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

- anonymous

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

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- anonymous

Can I divide each term by 1/dy to separate?

- anonymous

This is in the form m+ny'=0

- anonymous

see if its exact

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## More answers

- anonymous

do the partial with respect to y for the m and then do the partial with respect to x in the n portion

- anonymous

Start by isolating y' .. then integrate both sides

- anonymous

if they are equal then it is exact

- anonymous

But to isolate y' I would need to divide each term by y' correct?

- anonymous

wait...is this cal 1 or differential equations and linear algebra

- anonymous

Calc II differential equations

- anonymous

http://1337.is/~gaulzi/tex2png/view.php?png=201103312034348813.png would look like this isolated

- anonymous

Well you can try implicit differentiation.

- anonymous

take the derivative of the first term with respect to x

- anonymous

kristin wouldn't it be -x^2?

- anonymous

sorry yes, thats correct, minor typo :)

- anonymous

Ok no problem. Thanks

- anonymous

I can toss up for you the answer if you want :)

- anonymous

y' is dy/dx correct? If i were to rewrite it

- anonymous

yeah

- anonymous

Ok so if I were to solve y'+y=10 would I just replace y' with dy/dx and then move it all around?

- anonymous

I'm trying to figure out the steps

- anonymous

it can be done by the separable variables method

- anonymous

after some modification you can get
dy=-x^2/(x^3+8) dx
just integrate both sides

- anonymous

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

- anonymous

I know the answer I'm trying to figure out the steps, could you show me the steps kristin? Plz.

- anonymous

I can show it to you

- anonymous

did you integrate like she said?

- anonymous

I know I need to integrate each piece, I just need to know how to get each piece by itself

- anonymous

do a u substitution

- anonymous

u=x^3+8
du=3xdu

- anonymous

du=3x^2 sorry

- anonymous

you should get: integral( -1/3(1/u) dx

- anonymous

just take u=x^3+8 --> du=3x^2dx
substitute in the integral you will get
\[y=-1/3\int\limits_{}^{}(1/u)du=-1/3\ln \left| u \right|\]

- anonymous

now just substitute for u=x^3+8
\[y=-1/3\ln \left| x^3+8 \right|+c\]

- anonymous

Oh right I got that. Thanks

- anonymous

np

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