Respuesta :
Answer:
[tex]y'=\frac{-y}{x}[/tex]
Step-by-step explanation:
[tex]x^2y^2-xy=4[/tex]
We will need to differentiate both sides.
Keep in mind the following:
* [tex]\frac{d}{dx}x^2=2x[/tex] by power rule ([tex](x^n)'=nx^{n-1}[/tex])
* [tex]\frac{d}{dx}y^2=2yy'[/tex] by power rule and chain rule.
* [tex]\frac{d}{dx}x=\frac{dx}{dx}=1[/tex]
* [tex]\frac{d}{dx}y=\frac{dy}{dx}=y'[/tex]
* [tex]\frac{d}{dx}(xy)=x\frac{d}{dx}y+y\frac{d}{dx}x=x\frac{dy}{dx}+y\frac{dx}{dx}[/tex]
[tex]=x\frac{dy}{dx}+y(1)=x\frac{dy}{dx}+y=xy'+y[/tex] by product rule and some already mentioned things. This is by product rule ([tex](uv)'=uv'+vu'[/tex].
* [tex]\frac{d}{dx}(x^2y^2)=x^2\frac{d}{dx}(y^2)+y^2\frac{d}{dx}(x^2)[/tex]
[tex]=x^2(2y)y'+y^2(2x)[/tex] by product rules and already mentioned things above.
[tex]=x^22yy'+2xy^2[/tex].
Let's put it all together by differentiating both sides:
[tex]x^2y^2-xy=4[/tex]
[tex](x^2y^2-xy)'=(4)'[/tex]
[tex](x^2y^2)'-(xy)'=0[/tex] I used the constant rule on the right hand side.
I also used difference rule. That is (f-g)'=f'-g'.
Now let's apply those mentioned things:
[tex](x^22yy'+2xy^2)-(xy'+y)=0[/tex]
Distribute:
[tex]x^22yy'+2xy^2-xy'-y=0[/tex]
Put [tex]y'[/tex] terms together:
[tex]x^22yy'-xy'+2xy^2-y=0[/tex]
Factor the [tex]y'[/tex] out of the terms containing the [tex]y'[/tex]:
[tex]y'(x^22y-x)+2xy^2-y=0[/tex]
Let's isolate the term containing the [tex]y'[/tex].
We will do this by adding [tex]y[/tex] on both sides and subtracting [tex]2xy^2[/tex] on both sides:
[tex]y'(x^22y-x)=y-2xy^2[/tex]
Now divide both sides the thing being multiplied by [tex]y'[/tex].
[tex]y'=\frac{y-2xy^2}{x^22y-x}[/tex]
I really don't like the coefficient of [tex]x^2y[/tex] not being in front so I'm going to rearrange that part using the commutative property of multiplication.
[tex]y'=\frac{y-2xy^2}{2x^2y-x}[/tex]
Let's see if this can be simplified.
I'm going to factor out what I can on both top and bottom.
The top terms contain a factor of [tex]y[/tex].
The bottom terms contain a factor of [tex]x[/tex].
[tex]y'=\frac{y(1-2xy)}{x(2xy-1)}[/tex]
We see that the other factor in top and bottom are opposite factors. I'm talking about the [tex]1-2xy[/tex] and the [tex]2xy-1[/tex].
So if we factor out -1 on top we get:
[tex]y'=\frac{-y(2xy-1)}{x(2xy-1)}[/tex]
Now we can cancel the common factor across the division there:
[tex]y'=\frac{-y}{x}[/tex]
Answer:
dy/dx = -y/x
Step-by-step explanation:
(x²)(2yy') + y²(2x) - [xy' + y] = 0
2x²yy' - xy' = y - 2xy²
y'(2x²y - x) = y - 2xy²
dy/dx = [y(1-2xy)]/[x(2xy-1)]
dy/dx = -y/x
