# General Solution to a D.E

1. Sep 28, 2004

### splitendz

Hello,

I'm having troubles finding the general solution to the following differential equation:

[x + 1]dy - [x^2 - y - 1]dx = 0

Any help would be great,

Cheers.

2. Sep 28, 2004

### Tide

Can you find an integrating factor?

3. Sep 28, 2004

### splitendz

No. I re-arrange the equation to dy/dx = (x^2 - y - 1 ) / (x + 1) and then dividing through by the highest power of x I obtain: (1 - y/x^2 - 1/x^2) / (1/ x + 1 / x^2)
which indicates that it is non-homogenous as the term 1/x is not of the form (y/x)^n. I cannot move any futher. ;(

4. Sep 28, 2004

### Tide

If
$$f(x, y) = \frac {1}{3} x^3 - (x+1) y$$
$$\frac {\partial f}{\partial x} dx + \frac {\partial f}{\partial y} dy = df = 0$$
if I did that correctly. It's late - check my results!

5. Sep 28, 2004

### splitendz

I'm not quite sure what you've done? According to my textbook the correct answer is: y = (x^3 - 3x + c) / 3(x + 1).

By the way, how did you insert those images to represent mathematical symbols??

6. Sep 28, 2004

### Tide

Oops! Yes, I left out the -x term in the numerator I wrote for f.

I simply found a function f(x, y) such that
$$\frac {\partial f}{\partial x} = x^2 - y - 1$$
and
$$\frac {\partial f}{\partial y} = -x -1$$
and this gives the same answer as the text.

Click on one of the equations here and download the pdf file linked in the popup to find out how to make the equations.

7. Sep 28, 2004

### Zurtex

Hmm I'll give it a quick try before I go to lessons:

$$(x+1) \frac{dy}{dx} - x^2 + y + 1 = 0$$

$$(x+1) \frac{dy}{dx} + y = x^2 - 1$$

Note here we need not bother with the integrating factor as it is alread of the form vu' + v'u

$$\frac{d}{dx} [(x + 1)y] = x^2 - 1$$

$$(x + 1) y = \int x^2 - 1 dx$$

I don't think I need to go any further? I may of made a mistake but I need to rush of to lessons.

8. Sep 28, 2004

### splitendz

Thanks. Is there any other method for solving this type of question without using partial derivatives? I've never covered them before.

I think the point of this exercise that I'm working through is to test the students knowledge of first order differential equations by using only the provided textbook methods for variable separable equations, initial value problems, homogeneous equations, and linear first order equations... Briefly the methods covered are: solving a D.E by separating the variables and then integrating OR making the substitution v = y/x for D.E's in the form dy/dx = F[y/x] OR finding the integrating factor for D.E's in the form dy/dx + p[x]y = q[x].

Thanks again :)

9. Sep 28, 2004

### splitendz

Thanks a lot guys :)