find y" of x+xy+y=2
so i did the first differncition as
1+y+x(dy/dx)+1(dy/dx) = 0
y= 1+y/x+1
y"= ?
somebody solve this for me ?
y" = ( 1+y)-(x+1)/ (x+1)^2
i think this is not right .
what is the answear. show me with working !
So you've already taken the first differentiation which is,
$\displaystyle 1+y+x\left[ \frac{dy}{dx}\right] +\frac{dy}{dx}=0\qquad \qquad (1)$
from the given equation we know that,
$\displaystyle y=\frac{(2-x)}{(x+1)}$
for later use, let's isolate $\displaystyle \frac{dy}{dx}$ from (1),
$\displaystyle \frac{dy}{dx}=\frac{-(1+y)}{(x+1)}=\frac{-\left( 1+\frac{(2-x)}{(x+1)}\right) }{(x+1)}=\frac{-3}{(x+1)^2}\qquad \qquad (2)$
to find $\displaystyle y''$, differentiate the equation (2) w.r.t x,
Hello, arsenal12345!
$\displaystyle \text{Differentiate implicitly. \;Find }y"\text{of }x+xy+y\:=\:2$
Differentiate: .$\displaystyle 1 + y + xy' + y' \:=\:0$
n . . . . . . . . . . . . . . $\displaystyle (x+1)y' \:=\:-(y+1) $
. . . . . . . . . . . . . . . . . . . . $\displaystyle y' \:=\:-\frac{y+1}{x+1}$ .[1]
Differentiate: . .$\displaystyle y" \:=\:-\frac{(x+1)y' - (y+1)1}{(x+1)^2} $
Substitute [1]: .$\displaystyle y" \:=\:-\frac{(x+1)\left(-\frac{y+1}{x+1}\right) - (y+1)}{(x+1)^2}$
. . . . . . . . . . . . $\displaystyle y" \:=\:\frac{y+1 + y+1}{(x+1)^2} \:=\:\frac{2(y+1)}{x+1)^2} $