/sci/ - Science & Math

[Anonymous No.5534985]

(pic unrelated)

I need help in implicit differentiation, I know the idea behind it but I'm not sure how to solve this.

$$\dfrac{dy}{dx} x^2y-y^3 = 6$$

Any help would be greatly appreciated! (I'm having a test tomorrow...)

[Anonymous No.5534994]

Second try at latex

<span class="math">\frac{dy}{dx} x^{2}y-y^{3} = 6[/spoiler]

[Anonymous No.5535027]

>>5534994

So the idea is, you know what's going on with x and you know y relies on x. So simply say dy/dx is a placeholder for an unknown value and voila

[Anonymous No.5535091]

>>5534994
As written, that's a first order differential equation.

It's possible you meant:

d/dx(x^2y) - y^3 = 6

But far more likely:

d/dx(x^2y - y^3) = 6

If so, differentiation is linear so:
d/dx(x^2y) - d/dx(y^3) = 6

Apply the product rule:
y(d/dx[x^2])+x^2(d/dx[y]) - d/dx(y^3) = 6
2xy + (x^2)y' - d/dx(y^3) = 6

And the chain rule
2xy + (x^2)y' - (3y^2)d/dx(y) = 6
2xy + (x^2)y' - (3y^2)y' = 6

Isolate y'
(x^2-3y^2)y' = 6-2xy
y' = (6-2xy)/(x^2-3y^2)