Solving an IVP for y(x)

**fractal5** · July 4th 2012, 01:12 PM

$\frac{d^2y}{dx^2} = \frac{1}{a} \sqrt{1 + {(\frac{dy}{dx})^2}}$ ; y(0)=a>0, y'(0)=0

let w = y'

$\frac{dw}{\sqrt{1+w^2}} = \frac{1}{a}dx$

$ln|w + \sqrt{w^2 + a} | = \frac{x}{a} + c$

$ln|y' + \sqrt{y'^2 + a} | = \frac{x}{a} + c$

for the given values c = 0

$y' + \sqrt{y'^2 + a} = e^\frac{x}{a}$

From here I don't know how to proceed. How do you reduce it such that I can get y(x) = ....

Thanks.

**Prove It** · July 4th 2012, 10:36 PM

Originally Posted by fractal5

$\frac{d^2y}{dx^2} = \frac{1}{a} \sqrt{1 + {(\frac{dy}{dx})^2}}$ ; y(0)=a>0, y'(0)=0

let w = y'

$\frac{dw}{\sqrt{1+w^2}} = \frac{1}{a}dx$

$ln|w + \sqrt{w^2 + a} | = \frac{x}{a} + c$

$ln|y' + \sqrt{y'^2 + a} | = \frac{x}{a} + c$

for the given values c = 0

$y' + \sqrt{y'^2 + a} = e^\frac{x}{a}$

From here I don't know how to proceed. How do you reduce it such that I can get y(x) = ....

Thanks.

I would not do $\displaystyle \begin{align*} \int{\frac{dw}{\sqrt{1 +w^2}}} = \ln{\left|w + \sqrt{w^2 + a}\right|} \end{align*}$ .

Instead, I'd make the substitution $\displaystyle \begin{align*} w = \sinh{t} \implies dw = \cosh{t}\,dt \end{align*}$ which would make

$\displaystyle \begin{align*} \int{\frac{dw}{\sqrt{1 + w^2}}} &= \int{\frac{\cosh{t}\,dt}{\sqrt{1 + \sinh^2{t}}}} \\ &= \int{\frac{\cosh{t}\,dt}{\sqrt{\cosh^2{t}}}} \\ &= \int{\frac{\cosh{t}\,dt}{\cosh{t}}} \\ &= \int{1\,dt} \\ &= t + C \\ &= \sinh^{-1}{w} + C \end{align*}$

So the solution to your DE becomes

$\displaystyle \begin{align*} \sinh^{-1}{w} &= \frac{x}{a} + c \\ w &= \sinh{\left(\frac{x}{a} + c\right)} \\ \frac{dy}{dx} &= \sinh{\left(\frac{x}{a} + c\right)} \end{align*}$

I'm sure you can go from here.

**fractal5** · July 4th 2012, 11:03 PM

Genius! Thanks a lot, I would've never thought to use the sinh function.

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