# Elimination of arbitrary constants 2

#### bergausstein

##### Active member
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

#### MarkFL

Staff member
What do you get in both problems when you differentiate with respect to $x$?

#### bergausstein

##### Active member
for the first problem I get

$m^2-my'+h=0$ is this correct?

#### MarkFL

Staff member
for the first problem I get

$m^2-my'+h=0$ is this correct?
No, $m$ and $h$ are both constants, right?

#### bergausstein

##### Active member
but only m is to be eliminated not h. I'm confused.

$m^2-my'+x-y+h=0$ is this right?

#### MarkFL

Staff member
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
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
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 $\displaystyle y=xy'+\frac{h}{y'}$