Thread: HELP: Find the solution of this ODE

Find the solution of this ODE

Originally Posted by goolai

Find the solution of this ODE

First let's set out some notation:

Let:
M = 2xy/(x^2 + y^2)^2 dx
N = 1 + (y^2 - x^2)/(x^2 + y^2)^2 dy
My = partial derivative with respect to y of M
Nx = partial derivative with respect to x of N
Mdx = integral with respect to x of M

First we find My and Nx

My = [2x(x^2 + y^2) - 8xy^2(x^2 + y^2)]/(x^2 + y^2)^4
= (2x^5 - 4x^3y^2 - 6xy^4)/(x^2 + y^2)^4 .........after a lot of simplifying

Nx = [-2x(x^2 + y^2)^2 - 4x(x^2 + y^2)(y^2 - x^2)]/(x^2 + y^2)^4
= (2x^5 - 4x^3y^2 - 6xy^4)/(x^2 + y^2)^4 .........after a lot of simplifying

Since My = Nx, we will use the method of exact equations.
Here are the steps in the method, since it may be hard to follow by just looking at the formulas.

Step 1: Integrate M with respect to x, adding h(y) instead of C as the arbitrary constant. (Since when differentiating with respect to x, you would treat any function in y only as a constant). You obtain the function Mdx by doing this.

Step 2: Differentiate Mdx with respect to y, adding h'(y) as the derivative of h(y).

Step 3: Equate the derivative of Mdx with N. The function should look almost the same except for some function in y or a constant. This extra part of N would therefore be equated to h'(y).

Step 4: Integrate h'(y) to obtain h(y). Plug it into Mdx and that's your solution.

Now let's do it. Fasten your seat belts.

Mdx = int (M) dx = -y/(x^2 + y^2) + h(y)

d/dy(Mdx) = [2y^2 -(x^2 + y^2)]/(x^2 + y^2)^2 + h'(y)
= (y^2 - x^2)/(x^2 + y^2)^2 + h'(y)

Set d/dy(Mdx) = N
=> (y^2 - x^2)/(x^2 + y^2)^2 + h'(y) = 1 + (y^2 - x^2)/(x^2 + y^2)^2

We see that the constant 1 is the odd one out, so h'(y) = 1
=> h(y) = y + C

=> solution: -y/(x^2 + y^2) + y = C

And you can check this answer. If you differentiate it with respect to x you get M, if you do i twith respect to y, you get N. I skipped a lot of steps in working out becuase typing all these formulas was getting to me. But if there is anything you don't understand you can say so, and I'll explain it better. The method is there, so if you have problems it should only be with computation.

Note, if My did not equal Nx, we could not use this method. We would use a method similar to the one we do in solving first order differential equations. That is, multiply the equation by a function that causes My = Nx, and then proceed as above.