Differentiate $y^2-2xy+3y=7x$ w.r.t. $x$. Hence show that $\frac{d^2y}{dx^2}(2y-2x+3)=\frac{dy}{dx}(4-2\frac{dy}{dx}).$

Question

Differentiate $y^2-2xy+3y=7x$ w.r.t. $x$. Hence show that $\frac{d^2y}{dx^2}(2y-2x+3)=\frac{dy}{dx}(4-2\frac{dy}{dx}).$

I differentiated $y^2-2xy+3y=7x$ w.r.t. $x$ and got:

$$\frac{dy}{dx}=\frac{7+2y}{2y-2x+3}$$

Differentiating one more time, I get:

$$\frac{d^2y}{dx^2}=\frac{-4(x+2)\frac{dy}{dx}+4y+14}{(2y-2x+3)^2}$$

Multiplying by $(2y-2x+3)$ I get:

$$\frac{d^2y}{dx^2}(2y-2x+3)=\frac{-4(x+2)\frac{dy}{dx}+4y+14}{(2y-2x+3)}$$

Now if I understand correctly the question wants me to show that $\frac{d^2y}{dx^2}(2y-2x+3)=\frac{dy}{dx}(4-2\frac{dy}{dx})$

So from what I already got, I guess I need to show that

$$\frac{dy}{dx}(4-2\frac{dy}{dx})=\frac{-4(x+2)\frac{dy}{dx}+4y+14}{(2y-2x+3)}$$

But I don't know how to do this and I'm not sure if I've been following the question correctly. I reckon I might be over-complicating things too. Grateful for any hints/guidance.

Answer 1

Rule of thumb: Try to never differentiate quotients if you can avoid it; they're gross.
You correctly found $$\frac{dy}{dx}=\frac{7+2y}{2y-2x+3}$$ which can equally be written as $$\frac{dy}{dx}(2y-2x+3)=7+2y$$ Differentiating both sides with respect to $x$, the product rule gives $$\frac{d^{2}y}{dx^{2}}(2y-2x+3)+\frac{dy}{dx}\left(2\frac{dy}{dx}-2\right)=2\frac{dy} {dx}$$ which rearranges to what you want.

Answer 2

By starting with $y^{2} - 2 x y + 3 y = 7 x$ then differentiation leads to: \begin{align} 2 \, y \, y' - 2 \, x \, y' - 2 y + 3 y' = 7. \end{align} Differentiating a second time leads to \begin{align} 2 \, y \, y'' + 2 (y')^{2} - 2 \, x \, y'' - 4 y' + 3 y'' = 0 \end{align} The second equation can be reformed into \begin{align} (2 y - 2 x + 3) \, y'' = (4 - 2 y' ) \, y' \end{align} which is the desired result.

Answer 3

$y^2−2xy+3y=7x$
differentiate both sides wrt $x$,we get
$2y\frac{dy}{dx}-2x\frac{dy}{dx}-2y\frac{dx}{dx}+3\frac{dy}{dx}=7$
$(2y-2x+3)\frac{dy}{dx}=7+2y$

Now without taking $(2y-2x+3)$ into denominator of $7+2y$,differentiate both sides wrt $x$,using product rule

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

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

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

$(2y-2x+3)\frac{d^2y}{dx^2}=(4-2\frac{dy}{dx})\frac{dy}{dx}$