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

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- Jun 9th 2011, 07:44 AMrainerThis is a straight line?
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 - Jun 9th 2011, 07:59 AMAckbeet
The only way for a fraction to be zero is when?

- Jun 9th 2011, 08:02 AMProve It
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 $\displaystyle \displaystyle y$? - Jun 9th 2011, 09:01 AMHallsofIvy
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.

Quote:

$\displaystyle \begin{align*} \int{(1 + Y)^{-\frac{3}{2}}\,\frac{dY}{dx}\,dx} &= \int{0\,dx} \\ \int{(1 + Y)^{-\frac{3}{2}}\,dY} &= C_1 \\ \frac{(1 + Y)^{-\frac{1}{2}}}{-\frac{1}{2}} + C_2 &= C_1 \\ -2(1 + Y)^{-\frac{3}{2}} &= C_1 - C_2 \\ (1 + Y)^{-\frac{3}{2}} &= -\frac{1}{2}(C_1 - C_2) \\ 1 + Y &= \left[-\frac{1}{2}(C_1 - C_2)\right]^{-\frac{2}{3}} \\ Y &= \left[-\frac{1}{2}(C_1 - C_2)\right]^{-\frac{2}{3}} - 1 \end{align*}$

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 $\displaystyle \displaystyle y$?

- Jun 9th 2011, 10:33 AMzg12
Is there not a way to delete a post?

- Jun 9th 2011, 10:48 AMJester
- Jun 9th 2011, 11:11 AMAckbeet
- Jun 9th 2011, 05:25 PMrainer
- Jun 10th 2011, 03:41 AMProve It
- Jun 10th 2011, 04:50 AMAckbeet
- Jun 10th 2011, 05:24 AMProve It
- Jun 10th 2011, 06:46 AMJester