The equation:
$e^{x}+2xy^{2}-y^{5}=0$
defines a function $y$ of $x$ in an implicit way.
Calculate:
$y''(0)$
I have tried solving this, and got that the derivative of $y$ in respect to $x$ is:
$\dfrac{e^{x}+2y^{2}}{y\cdot(5y^{3}-4x)}$
Now if I find another derivative with respect to $x$, I get a function that contains both $x$ and $y$. If I try to manipulate algebraically, I get a differential equation.
This question should be fairly simple. What am I missing?
$$\frac{d}{dx}\left[e^x + 2xy^2 - y^5\right] = e^x + 2y^2 + 4xy\frac{dy}{dx} - 5y^4 \frac{dy}{dx}, \tag{1}$$ so $$\frac{dy}{dx} = \frac{e^x + 2y}{5y^4 - 4xy}. \tag{2}$$
Now take the derivative of $(1)$ again:
$$\begin{align} \frac{d^2}{dx^2}\left[e^x + 2xy^2 - y^5\right] &= e^x + 4y \frac{dy}{dx} + \frac{d}{dx}\left[4xy - 5y^4\right] \frac{dy}{dx} + (4xy-5y^4)\frac{d^2y}{dx^2} \\ &= e^x + 4y \frac{dy}{dx} + \left(4y + 4x\frac{dy}{dx} - 20y^3 \frac{dy}{dx}\right)\frac{dy}{dx} + (4xy-5y^4)\frac{d^2 y}{dx^2} \\ &= e^x + 4\left(2y + (x - 5y^3)\frac{dy}{dx}\right)\frac{dy}{dx} + (4xy - 5y^4)\frac{d^2 y}{dx^2}. \tag{3} \end{align}$$
Hence $$\frac{d^2 y}{dx^2} = \frac{e^x + 4(2y + (x-5y^3) \frac{dy}{dx})\frac{dy}{dx}}{5y^4-4xy}. \tag{4}$$ All that remains is to substitute $(2)$ into $(4)$ and simplify. However, as the result is quite messy, and since what we want is simply $y''(0)$, it is easier to first evaluate $y(0)$ and $y'(0)$. We note that at $x = 0$,
$$0 = e^0 + 2(0)^2 - y^5 = 1 - y^5, \tag{5}$$
hence $y(0) = 1$. Then from $(2)$, $$y'(0) = \frac{e^0 + 2(1)}{5(1^4) - 4(0)(1)} = \frac{3}{5}. \tag{6}$$ Finally from $(4)$ and $(6)$, $$y''(0) = \frac{e^0 + 4(2(1) + (0-5(1^3))\frac{3}{5})\frac{3}{5}}{5(1^4)-4(0)(1)} = -\frac{7}{25}.$$