Slope Fields

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April 26th 2009, 05:57 PM
mamori

Slope Fields

I can't really show the slope field, but I don't believe the image is needed for this problem. But either way, the slopes end up looking something like this:

$\smile$
$) ($
$\frown$

In other words, a hyperbola.

I have two questions concerning this:

1) Consider the differential equation $dy/dx = x/y$ , where y (does not equal) 0. (sorry, I don't know the latex/symbol for that)

Find the particular solution y=f(x) to the differential equation with the initial condition f(3) = -1, and state its domain.

I tried to solve it, but got y=x, which I am sure is wrong. What am I doing wrong?

b) I didn't really learn slope fields too well either. Is the image that's shown through the slope fields the original function or the graph of the derivative?

Thank you in advance once again!
April 26th 2009, 06:05 PM
TheEmptySet

Quote:

Originally Posted by mamori

I can't really show the slope field, but I don't believe the image is needed for this problem. But either way, the slopes end up looking something like this:

$\smile$
$) ($
$\frown$

In other words, a hyperbola.

I have two questions concerning this:

1) Consider the differential equation $dy/dx = x/y$ , where y (does not equal) 0. (sorry, I don't know the latex/symbol for that)

Find the particular solution y=f(x) to the differential equation with the initial condition f(3) = -1, and state its domain.

I tried to solve it, but got y=x, which I am sure is wrong. What am I doing wrong?

b) I didn't really learn slope fields too well either. Is the image that's shown through the slope fields the original function or the graph of the derivative?

Thank you in advance once again!

The equation is seperable so we get

$ydy=xdx \implies \frac{1}{2}y^2=\frac{1}{2}x^2+c$

Now we use the point (3,-1) to find the value for c

$\frac{1}{2}(-1)^2=\frac{1}{2}(3^2)+c \iff -4=c$

Plugging in the value and a little algebra gives

$x^2-y^2=8$

To generate a slope field you plug ordered pairs into the ODE

remember that $\frac{dy}{dx}$ is the slope at each point
April 26th 2009, 06:10 PM
mamori

Hi TheEmptySet! That was a really nice explanation! But I'm now sure how you got:

http://www.mathhelpforum.com/math-he...9bb2e23b-1.gif

Could you please explain that to me?
April 26th 2009, 06:14 PM
TheEmptySet

Quote:

Originally Posted by mamori

Hi TheEmptySet! That was a really nice explanation! But I'm now sure how you got:

http://www.mathhelpforum.com/math-he...9bb2e23b-1.gif

Could you please explain that to me?

(Worried)

By integrating both sides (Surprised) (Power rule)

$\int ydy=\frac{1}{2}y^2$ and

$\int xdx =\frac{1}{2}x^2$

The c is just the arbitary constant of integration.
April 26th 2009, 06:20 PM
mamori

*headdesks* I completely forgot to integrate when I knew I had to! Sorry about that! I seem to have a really bad habit of forgetting to integrate. (Worried) I was wondering what state it's domain means, but does that just mean find the question?

And finally, I just wanted to triple check this part. Slope fields generate an image of the original equation and not the derivative, correct?

Thank you so much for all your help!

[Edit]
I'm going to bed now, so I won't reply anymore tonight. But thank you so much for your help and explanation!