awfawef
*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)
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awfawef
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I already solved the first part.
C=8y^2 -(x^2/2)
myininaya
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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
awfawef
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so y=sqrt((x^2/16) + c) and i plugged in (-4,1) and got y=sqrt((x^2/16)
is that correct?
awfawef
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c= 0 for the second question?
myininaya
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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
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\[C=8-\frac{16}{2}=8-8=0\]
awfawef
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yes that's what i did
myininaya
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I can't do arithmetic somethings lol
But you have \[0=8y^2-\frac{x^2}{2} \]
myininaya
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that is the equation satisfying the point (-4,1)
awfawef
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hmm it the system doesn't seem to accept it.. it says it isn't a linear equation
awfawef
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doesn't that mean i have to solve for y=...
myininaya
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Your differential equation was:
\[\frac{dy}{dx}=\frac{x}{16y} \text{ \right ?} \]
awfawef
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yup
myininaya
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Ok solve for y
make sure you keep in mind we want to go through the point (-4,1)
awfawef
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ok so y=x/4
awfawef
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but it still doesn't work??
awfawef
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oh wait y=-x/4 works!!
myininaya
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We wanted y=-x/4
That is why I said to keep in mine we want it going through (-4,1) :)
myininaya
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mind*
myininaya
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The equation y=x/4 doesn't satisfy the point (-4,1)
but y=-x/4 does
myininaya
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Like 1=-(-4)/4
awfawef
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ohh i get it now
awfawef
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thanks a lot!!
awfawef
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i'm going to do the last part now
awfawef
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c=128 for the last part right?
awfawef
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oh i got the last part too!!
awfawef
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thanks a lot again!!!
myininaya
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yes C=128 for the last one