2010
y x ( ) ...
5 !!!!!
:
$ \frac {2x}{dx} = \frac {1+y^2}{2ydy}$
$ \frac { dx}{ x} = \frac{2y}{1+y^2}dy$
$ \int\frac { dx}{ x} =\int\frac{2y }{1+y^2}dy $
$ \ln|x|=\ln(1+y^2 ) +lnc $
$ \ { cx} = 1+y^2 $
$when x=2, y=3$
$c=5 $
$ 5x = 1+y^2 $
\
\frac {2x}{dx} = \frac {1+y^2}{2ydy}
\frac { dx}{ x} = \frac{2ydy}{1+y^2}
\int\frac { dx}{ x} =\int\frac{2ydy}{1+y^2}
\ln|x|=\ln(1+y^2 ) +lnc
\ { cx} = 1+y^2
when x=2, y=3
c=5
\ 5 = 1+y^2 x
; 06-02-2016 01:57 PM
(( ))
$2xy\frac{dy}{dx}=1+y^2$
$\frac{2ydy}{1+y^2}=\frac{dx}{x}$
$\int {\frac{2ydy}{1+y^2}}=\int {\frac{dx}{x}}$
$ln|1+y^2|=ln|x|+c$
$ when x=2 , y=3 $
$ln|1+9|=ln|2|+c$
$c=ln5$
$ln|1+y^2|=ln|x|+ln5$
$y^2=5x-1$
$ln|1+y^2|=ln|x|+c$
$e^{ln|1+y^2|}=e^{ln|x|+c}$
$|1+y^2|=e^{c}|x|$
$|1+y^2|=C|x|$
$x=\frac{1}{C}(1+y^2)$
$ where\ C=-+e^{c}$
$C=5$
$x=\frac{1}{5}(1+y^2)$
; 05-31-2016 07:51 PM
; 05-30-2016 07:15 AM
: 1 (0 1 )
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