# 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

http://stuff.bsodmike.com/Integral%20for%20an%20Exact%20Equation%20(1st%20Order%20ODEs).jpg [Broken]

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.

Last edited by a moderator: May 3, 2017