Question

(x-y)^3=A(x+y), prove that (2x+y)dy/dx = x + 2y

Answer 1

If $x+y=0=>x=-y$

$$(x+x)^3=0=>x=0=>y=0$$

$$x=-y=>\dfrac{dy}{dx}=-1$$

$$(2(0)+0)(-1)=0+2(0), True$$

If $x+y\not=0$

$$\dfrac{(x-y)^3}{x+y}=A$$

Differentiate both sides with respect to $x$

$$\dfrac{d}{dx}\big(\dfrac{(x-y)^3}{x+y}\big)=\dfrac{d}{dx}(A)$$

Use the Chain Rule

$$\dfrac{3(x-y)^2(1-\dfrac{dy}{dx})(x+y)-(1+\dfrac{dy}{dx})(x-y)^3}{(x+y)^2}=0$$

$$(x-y)^2(3x+3y-3x\dfrac{dy}{dx}-3y\dfrac{dy}{dx}-x+y-x\dfrac{dy}{dx}+y\dfrac{dy}{dx})=0$$

If $x=y,$ $\dfrac{dy}{dx}=1$

$$(2x+x)(1)=x+2x, True$$

Or

$$2x+4y-4x\dfrac{dy}{dx}-2y\dfrac{dy}{dx}=0$$

$$(2x+y)\dfrac{dy}{dx}=x+2y, True$$

Therefore

$$(2x+y)\dfrac{dy}{dx}=x+2y, \text {True, for each } x, y$$