Question

solve diff.eq.

dy/dx=xy^2

Answer 1

this is separable:$$dy/dx=xy^2\\1/y^2dy=x\,dx\\\int1/y^2dy=\int x\,dx\\\dots$$

Answer 2

also please get in the habit of awarding the most helpful answers to your questions medals, otherwise people may avoid helping you :p

Answer 3

then that becomes ln[y^2]=.5x+c

Answer 4

throw in the e

y^2=e^(.5x^2+c)

Answer 5

then square root

y=ce^(.5x)?

Answer 6

nahh! try integrating \(1/y^2\) again... it's just the power rule...

Answer 7

$$\int y^{-2}\,dy=\frac1{-2+1}y^{-2+1}=-y^{-1}=-\frac1y$$

Answer 8

\[\int\limits\limits y ^{-2}dy=\int\limits\limits xdx+c or \frac{ y ^{-1} }{1 }=\frac{ x ^{2} }{2 }+c\]

Answer 9

...ok got that after writing every step

Answer 10

im stuck at -1/y=.5x^2+c, do i integrate again?

Answer 11

correction
\[\it is \frac{ y ^{-1} }{ -1 }\]

Answer 12

so thats where i stop?

Answer 13

@freddy_eighty7 $$\frac1{y^2}dy=x\,dx\\\int\frac1{y^2}dy=\int x\,dx\\-\frac1y=\frac12 x^2+C\\y=-\frac1{\frac12x^2+C}=-\frac2{x^2+C}$$

Answer 14

\[x ^{2}y+2cy=-2 or y \left( x ^{2}+C \right)=-2 wher C=2c\]