Question

solve (y-1)dx + (x-2)dy = 0

Answer 1

This is a separable DE, meaning that we can put x on one side and y on the other.

Answer 2

i have solved it but don't know whether it is correct or not

Answer 3

All you'd have to do is to integrate both sides.
Feel free to post your answer for checking!

Answer 4

( dy(x))/( dx) = (-y(x)+1)/(x-2)
Divide both sides by 1-y(x):
(( dy(x))/( dx))/(-y(x)+1) = 1/(x-2)
Integrate wrt x
integral (( dy(x))/( dx))/(-y(x)+1) dx = integral 1/(x-2) dx
-log(-y(x)+1) = log(x-2)+c, where c is a constant.
then
y(x) = (e^(-c)-x+2)/(-x+2)
answer i am getting is
y(x) = (-x+c)/(-x+2)

Answer 5

i am also having trouble with (x+2u+1)u'=x+2u+4 , let x+2u=y

Answer 6

can u check this one also @ganeshie8 plz

Answer 7

is the correct @ganeshie8 @mathmate

Answer 8

I usually derive it this way:
Given
\((y-1)dx + (x-2)dy = 0\)
transpose first term to the right, change signs, and cross multiply:
\(\large \frac{dy}{y-1}=-\frac{dx}{x-2}\)

Answer 9

IS THE PREVIOUS ONE CORRECT @mathmate N ALSO IN THIS QUESTION THEY HAVE MENTIONED X+2U=Y SO AM STUCK WITH THAT

Answer 10

Yes, I put your solution into the original equation and it checks out (i.e. =0).
Good job!

Answer 11

integrating w.r.t x:
we get,
integral (( dy(x))/( dx))/(y(x)-1) dx = integral -1/(x-2) dx
by evaluating
(y(x)(x))/(y(x)-1) = -log(-x+2)+c
the we get
| y(x) = (-log(-x+2)+c+y(x)(x))/(-log(-x+2)+c)
but what should i do with X+2U=Y consideration will the answer change with that

Answer 12

(x+2u+1)u'=x+2u+4 , let x+2u=y
Then
u'=(y'-1)/2
(y+1)(y'-1)/2=y+4
(y+1)y'-y-1=2(y+4)
(y+1)y'=3(y+3)

Answer 13

thanks @mathmate

Answer 14

i have one last question solve xu'=(u-x)^3+u u(1)=2 take v=u-x

Answer 15

I was thinking, it might be even simpler to put y=x+2u+1

Answer 16

so when i solve is it like i complete with the solution which i have posted first then apply x+2u=y then u'=(y'-1)/2
(y+1)(y'-1)/2=y+4
(y+1)y'-y-1=2(y+4)
(y+1)y'=3(y+3) is that right

Answer 17

It would be a similar process of substitution:
xu'=(u-x)^3+u u(1)=2 take v=u-x
u'=v'+1
x(v'+1)=v^3+v+x
xv'=v^3+v
Can you take it from here?

Answer 18

can u join us @zepdrix

Answer 19

solve xu'=(u-x)^3+u u(1)=2 take v=u-x