Question

Solve xy' + y + y^2 = 0 for y(1) = 2
---------
I isolated y' such that: y' = (-y^2-y)/x
But I'm not sure what to substitute from here
I could do v = -y^2- y and v' = -2yy' - y' but I'm not sure this would help

Answer 1

just for fun
x dy/dx = -y^2 - y
- dy /[y(y+1) ] = dx/x
-1 ln [ y/(y+1) ] = ln x + ln c = ln xc
ln [ y/(y+1) ]^-1 = ln xc
x c = (y+1)/y
if x = 1 , y = 2
c = 3/2
3 x = 2(y+1)/y

Answer 2

I'm not sure if your integration is correct; I got -ln|1/y + 1|
However, thank you. Is substitution even possible for this problem?

Answer 3

I used my 1955 integration tables
int { dx/[x(ax+b)] } = (1/b)ln [ x/(ax+b) ] + constant

I did it that way because I did not figure out how o do it with substitution.

Answer 4

use partial fractions.

1/(y(y+1)) = 1/y - 1/(y+1)