Arithmetic simplification

Question

I am asked to find $\frac{d^2y}{dx^2}$, and prove that $\frac{d^2x^2+y^2=a^2}{dx^2}$=$-\frac{a^2}{y^3}$,

This is how I have proceeded:

$2 y \frac{dy}{dx}+2 x=2 a \frac{da}{dx}$ => $\frac{dy}{dx}\to \frac{a \frac{da}{dx}-x}{y}$

$\frac{d^2\left(x^2+y^2=a^2\right)}{dx^2}$= $2 y \frac{d^2y}{dx^2}+2 \left(\frac{dy}{dx}\right)^2+2=2 a \frac{d^2a}{dx^2}+2 \left(\frac{da}{dx}\right)^2$

$\left\{\left\{\frac{d^2y}{dx^2}\to \frac{a \frac{d^2a}{dx^2}+\left(\frac{da}{dx}\right)^2-\left(\frac{dy}{dx}\right)^2-1}{y}\right\}\right\}$

$\left\{\left\{\frac{d^2y}{dx^2}\to \frac{a \frac{d^2a}{dx^2}-\frac{\left(a \frac{da}{dx}-x\right)^2}{y^2}+\left(\frac{da}{dx}\right)^2-1}{y}\right\}\right\}$

Factoring out

$\frac{1}{y}$

Know this is how fare I get:

$\frac{d^2y}{dx^2}\to \frac{-a^2 \left(\frac{da}{dx}\right)^2+a y^2 \frac{d^2a}{dx^2}+y^2 \left(\frac{da}{dx}\right)^2+2 a x \frac{da}{dx}-x^2-y^2}{y^3}$

I see that $-x^2-y^2=-a^2$, but I am not sure how this would help me? Could someone please solve this step by step so that I am able to see how to do this?

Answer 1

If $F(x,y)=0$ defines $y$ as a function of $x$ then we know that $$y'=\frac{-F_x}{F_y}$$ Here, we have $x^2+y^2=a^2$ so we can assume that $F(x,y)=x^2+y^2-a^2=0$ and so $$F_x=2x,~~F_y=2y$$ and so $y'=\frac{-x}{y}$. Therefore we get $y'y+x=0$. Take another differentiation from both sides with respect to $x$ we get, $(y')'y+y'(y)'+1=0$ which is equal to $$y''y+y'^2+1=0$$ so $$y''=\frac{-1-y'^2}{y}=\frac{-1-(x^2/y^2)}{y}=\frac{-a^2}{y^3}$$

Answer 2

We have $$x^2+y^2=a^2$$

Differentiating wrt $x,$ $$x+y\frac{dy}{dx}=0\ \ \ \ (1)\implies \frac{dy}{dx}=-\frac xy\ \ \ \ (2)$$

Again, differentiating $(1)$ wrt $x,$

$$1+y\frac{dy^2}{dx^2}+\left(\frac{dy}{dx}\right)^2=0$$

Using $(2),$ $$1+y\frac{dy^2}{dx^2}+\frac{x^2}{y^2}=0$$

Write $x^2=a^2-y^2$