How do I solve the differential equation x^2+(x^3+8)y'=0

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How do I solve the differential equation x^2+(x^3+8)y'=0

Mathematics
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Can I divide each term by 1/dy to separate?
This is in the form m+ny'=0
see if its exact

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Other answers:

do the partial with respect to y for the m and then do the partial with respect to x in the n portion
Start by isolating y' .. then integrate both sides
if they are equal then it is exact
But to isolate y' I would need to divide each term by y' correct?
wait...is this cal 1 or differential equations and linear algebra
Calc II differential equations
http://1337.is/~gaulzi/tex2png/view.php?png=201103312034348813.png would look like this isolated
Well you can try implicit differentiation.
take the derivative of the first term with respect to x
kristin wouldn't it be -x^2?
sorry yes, thats correct, minor typo :)
Ok no problem. Thanks
I can toss up for you the answer if you want :)
y' is dy/dx correct? If i were to rewrite it
yeah
Ok so if I were to solve y'+y=10 would I just replace y' with dy/dx and then move it all around?
I'm trying to figure out the steps
it can be done by the separable variables method
after some modification you can get dy=-x^2/(x^3+8) dx just integrate both sides
http://1337.is/~gaulzi/tex2png/view.php?png=201103312042017465.png these are the two answers you can get.. depends if you like to use ln or log
I know the answer I'm trying to figure out the steps, could you show me the steps kristin? Plz.
I can show it to you
did you integrate like she said?
I know I need to integrate each piece, I just need to know how to get each piece by itself
do a u substitution
u=x^3+8 du=3xdu
du=3x^2 sorry
you should get: integral( -1/3(1/u) dx
just take u=x^3+8 --> du=3x^2dx substitute in the integral you will get \[y=-1/3\int\limits_{}^{}(1/u)du=-1/3\ln \left| u \right|\]
now just substitute for u=x^3+8 \[y=-1/3\ln \left| x^3+8 \right|+c\]
Oh right I got that. Thanks
np

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