Originally Posted by
ckatg
Okay, its a separable equation.
First, you know x can't equal 3 because the denominator can't be 0. So if x-3=0, x=3, therefore x=/=3
So, you multiply both sides by dx and divide both sides by y to get
dy/y=dx/(x-3)
Then you integrate both sides.
integration of 1/y is ln|y|+C
integration of 1/(x-3) is ln|x-3|+C
So you get
ln|y|=ln|x-3|+C1
using the property of logarithms
ln|y|=ln|e^C1(x-3)|
multiply by e both sides
y=e^c1(x-3) (since e^c1 is constant)
y=c(x-3)
sub in initial conditions y=1, x=-1
1=c((-1)-3)
1=c(-4)
-1/4=c
plug in
y=(-1/4)(x-3)
y=-1/4x+3/4
therefore your final equation is y=-1/4x+3/4 where x=/=3