xy’ + y = 1 + y’ Solve for y’?

Question

I haven't gone over this so i don't know how to solve it.

Thanks for the help!

Answer 1

Hey there!

Here's the answer.

xy'+y=1+y' --> Write the problem.
xz+y=1+z --> Substitute z for y'.
xz=1+z-y --> Subtract y on both sides of the equation.
xz-z=1-y --> Subtract z on both sides of the equation.
z(x-1)=1-y --> Factor out the z in the left side of the equation.
z=(1-y)/(x-1) --> Divide x-1 on both sides of the equation.
y'=(1-y)/(x-1) Substitute y' for z.

So the answer is y'=(1-y)/(x-1).

Hope it helps!

Answer 2

Collect the y' terms in one sides and all the other terms in the other side,
y'(1-x) = y-1 => y' = (y-1)/(1-x)
If somehow you want to solve for y, here is my approach:
dy/(y-1) = -dx/(x-1)
ln|y-1| = -ln|x-1|+c'
(y-1)(x-1) = c, where c = lnc'
y = c(x-1)+1

Answer 3

xy’ + y = 1 + y’
Get y' on one side:
xy' - y' + y = 1
Factor:
y'(x -1) + y = 1
since ab' +a'b = (ab)', the left-hand side is
[y(x - 1)]' = 1
Now integrate both sides:
y(x -1) = x + C
and solve for y:
y = [x/(x-1)] + C(x -1)^-1

Answer 4

x dy / dx + y = 1 + dy / dx
(x - 1) dy / dx = 1 - y
∫ (1 / (1 - y) dy = ∫ [1 / (x - 1) ] dx
- log ( 1- y ) = log (x - 1) + log C
- log (1 - y ) = log [ C (x - 1) ]
log [ (1 - y)^(-1) ] = log [ C (x -1) ]
1 / (1 - y) = C (x + 1)
(1 - y) = 1 / [ C (x + 1) ]
1 - y = k / (x + 1) where k = 1 / C
y = 1 - k / (x + 1)