Question

dy/dx = 2 + 2y + x - xy

Transform into a seperable form. Please show all steps in working.

Answer 1

if it aint seperable by normal means then cant we imploy the y=vx trick?

Answer 2

dy/dx = 2 + 2y + x - xy

dy/dx = 1(2 + x) + y(-x + 2)

ugh!!!

Answer 3

64 so with this type of quesion you have to differentiate with respect to x and y then put them together, however in this case you had to use that trick, what does that do?

Answer 4

the v=y(x)/x defines v as a useful function of x to get it top seperate easier

Answer 5

v(x)*x = y(x)

v'(x)*x + v(x) = y'(x) or simply; v'x+v = y' and sub that in for y'

Answer 6

and sub vx in for y

Answer 7

that is if i read it right

Answer 8

These are the workings from my lecturer:

dy/dx = 2(1+y) + x(1+y)

dy/dx = (2+x) (1+y)

dy/1+y = (2+x) dx

That was his solution..

Answer 9

yeah, I was trying to get it to factor by grouping as well to begin with

Answer 10

what happened to the negative?

Answer 11

yeah i know he might have an error.

Answer 12

2(1+y) + x(1+y)

2 + 2y + x + xy !=! 2 + 2y + x - xy

Answer 13

another way to find a solution is:

y' = 2 + 2y + x - xy

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

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

e^{(2x-x^2/2)} y = \(\int 2 e^{(2x-x^2/2)} + x e^{(2x-x^2/2)}dx \)

but that might be stretching it a bit; and we would get to do some constant coeffs stuff

Answer 14

heh, i forgot to include the negative ... i really do need to practice these more often

Answer 15

y' = 2 + 2y + x - xy

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

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

y' + (-2+x) y= 2 + x\[e^{\int-2+x\ dx}(y' + (-2+x) y= 2 + x)\]
\[e^{\int-2+x\ dx}y= \int(2+x)e^{\int-2+x\ dx}dx\]
\[y=\frac{ \int(2+x)e^{\int-2+x\ dx}dx+C}{e^{\int-2+x\ dx}}\]

Answer 16

I will keep going, and i will go back and ask him to explain his solution if i see you on here i will let you know thanks for having a go.

Answer 17

yep, good luck ;)