Find the value of $d^2y/dx^2$ when $ x=1$ [closed]

Solution 1:

I'll do an easier problem, see if you can understand it:

Question. Find $d^2y/dx^2$ when $x=1$ for $$ xy = 3\tag1$$

Solution. Implicit differentiation once gives $$ y +x\frac{dy}{dx} = 0 \tag2$$ Doing it a second time gives $$ \frac{dy}{dx} +\frac{dy}{dx} + x\frac{d^2y}{dx^2} = 0 $$ which simplifies to $$2\frac{dy}{dx} + x\frac{d^2y}{dx^2} = 0\tag3$$ Putting $x=1$ in equation (1), we find that $y=3$.

Putting $x=1,y=3$ in equation (2), we find that $\frac{dy}{dx}=-3$.

Finally, using $x=1,\frac{dy}{dx}=-3$ in equation (3), we get the final answer $$ \frac{d^2y}{dx^2} = 6.$$

Can you understand the steps? The same idea works for your problem.

For your problem, I'll give you the equation after implicitly differenting twice, so that you can check: after simplifying, its

$$ -x^2\frac{d^2y}{dx^2} + 6xy \frac{d^2y}{dx^2} +6x\left(\frac{dy}{dx}\right)^2-4x \frac{dy}{dx} +12 y \frac{dy}{dx} - 2y + 2 = 0$$