find the general solution of the following differential equation: dy/dx + y/x = y^3

- anonymous

find the general solution of the following differential equation: dy/dx + y/x = y^3

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- anonymous

find your \[\mu\]

- anonymous

integrate 1/x

- anonymous

e is going to be raised by this

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## More answers

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

e^lnx= just x

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

god bless you.. but i really suck at math
and i have two exams tomorrow
so i cant study for them both

- anonymous

i truly appreciate your help, but if any one of you guys could do the whole thing, you would save my life

- anonymous

This is a bernoulli's equation. You have to make the substitution v=y^(-2). v' will then equal=-2y^(-3)dy/dx. Sub in dy/dx from the original equation. Things will cancel only leaving you with v and x. Solve this new equation as a linear 1st order and then sub back in your v's with y's.

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

can you please solve the problem step by step
because im so tired and i still have so much to study
I HONESTLY APPRECIATE YOUR HELP SO MUCH GUYS :) thank you

- anonymous

mathtio is tryign to solve this as a linear first order, but you can't because the right side is in terms of y and not x

- anonymous

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

- anonymous

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

- anonymous

sorry for the confusion.

- anonymous

its ok thank you for trying.. its much appreciated

- anonymous

this thing is not linear

- anonymous

can you help me?

- anonymous

instead of waiting for every step you can at least try doing it, the problem is pretty straight forward once you get started on it

- anonymous

yes but my problem is that i dont get it at all.. I have always had problem with math.. i wish i could understand math half as well as you can..
if i knew how to solve it i wouldnt be asking for help :S

- anonymous

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

- anonymous

Yes you can. The y's can all be substituted nicely with v's. And, what part of my solution don't you understand? You can't just look at a problem and say you don't get it. If you don't even understand the very basics on how to start, even if I posted step by step solutions you wouldn't get it

- anonymous

i just need a solution, thats all.

- anonymous

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

- anonymous

thanks i guess

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