstein

Active member
Jul 30, 2013
191
but only m is to be eliminated not h. I'm confused.

$m^2-my'+x-y+h=0$ is this right?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
but only m is to be eliminated not h. I'm confused.

$m^2-my'+x-y+h=0$ is this right?
I am going to assume that $m$ and $h$ are both parameters, and thus are constants. We are given:

\(\displaystyle y=mx+\frac{h}{m}\)

And so differentiating with respect to $x$, we obtain:

\(\displaystyle y'=m\)
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191
To eliminate an arbitrary constant via differentiation, I would solve for that constant first:
\begin{align*}
y&=mx+ \frac{h}{m} \\
my&=m^{2}x+h \\
0&=m^{2}x-my+h \\
m&= \frac{y \pm \sqrt{y^{2}-4xh}}{2x}.
\end{align*}
Now differentiate both sides w.r.t. $x$, holding $y=y(x)$.
 

bergausstein

Active member
Jul 30, 2013
191
I am going to assume that $m$ and $h$ are both parameters, and thus are constants. We are given:

\(\displaystyle y=mx+\frac{h}{m}\)

And so differentiating with respect to $x$, we obtain:

\(\displaystyle y'=m\)
now I see it.

the answer is $\displaystyle y=xy'+\frac{h}{y'}$