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*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)

Calculus1
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I already solved the first part. C=8y^2 -(x^2/2)
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
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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c= 0 for the second question?
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?
\[C=8-\frac{16}{2}=8-8=0\]
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

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