would you please solve this equation for me? it is(x+y)dx-xydy=0?

Answer 1

It's not clear what you are asking. Are you trying to integrate it ?
I think this is what you are looking for:

⌠(x+y)dx - ⌠xydy = ½x² + y + c - ½xy² + c Now add the two constants to get another constant.
⌠(x+y)dx - ⌠xydy = ½x² + y - ½xy² + c
⌠(x+y)dx - ⌠xydy = ½(x² - ½xy²) + y + c

Or maybe it's this:
(x+y)dx - xydy = 0
(x+y)dx = xydy
(x+y)/xy = dy/dx
dy/dy = [(xy)(1+y) - (x+y)(y)] / (xy)²
dy/dy = [(xy+xy² - xy - y²] / (xy)²
dy/dy = [xy² - y²] / (xy)²
dy/dy = [xy² - y²] / x²y²
dy/dy = [x - 1] / x²

Answer 2

dy/dx = 1/x + 1/y

You can find higher order derivatives by repeated differentiating of this with respect to x

y'' = -1/x^2 -(1/y^2)y'

Then substitute for y'
using y' = 1/x + 1/y

The derivatives become more complicated but are straight forward to compute.

We are computing derivatives since we want a taylor series expansion about a value of x

Suppose x=1

y(x) = y(1) + y'(1)(x-1) + y''(1)/2 (x-1)^2 + ...

where the nth term is the nth derivative of y evaluated at 1
divided by n! and this multiplies (x-1)^n

I got this idea by using Mupad Pro which is a computer algebra program to get a series solution about x=1

It gave the sum of terms up to and including (x-1)^5
with very involved coeefficients

Answer 3

i'm not going to solve it for you - but you have to separate the dy and the dx so that they are on opposite sides of the =, once you do that you can integrate both sides.

Answer 4

we know d/dx(u/v) = (vdu/dx - udv/dx) v^2
= (du/dx)/v -(dv/dx)/v^2

let u = y and v = x

so d/dx(y/x) = dy/dx/x - dv/dx/x^2
or x^2d/dx(y/x) = xdy/dx - 1
so x^2yd/dx(y/x) = xydy/dx - y
so xydy - y dx = x^2yd(y/x) ...1

from given expression

xdx = xydy -ydx = x^2 d(y/x)

devide by x^2
dx/x = d(y/x)

integrate both sides ln x = y/x + C
or y/x = ln x - C
y = x(ln x -C)

Edit: this has an error. I shall visit in a day or so

Answer 5

wow.. do your own homework