Question

Please help me solve the following ODE (explicit soln):

dy/dx = (y-1)^2 , y(0) = 1

I have worked through it and my answer is y = -1/x but my book is showing the answer as y = 1

):

Answer 1

\[\frac{ dy }{ dx }=(y-1)^{2} , y(0) = 1\]

Answer 2

Applying separating variables method

Answer 3

\[\frac{ dy }{ (y-1)^2 }=dx\]

Answer 4

\[\int\limits (\frac{ dy }{ (y-1)^2 }=dx)\]

Answer 5

\[\int\limits\limits \frac{1}{(y-1)^2} = \int\limits\limits dx\]

Answer 6

\[-\frac{1}{y-1}=x+c\]

Answer 7

ok we need an alternative approach because as we can see this can not be solve by implicit ODE

Answer 8

so let's try the explicit ODE...

Answer 9

What is the explicit ODE?

Answer 10

I think the explicit form is \[y=-\frac{ 1 }{ x }-1+c\] which doesn't help me with the initial condition either. So I tried only simplifying to the form \[-1=x(y-1)+c\] to get c = -1, which if I plug into my explicit form, makes the answer \[y=-\frac{ 1 }{ x }\]

Answer 11

That still doesn't agree with my book's answer of y=1

Answer 12

yes because solving this explicitly requires recalling the exponential function in order to get the right answer

Answer 13

if the explicit method does not give a 1 then we will consider that an error in the book

Answer 14

by the y(0)=1 which means x=0

Answer 15

right, so using the explicit method with that initial condition makes it undefined...right?

Answer 16

no using implicit method makes it undefined

Answer 17

Then what is the explicit method?

Answer 18

ok i got y = 1

Answer 19

sorry i was trying this on a piece of paper

Answer 20

here is how you do it

Answer 21

The suspense is killing me!
(:

Answer 22

sorry i was on the phone

Answer 23

ok we integrate the 1/(y-1)^2 by expanding the value 1/(y^2-2y +1), then we integrate RHS we get x+c

Answer 24

we will end up with y=quberoot(3x-1+c)+1, which yields to the fact that y=1 for x=0 and y(0)=1

Answer 25

i need to restart

Answer 26

the pc brb

Answer 27

LHS still integrates to -1/(y-1)