Question

dy/dx=2/x-1/x^2 where y=2 when x=-1

Answer 1

y=2ln(x)+1/x +c

put x=-1
get c

Answer 2

\[-1=e^{(i (2n+1)\pi)}\]\[\ln(-1)=i(2n+1)\pi\]
2=2i(2n+1) pi -1+c
c=3-2i(2n+1)pi
\[y=2\ln(x)+\frac1x+3-2i(2n+1)\pi\] where\[n=(.....,-1,0,1,2,3...)\]

Answer 3

did u get that?????

Answer 4

yes but the book says the answer should be y=ln(x)^2+(1/x)-2

Answer 5

2ln(x)=ln(x)^2....and is the point x=-1 given in your book???

Answer 6

my mistake you are right thank you