Question

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

find y(x)

Answer 1

i think i made a mistake somewhere

Answer 2

just a typo there, integral w+1 instead of integral w

Answer 3

2

Answer 4

thanks @hartnn ,
im still not getting what wolfram get

Answer 5

wolf is getting something more complicated than your aolution

Answer 6

yeash its just the i factor

Answer 7

isn't your solution acceptable ?
i do not see any mistake there

Answer 8

oh wait!

Answer 9

shouldn't there be two solution sets?

Answer 10

w^2/2+w =-x+c
w^2+2w =2(-x+c)
w^2+2w+1=2(-x+c)+1
(w+1)^2 =-2x+2c+1
(w+1)^2 =2(-x+C)
w+1 =√(2(-x+C))

Answer 11

w^2/2+w =-x-c
w^2+2w =-2(x+c)
w^2+2w+1=-2(x+c)+1
(w+1)^2 =-2x-2c+1
(w+1)^2 =-2(x+C)
w+1 =√(-2(x+C))
x-y+1 =i√2√(x+C)
\[ y =x-i√2\sqrt{x+C}+1\]

Answer 12

\((w+1)^2= 2(C-x) \\ w+1 = \pm \sqrt 2 \sqrt {C-x} \\ y =x+1 \mp \sqrt{2}\sqrt {C-x}\)
if we purposely bring the imaginary term,
\(y =x+1 \mp i\sqrt{2}\sqrt {x-C}\)

Answer 13

if you look at the wolf solution
you will notice that the \(\huge x^2\) term inside sqrt cancels out!

giving you the form of \(\sqrt {x-C}\) only!
you're correct :)

Answer 14

\(\sqrt {c_1 -2 (-x^2/2) -2(-2x)-(x^2+2x+1)}
\\ \sqrt {c_1 +x^2 +4x-x^2-2x-1} \\ \sqrt {c_2+2x} \\ \sqrt 2 \sqrt {x+c_2/2} = \sqrt 2 \sqrt {x-C} \)

Answer 15

hmm yeah,
I wasn't really expecting \(i\) in the solution,
none other problem in this book have solutions with \(i\),
is there some other method or some other form, that might appear simpler?

Answer 16

or do i have to find a domain for C to make the solution real valued only

Answer 17

i don't see how we get a more simplified solution than
\( y =x+1 \mp \sqrt{2}\sqrt {C-x}\)
also, i never really cared about the domain of constant...is it required to show that in your work ?

Answer 18

i dont know, i just wasn't expecting a complex answer,

Answer 19

ah!, if i solve it like an exact equation i get a different looking answer,

Answer 20

if i was going to use the method i tried first , maybe i should leave the solution implicitly , as
\[(x-y)^2=-2x+c\]

Answer 21

Using the exact method , i got

f(x,y) = x^2/2-xy+2x+y^2/2-y = k
(x-y)^2 +4x-2y = k

Answer 22

if i was going to use the method i tried first , maybe i should leave the solution implicitly , as
\[(x-y+1)^2=-2x+c\]

___

Using the exact method , i got

\[ f(x,y) = x^2/2-xy+2x+y^2/2-y = k\]\[\qquad\qquad\qquad\qquad (x-y)^2 +4x-2y = k\]

And these plot the same thing, so they are the same,

Answer 23

yes, but the question specifically asked to fin y

Answer 24

sorry about that , thanks again!

Answer 25

no problem! welcome ^_^