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find your \[\mu\]

integrate 1/x

e is going to be raised by this

can you please solve it step by step because im really bad at math

in the end you are just going to end up with x, so multiply both sides of the equation by this

e^\[\int\limits_{}\] 1/x

e^lnx= just x

x(dy/dx)+y=y^3x

x(dy/dx)+y is just (xy)(dy/dx)

from here you can separate and put your y's with dy and x's with dx

this will give you (dy/y^2)=dx, just integrate now

the final answer should be y=1/-(x+c)

or play with the constants and you should get y=1/(x+c)

I don't think that's the solution mathtio, I'm not quite sure what you're doing

you are right. I forgot about the exponent in the y.

attention to detail.....spaceknight should be able to help you better than i can.

sorry for the confusion.

its ok thank you for trying.. its much appreciated

this thing is not linear

can you help me?

@spaceknight, am afraid you cannot get rid of y even if you sub with v.

i just need a solution, thats all.

If you want the solution, you can use wolframalpha
y^2=-1/(-Cx^2-2x)

thanks i guess