Simple differential equation -

[mikehibbert]

Homework Statement

dy/dx = 2x / (y+1)

Homework Equations

n/a

The Attempt at a Solution

I have to solve this - and i know it's separation of variables, but I always get y on both sides when I try to solve and end up with:

y = 2(x^2) / (y+1)

which is clearly wrong.

Where am I messing up? :S

 

[jeffreydk]

Sorry the latex doesn't seem to be working, but since it's seperable, go...


integral[ (y+1) dy] = integral[ 2x dx]

and when you integrate that you should be on your way.

 

[mikehibbert]

that's what i did - but that came out as the solution i gave in my original post - which isn't right is it?

 

[tiny-tim]

Hi mikehibbert!

that's what i did - but that came out as the solution i gave in my original post

No, it doesn't … try again!

 

[mikehibbert]

argh I've literally been thinking about this for like two hours, it's ridiculous.

integrating (y+1) wrt y gives: (y^2)/2 + y

and integrating 2x wrt x gives: x^2

giving (y^2)/2 + y = x^2

and rearranged gives y = 2x / (y+1)

i still get it no matter how many times i think of it :(

arrgghhh

 

[tiny-tim]

giving (y^2)/2 + y = x^2

and rearranged gives y = 2x / (y+1)

I assume you mean y = 2x² / (y+1) …

but shouldn't there be a 2y in there somewhere?

and what about a constant of integration?

 

[mikehibbert]

yes i did mean y = 2(x^2) / (y+1) sorry.

and there should be a +c on the end, granted.

but my solution is still wrong isn't it?

you can't have a y on both sides?

 

[mikehibbert]

oh and i can't see where a 2y would come from??

the LHS integral is (y^2) / 2 + y

which factorizes to 0.5y(y+1) ?

then re-arranges to what i said earlier.

 

[tiny-tim]

the LHS integral is (y^2) / 2 + y

which factorizes to 0.5y(y+1) ?

Nope … 0.5y(y+2)

 

[mikehibbert]

grrr I'm literally an idiot...

 

[mikehibbert]

i swear I'm going to kill the person than wrote this question...

now I'm getting y = 2(x^2) + x^2 + c

which doesn't really look right to me :S

 

[tiny-tim]

i swear I'm going to kill the person than wrote this question...

now I'm getting y = 2(x^2) + x^2 + c

which doesn't really look right to me :S

D'oh!

Get some sleep … :zzz:

 

[mikehibbert]

i'm gunna have to.
gotta be handed in at 10am tomo grrr.
i give in :P
the easiest question on the sheet has beaten me!

 

[tiny-tim]

It'll all be clear at breakfast!

 

[jeffreydk]

Just rewrite it as y^2+y-2x^2-C=0, and you want to factor it. So use the quadratic formula and you'll be able to get the two solutions for the differential equation.

 

[HallsofIvy]

i swear I'm going to kill the person than wrote this question...

I don't think that is the right person to kill!

now I'm getting y = 2(x^2) + x^2 + c

which doesn't really look right to me :S

You have, correctly, (1/2)y²+ y= x²+ C and you want to solve for y. That is a quadratic equation:
y²+ y- (x²+ C)= 0. Use the quadratic formula.

 

[tiny-tim]

y^2+y-2x^2-C=0

y²+ y- (x²+ C)= 0

Hey guys!

Why do you keep writing the wrong equations?

 

[Geekchick]

Hi!
I'd like to venture a guess (I'm currently in the middle of calculus 1 and haven't really done any differential equations)

For the integral of (y+1) I have [(y^2)/2] + y

For the integral of 2x I have x^2

then I set them equal and solved for x to get

x = the +-squareroot of {[y^2+2y]/2} I thought it looked nicer to give them a common denominator.

Anyway since originator of the post has already turned in the assignment I didn't see what it would hurt to post a possible answer, but anyway could someone let me know if I am anywhere close or if not post the right answer.

Thank!

 

[tiny-tim]

… x = the +-squareroot of {[y^2+2y]/2} I thought it looked nicer to give them a common denominator.

Hi Geekchick!

i] You must include a constant of integration.

ii] Don't use √s … as HallsofIvy said, this is a quadratic equation … so leave it as such … it tells you what shape the curve is.

 

[Geekchick]

alright...Best I can come up with is y=-1+-square root of (2x^2+C)