Question

dy/dx=2y^2
and if y=-1 when x=1,
then when x=2, y=?

thanks

Answer 1

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

y(1) = -1
-1 / (1 + C) = -1
C = 0

y = -1 / x
y = -1 / 2

Answer 2

shoot its 2x

Answer 3

-1/y = 2x + C
y = -1 / (2x + C)

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

y = -1 / (x - 1)
y = -1 / (2 - 1)

y = -1

Answer 4

isn't y=-1/3?

dy/y^2=2 dx
y^-2=2dx
-1/y=2x+c
c=-1

-1/y=2x-1
y=-1/3

Answer 5

yea i found it to be -1/3

Answer 6

THANKS ANYWAY

Answer 7

Sorry, it is y = -1 / (2 - -1) = -1 / 3

Answer 8

I forgot the second negative

Answer 9

thanks