What is the General Solution to a Differential Equation?

[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.

 

[Tide]

Can you find an integrating factor?

 

[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. ;(

 

[Tide]

If
f(x, y) = \frac {1}{3} x^3 - (x+1) y
then your equation becomes
\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!

 

[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??

 

[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.

 

[Zurtex]

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.

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.

 

[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 :)

 

[splitendz]

Thanks a lot guys :)