First order linear differential equation

1. Nov 30, 2008

JoeTrumpet

1. The problem statement, all variables and given/known data
Solve this differential equation:
(y^2 +1)*dx + (2xy + 1)*dy = 0

2. Relevant equations
dy/dx + P(x)*y = Q(x)
u(x) = e^(integral of P(x)dx)
(d/dx)(u(x)*y) = Q(x)*u(x)
y = (integral of (Q(x)*u(x)dx))/(u(x)

3. The attempt at a solution
I tried dividing by dx then distributing and rearranging to get it into the right form, but run into problems:
y^2 + 1 + 2xy*dy/dx + dy/dx = 0
dy/dx + y/2x = (-1/(2xy))(dy/dx) -1/(2xy)

this is the closest I could get it to the right form. It would give me u(x) = x^(1/2), but I wouldn't be able to integrate the right side as it would have both x and y. Is there a way to get past this, or did I just rearrange poorly? I just integrated it anyway and the part that was integrated with respect to x I held y as constant, and vice versa, but I'm sure it's wrong so I won't show how I did that. This is for a calc II class so it should be doable without any advanced tricks.

thanks!

2. Nov 30, 2008

Dick

That's an exact differential. You should show us how you did it wrongly by integrating. Because you should be able to do it that way.

3. Nov 30, 2008

JoeTrumpet

that's interesting. We haven't done exact differentials as far as I know, but since I don't know what that means perhaps we have!

dy/dx + y/2x = (-1/(2xy))(dy/dx) -1/(2xy)
using u(x) = x^(1/2)

integral of ((d/dx)(y*x^(1/2))dx) = integral of (-1/(2*x^(1/2)*y)(dy/dx)(dx)) - integral of (-1/(2*x^(1/2)*y)(dx))

I integrated with respect to y for the first one on the right side, treating x as a constant, and integrated with respect to x on the right side, holding y constant, to get:

x^(1/2)*y = -1/(2*x^(1/2))*ln(abs(y)) - (x^(1/2))/y + C
y = -1/(2*x)*ln(abs(y)) - 1/y + C/(x^(1/2))

4. Nov 30, 2008

rock.freak667

Another way to about doing this is to consider what d/dx( xy2) works out to be i.e. d(xy2)

5. Dec 1, 2008

Dick

Not quite. You are looking for a solution of the form f(x,y)=C. The differential of f is (df/dx)*dx+(df/dy)*dy (the derivatives are partial). So you are looking for a function such that df/dx=(y^2 +1) and df/dy=(2xy + 1). Like rock.freak667 points out, you could probably guess the answer.

6. Dec 1, 2008

JoeTrumpet

Oh, I think I know what to do then. I can flip my strategy around and solve for x instead:

(y^2 +1)*dx + (2xy + 1)*dy = 0
dividing by dy, distributing, and rearranging:
dx/dy + 2x/y = - 1/y^2 - (1/y^2)*dx/dy

therefore u(y) = e^(2*lny) = y^2

so

(d/dy)((y^2)*x) = -1 - dx/dy
integrating with respect to y
(y^2)*x = -y - x + C

Is this correct? Also, a question regarding my last integration with respect to y: when I integrate (dx/dy)*dy, I'm told that in situations like this "the dy's don't actually cancel out, but you can treat it as such." What *actually* happens then?

Thanks for the help!

7. Dec 2, 2008

Dick

You've got the right answer. f(x,y)=x*y^2+x+y=C. So (df/dx)*dx+(df/dy)*dy=0. I'm a little confused by your last question. If you are integrating dy, then C could be any function of x. But if you differentiate d/dx and d/dy, you can figure out what it is, right?

8. Dec 2, 2008

JoeTrumpet

Oh, sorry, for my last question I meant regarding the step where I integrated

integral of (-1-(dx/dy))dy

I integrated by "distributing" the dy and "cancelling" it out on the right side to get

int(-dy - dx)
-y - x

The step where I simplified (dx/dy)*dy to dx is what I was asking about. I remember hearing that technically that's not allowed, but it works out because it implies that you're actually doing something else, I think with substitution. I was wondering what implied substitution I was doing so I can more readily understand the integration (we've never worked with multiple variables before so it's a very new to me).

9. Dec 2, 2008

Dick

(dx/dy)*dy IS dx. Technically, it's not allowed to deal with 'infinitesimal' quantities at all, since they aren't 'real' numbers. But that's a technicality. Do it anyway. Everybody else does.

10. Dec 3, 2008

bsodmike

This would probably a more organised method of arriving at the solution. What is also nice, is that you can test for the condition that it is an exact differential.