Question

Solve the differential Eqn:
y'=(x+y-2)Whole square

Answer 1

\[y'=(x-y-2)^2\]

Answer 2

\[y' = (x + y - 2)^2 \]

Answer 3

dammit

Answer 4

lol

Answer 5

i'm lookin for solution...

Answer 6

it dosent look separable or linear

Answer 7

it dosent look homogenous

Answer 8

yeah u'r right

Answer 9

is it a Bernoulli equation?

Answer 10

if we can write the equation in this form \[y' + p(x)y = q(x)y^n\]it is a bernoulli equation

Answer 11

I don't think it is...

Answer 12

i guess we have to make a substitution

Answer 13

let \(w=x+y(x)\)
\(\text dw=\)

Answer 14

yep
you got it

Answer 15

@UnkleRhaukus still there?

Answer 16

Hah that was a trip down memory lane pretty fun:)

Answer 17

@shubham9495 still there?

Answer 18

hmm..

Answer 19

your sub.s works

Answer 20

w= x+y
w' = y' +1
y' = w' -1
w'-1 = (w-2)^2
w'-1 = w^2 -4w +4
w'= w^2 -4w +5
dw/(w^2 -4w +5) = dx

Answer 21

integrate
-arctan(2-w) = x
2-w = tan(-x)
w= -tan(-x) +2
x+y = -tan(-x) +2
y= -tan(-x) {{call this tan(x) now}} +2 -x

Answer 22

verify in original
tan^2(x) = tan^2(x)

Answer 23

what happened to the constant ?

Answer 24

I was thinking about that...

Answer 25

think it has to be a whole period , so it's just a phase shift...
not %100 on that however

Answer 26

the solution to a differential equation should have a constant

Answer 27

I was talking abut the constant of integration

Answer 28

Guess I should have left it in... rather than assume it was a whole period shift