Question

solve the IVP, y'=abs(y-1), y(0)=1

Answer 1

is this diff eqs?

Answer 2

yeah

Answer 3

you mean calculus or differential equations?

Answer 4

i know that abs(y-1)=sqrt((y-1)^2) but i cant see how to apply it and i have just started studying diff eq

Answer 5

it is an initial value problem from a differential equations class

Answer 6

have not taken the class yet, sorry

Answer 7

Suppose that y>=1, then
You can verify that
\[ y= 1 - e^x
\]
is a solution

Answer 8

Suppose that y<=1, then You can verify that

\[
y= 1- e^{-x}

\]
is a solution.

Answer 9

how does dy/dx = -e^x solve dy/dx=abs(y-1)??

Answer 10

you mean e^x + 1 ?

Answer 11

we suppose in each case that y >1 or y<1

Answer 12

ok so walk me through this, sorry. if y = 1 - e^x and y >1 then x<ln(1) = 0 right? and then dy/dx = -e^x but - e^x is negative for this interval and so isnt a solution to the diff eq

Answer 13

The first solution is not acceptable sinc
\[
y = 1 - e^x <1
\]
and the hypothesis was for y >1

Answer 14

Let concentrate on y < 1, so y -1 <0 and Abs(y-1)=1-y

Answer 15

i see how this solves the condition that y(0)=1 but do not see how it solves dy/dx=abs(y-1), it works if you ignore the abs but not otherwise

Answer 16

ok i follow

Answer 17

Use the initial condition to get c =0

Answer 18

\[ \frac {dy}{dx}= 1- y\\ \frac {dy}{1-y}= dx\\ -\ln(1-y) = x +c\\ e^{-\ln(1-y)}= e^{x+c}\\ \frac 1 {1-y} =e^{x+c}\\ 1-y= e^{-x +c}\\ y = 1 - e^{-x -c}\\



\]

Answer 19

very clear, thank you :)

Answer 20

YW