Anyone want to help with math?

[minime72706]

I know I should do my own homework but I suck at it and I'm trying.

so y'=(x+y)/x y(1)=1 Every time I try to do this I wind up getting y=(3/2)/x-x/2.

I get y'-(1/x)*y=1 solve using the integrating factor.


I have the answer for y(2) using Runge-Kutta so I know my answer is wrong.


Can anyone offer some guidance on why I suck at this? Thanks

 

[TecHNooB]

I'm on it! Gimme a sec..

 

[JM Aggie08]

i'm about sick n tired of ppl showin off they're fancy math problems.

i'll take ax^2+bx+c over that mess any day!

 

[Canai]

6

 

[Cawchy87]

Yikes, I haven't done this for a few years and totally forget how to do it. Bump! Hope you get an answer!

 

[TecHNooB]

Ok, this is a homogenous equation. To determine this, all you have to do is verify that:

f(tx,ty) = f(x,y)

To solve the equation, let v = y/x (also y = xv) and reduce to a seperable equation.

If you take the partial of y = xv with respect to x, you get dy/dx = x * (dv/dx) + v.

So.. replace all the y's with xv.

dy/dx = (x + xv) / x = x * (dv/dx) + v

[x * (1 + v)] / x = x * (dv/dx) + v

1 + v = x * (dv/dx) + v

1 = x * (dv/dx)

(1/x) * dx = dv

Integrate both sides

ln(x) + K = v

ln(x) + K = y/x ~(v = y/x)

y = x[ln(x) + K]

y(1) = 1

1 = ln(1) + K ------> K = 1

y = x[ln(x) + 1]

I hope this is the correct answer ~_~

 

[Capt Caveman]

Are you going to ask homework questions every night?

BAN

 

[CraKaJaX]

Originally posted by: TecHNooB
Ok, this is a homogenous equation. To determine this, all you have to do is verify that:

f(tx,ty) = f(x,y)

To solve the equation, let v = y/x (also y = xv) and reduce to a seperable equation.

If you take the partial of y = xv with respect to x, you get dy/dx = x * (dv/dx) + v.

So.. replace all the y's with xv.

dy/dx = (x + xv) / x = x * (dv/dx) + v

[x * (1 + v)] / x = x * (dv/dx) + v

1 + v = x * (dv/dx) + v

1 = x * (dv/dx)

(1/x) * dx = dv

Integrate both sides

ln(x) + K = v

ln(x) + K = y/x ~(v = y/x)

y = x[ln(x) + K]

y(1) = 1

1 = ln(1) + K ------> K = 1

y = x[ln(x) + 1]

I hope this is the correct answer ~_~

If that's correct, he said he has the initial condition y(2) = ...?

Plugging in 2, and assuming your y is correct, that would mean y = ~3.4?

 

[minime72706]

Originally posted by: TecHNooB


y = x[ln(x) + 1]

I hope this is the correct answer ~_~

Thats the right answer. y(2)=3.38 which is what I get with Runge-Kutta which is supposed to be within .01.

Thanks a lot. I just don't remember seeing it being to that way in class. The separation I get the substitution not so much, but I'll try to figure it out.

 

[Random Variable]

y'=1+y/x

y'-y/x = 1

now multilpy both sides by 1/x

(1/x)*y' -y/x^2 = 1/x

d/dx(y/x) = 1/x

y/x = ln(x) + c

y = x*ln(x) + cx

1= 1*ln(1)+c*1 =>c=1

y = x*ln(x) + x

 

[CycloWizard]

Originally posted by: minime72706
Thats the right answer. y(2)=3.38 which is what I get with Runge-Kutta which is supposed to be within .01.

Thanks a lot. I just don't remember seeing it being to that way in class. The separation I get the substitution not so much, but I'll try to figure it out.

The accuracy of Runge Kutta completely depends on which RK formula you're using and the discretization used. And the DE isn't homogeneous, since (x-y)/x=1-y/x. It's first order linear and can be solved using integration factors.

 

[BlackTigers]

a^2+b^2=c^2

Therefore, A+b=C

The answer is 42.

 

[robphelan]

Originally posted by: CycloWizard

Originally posted by: minime72706
Thats the right answer. y(2)=3.38 which is what I get with Runge-Kutta which is supposed to be within .01.

Thanks a lot. I just don't remember seeing it being to that way in class. The separation I get the substitution not so much, but I'll try to figure it out.

The accuracy of Runge Kutta completely depends on which RK formula you're using and the discretization used. And the DE isn't homogeneous, since (x-y)/x=1-y/x. It's first order linear and can be solved using integration factors.

yeah.. so take that!

 

[Fenixgoon]

man, i used to know ODE's like a mofo. oh how quickly things change in two years