Question

(dy)/(dx)=-((x^(2)+y^(2)))/(2xy),;y(1)=-2 10) $\frac{dy}{dx} = - \frac{(x^2+y^2)^2}{2xy}$ ; $y(1) = -2$

Name: dydx x2y22xyy1 2 10 fracdydx fracx2y222xy y1 2 88595
Uploaded: 2022-01-09T17:50:03-08:00
Duration: 3 min 34 s
Channel: Adriano Chikande
Description: dydx x2y22xyy1 2 10 fracdydx fracx2y222xy y1 2 88595

          (dy)/(dx)=-((x^(2)+y^(2)))/(2xy),;y(1)=-2
10) $\frac{dy}{dx} = - \frac{(x^2+y^2)^2}{2xy}$ ; $y(1) = -2$

$dydx x2y22xyy1 2 10 fracdydx fracx2y222xy y1 2 88595$

Added by Cheryl J.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Adriano Chikande

01/09/2022

Step 1

We can check this by replacing $x$ with $tx$ and $y$ with $ty$: $f(tx, ty) = - \frac{((tx)^2+(ty)^2)^2}{2(tx)(ty)} = - \frac{(t^2x^2+t^2y^2)^2}{2t^2xy} = - \frac{(t^2(x^2+y^2))^2}{2t^2xy} = - \frac{t^4(x^2+y^2)^2}{2t^2xy} = t^2 \left( - \frac{(x^2+y^2)^2}{2xy} Show more…

Show all steps