*Solve the differential equation dy/dx=x/16y.
*Find the equation of the solution through the point (x,y)=(−4,1).
*Find the equation of the solution through the point (x,y)=(0,−4). Your answer should be of the form y=f(x)

- anonymous

- katieb

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- anonymous

I already solved the first part.
C=8y^2 -(x^2/2)

- myininaya

ok replace your x and y with -4 and 1 respectively to find the C so that your graph for your equation will go through that point

- anonymous

so y=sqrt((x^2/16) + c) and i plugged in (-4,1) and got y=sqrt((x^2/16)
is that correct?

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## More answers

- anonymous

c= 0 for the second question?

- myininaya

You have \[C=8y^2-\frac{x^2}{2}\]
Replace x with -4 and y with 1 like this:
\[C=8(1)^2-\frac{(-4)^2}{2}\]
Is this what you did to solve for C?

- myininaya

\[C=8-\frac{16}{2}=8-8=0\]

- anonymous

yes that's what i did

- myininaya

I can't do arithmetic somethings lol
But you have \[0=8y^2-\frac{x^2}{2} \]

- myininaya

that is the equation satisfying the point (-4,1)

- anonymous

hmm it the system doesn't seem to accept it.. it says it isn't a linear equation

- anonymous

doesn't that mean i have to solve for y=...

- myininaya

Your differential equation was:
\[\frac{dy}{dx}=\frac{x}{16y} \text{ \right ?} \]

- anonymous

yup

- myininaya

Ok solve for y
make sure you keep in mind we want to go through the point (-4,1)

- anonymous

ok so y=x/4

- anonymous

but it still doesn't work??

- anonymous

oh wait y=-x/4 works!!

- myininaya

We wanted y=-x/4
That is why I said to keep in mine we want it going through (-4,1) :)

- myininaya

mind*

- myininaya

The equation y=x/4 doesn't satisfy the point (-4,1)
but y=-x/4 does

- myininaya

Like 1=-(-4)/4

- anonymous

ohh i get it now

- anonymous

thanks a lot!!

- anonymous

i'm going to do the last part now

- anonymous

c=128 for the last part right?

- anonymous

oh i got the last part too!!

- anonymous

thanks a lot again!!!

- myininaya

yes C=128 for the last one

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