Question

Solve y'+y^2=0

Answer 1

Seperable?

Answer 2

let z = y^2

Answer 3

i really know nothing or less than nothing about differential equations, but i would do this
\[y'=-y^2\]
\[\frac{y'}{y^2}=-1\]

\[\int \frac{y'}{y^2}=-\frac{1}{y}=\int -1dx=-x+c\]

Answer 4

@amistre64 that look okay or am i making stuff up?

Answer 5

im pretty sure your just making stuff up :) but im no expert in it either .... i was thinking a burnoodle tho

z' = 2y y'

y' = z'/(2y)

Answer 6

lol probably
but i can always check my answer
i get
\[y=\frac{1}{x+c}\]

Answer 7

the wolf says: "all signs point to yes"

Answer 8

ok good
sometimes it is easier to do something if you know nothing
what i was really thinking was \((\frac{1}{x})'-\frac{1}{x^2}=0\) for sure

Answer 9

Thanks

Answer 10

The technique @satellite73 has uses is called seperation of variables