Can someone please point out to me how the following reduces to y=mx+b:
$\displaystyle \frac{y''(x)}{(1+y'(x))^{\frac{3}{2}}}=0$
Thanks
Let $\displaystyle \displaystyle Y = \frac{dy}{dx}$, which means $\displaystyle \displaystyle \frac{dY}{dx} = \frac{d^2y}{dx^2}$ and the DE becomes
$\displaystyle \displaystyle \begin{align*} (1 + Y)^{-\frac{3}{2}}\,\frac{dY}{dx} &= 0 \\ \int{(1 + Y)^{-\frac{3}{2}}\,\frac{dY}{dx}\,dx} &= \int{0\,dx} \\ \int{(1 + Y)^{-\frac{3}{2}}\,dY} &= C_1 \\ \frac{(1 + Y)^{-\frac{1}{2}}}{-\frac{1}{2}} + C_2 &= C_1 \\ -2(1 + Y)^{-\frac{3}{2}} &= C_1 - C_2 \\ (1 + Y)^{-\frac{3}{2}} &= -\frac{1}{2}(C_1 - C_2) \\ 1 + Y &= \left[-\frac{1}{2}(C_1 - C_2)\right]^{-\frac{2}{3}} \\ Y &= \left[-\frac{1}{2}(C_1 - C_2)\right]^{-\frac{2}{3}} - 1 \end{align*}$
So $\displaystyle \displaystyle \frac{dy}{dx} &= C $ where $\displaystyle \displaystyle C = \left[-\frac{1}{2}(C_1 - C_2)\right]^{-\frac{2}{3}} - 1$.
Can you go from here to solve for $\displaystyle \displaystyle y$?
So either [tex]\frac{dY}{dx}= 0[//tex] or $\displaystyle (1+ Y)^{-\frac{3}{2}}= 0$. But that second equation is impossible.
That's essentially what Ackbeat said.
$\displaystyle \begin{align*} \int{(1 + Y)^{-\frac{3}{2}}\,\frac{dY}{dx}\,dx} &= \int{0\,dx} \\ \int{(1 + Y)^{-\frac{3}{2}}\,dY} &= C_1 \\ \frac{(1 + Y)^{-\frac{1}{2}}}{-\frac{1}{2}} + C_2 &= C_1 \\ -2(1 + Y)^{-\frac{3}{2}} &= C_1 - C_2 \\ (1 + Y)^{-\frac{3}{2}} &= -\frac{1}{2}(C_1 - C_2) \\ 1 + Y &= \left[-\frac{1}{2}(C_1 - C_2)\right]^{-\frac{2}{3}} \\ Y &= \left[-\frac{1}{2}(C_1 - C_2)\right]^{-\frac{2}{3}} - 1 \end{align*}$
So $\displaystyle \displaystyle \frac{dy}{dx} &= C $ where $\displaystyle \displaystyle C = \left[-\frac{1}{2}(C_1 - C_2)\right]^{-\frac{2}{3}} - 1$.
Can you go from here to solve for $\displaystyle \displaystyle y$?