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

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

c= 0 for the second question?

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

yes that's what i did

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

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

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

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

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

yup

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

ok so y=x/4

but it still doesn't work??

oh wait y=-x/4 works!!

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

mind*

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

Like 1=-(-4)/4

ohh i get it now

thanks a lot!!

i'm going to do the last part now

c=128 for the last part right?

oh i got the last part too!!

thanks a lot again!!!

yes C=128 for the last one