(x^2)y - x(y^2) = 4
I came out with...
y' = [(y^2) - 2xy] / [(x^2)-2xy)]
The answer in the back isnt this...
How do you do this?
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(x^2)y - x(y^2) = 4
I came out with...
y' = [(y^2) - 2xy] / [(x^2)-2xy)]
The answer in the back isnt this...
How do you do this?
$\displaystyle x^2y-xy^2=4$Quote:
Originally Posted by Morgan82
So we have to differentiate implicitly,
$\displaystyle
2x(y)+x^2y'-1\cdot y^2-x(2y)y'=0
$
And solve for y' to get
$\displaystyle x^2y'-2xyy'=y^2-2xy$
$\displaystyle y'(x^2-2xy)=y^2-2xy$
$\displaystyle y'=\frac{y^2-2xy}{x^2-2xy}$
The slope is defined as long as the denominator doesn't equal 0.
It is not defined when x=0, it is not defined when $\displaystyle y=\frac{x}{2}$
I agree with you ... what "answer" was in the back of the book?Quote:
Originally Posted by Morgan82
$\displaystyle x^2y' + 2xy - 2xy y' - y^2 = 0$
$\displaystyle x^2y' - 2xy y' = y^2 - 2xy$
$\displaystyle y'(x^2 - 2xy) = y^2 - 2xy$
$\displaystyle y' = \frac{y^2-2xy}{x^2-2xy}$
