Question

variable saparable(separation of variable)

1.) (2x+y)dy + (2x+y+6)dx=0

2.)(5t+1)t ds + (25t-1)sdt=0

I don't the solution of this problem I think it a "repeated expression". Can someone help me on this? thanks a lot. :)

Answer 1

I noticed your equations when you posted them several days ago. I do not see how they can be solved by separation of variables. There are too many terms that involve both x and y.

Both problems are too difficult for me to find a closed-form solution.

Answer 2

2) (5t+1)t ds + (25t-1)sdt=0

Divide by (5t+1)ts :

ds/s + (25t-1)/[(5t+1)t]dt=0



1) (2x+y)dy + (2x+y+6)dx=0

Change of variables:

u = 2x+y

v = x

Then

y = u - 2 v

x = v

dy = du - 2 dv

dx = dv


1) becomes:

u (du - 2 dv) + (u+6)dv = 0 ------>

u du + (6 - u)dv = 0

Divide by (6-u):

u/(6-u) du + dv = 0