- Thread starter
- #1

#### bergausstein

##### Active member

- Jul 30, 2013

- 191

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!

- Thread starter bergausstein
- Start date

- Thread starter
- #1

- Jul 30, 2013

- 191

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!

- Admin
- #2

- Thread starter
- #3

- Jul 30, 2013

- 191

for the first problem I get

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

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

- Admin
- #4

No, $m$ and $h$ are both constants, right?for the first problem I get

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

- Thread starter
- #5

- 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?

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

- Admin
- #6

I am going to assume that $m$ and $h$ are both parameters, and thus are constants. We are given:but only m is to be eliminated not h. I'm confused.

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

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

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

\(\displaystyle y'=m\)

- Admin
- #7

- Jan 26, 2012

- 4,198

\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 $y=y(x)$.

- Thread starter
- #8

- Jul 30, 2013

- 191

now I see it.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:

\(\displaystyle y'=m\)

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