Question

Solve the initial-value problem y'+((y+1)(y-1)(y-2))/(x+1)=0, y(1)=0.

Answer 1

Here's the work:

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

A/(y+1)+B/(y-1)+C/(y-2)=1

How do I find A, B, C?

Answer 2

many way to find a,b,c
but one of them
is
put y=0,-1,-2
3 equation and find a,b,c from the system

Answer 3

\[\frac{ a }{ y+1 }+\frac{ b }{ y-1 }+\frac{ c }{ y-2}=\frac{1 }{ (y+1)(y-1)(y-2)}\\\frac{ a(y-1)(y-2) +b(y+1)(y-2)+c(y-1)(y+1) }{ (y+1)(y-1)(y-2)} \]

Answer 4

\[a(y^2-3y-2)+b(y^2-y-2)+c(y^2-1) =\\y^2(a+b+c) +y(-3a-b) +(-2a-2b-c)\\so\\(a+b+c)=0\\(-3a-b)=0\\(-2a-2b-c)=1\]

Answer 5

Do you know the partial fractions method?

Answer 6

yes

Answer 7

Can you tell me how to apply it to this problem?

Answer 8

@ the end
you have value of a,b,c
so
a ln|y+1| +b ln|y-1| +c ln|y-2| = - ln |x+1 | +const

Answer 9

apply y(0)=1 to find const
then
you have the solution

Answer 10

I got it! Thanks!