Question

which of the following is the solution to the differential equation dy/dx=y^2 where y(-1)=1?
a)y=1/x for x cannot be = 0
b)y=-1/x for x<0
c)y=-1/x for x>0
d)y=1/x for x>0
e)y=1/x for x<0

Answer 1

\[\frac{dy}{dx}=y^2\\
\frac{1}{y^2}~dy=dx\]
Integrating both sides yields
\[-\frac{1}{y}=x+C~~\iff~~y=-\frac{1}{x+C}\]
Now, given that \(y(-1)=1\), you have
\[1=-\frac{1}{-1+C}~~\Rightarrow~~C=0\]

Answer 2

what would the answer be

Answer 3

I'd go for b

Answer 4

thank you:)