Question

Find y' for this equation (1-x) (dy/dx) - xy = 1 ???
SOS please.

Answer 1

Lol SOS

y' is nothing but dy/dx itself. So, rearrange the whole equation above to write dy/dx implicitly.

Answer 2

do you mean find y?

Answer 3

it's a linear DE.

Answer 4

Yeah, am guessing that you mean to solve this linear D.E, and find the *y* actually..

Answer 5

yeah...

Answer 6

\[(1-x)\frac{\text dy}{\text dx}-xy=1\]
\[\frac{\text dy}{\text dx}-\frac{xy}{(1-x)}=\frac 1{(1-x)}\]
\[y'(x)+\frac {-x}{1-x}y=\frac 1{(1-x)}\]

Answer 7

ok what's the answer?
is dy/dx = (1+xy)) / (1-x) correct?

my steps are simple--> divide both sides by (1-x)
thus dy/dx - xy/(1-x) = 1 / (1-x)
then dy/dx = 1/(1-x) + xy / (1-x)
then dy/dx = (1+xy) / (1-x) [ common denominator, simple alegbra]
I'm newbie to implicit differentiiation. please correct me.

Answer 8

\[(1-x)y'(x)-xy=1\]
\[y'(x)+\frac {-x}{1-x}y=\frac 1{(1-x)}\]\[y'(x)=\frac 1{(1-x)}+\frac {x}{1-x}y\]

Answer 9

what you have done looks right

Answer 10

thank a lot. It does not appear to involve differentiation at all.

Answer 11

is the question asking for \(y'\) or \(y\)

Answer 12

What's the difference if it is y' or y? I am not sure?
Thanks a lot.

Answer 13

well to you think you are being asked to isolate y', which has been done and dosent require any integration, or to find y which requires a few integrations