# Initial Value Problem

1. Jul 12, 2007

### VinnyCee

1. The problem statement, all variables and given/known data

Solve the initial value problem for y.

$$\frac{d^2y}{dx^2}\,=\,\frac{1}{a}\sqrt{1\,+\,\left(\frac{dy}{dx}\right)^2}$$

$$y(0)\,=\,a$$ & $$y\prime(0)\,=\,0$$

a is a non-zero constant.

2. Relevant equations

The calculus.

3. The attempt at a solution

$$y\prime\,=\,sinh\,x$$ $$\longrightarrow$$ $$dy\prime\,=\,cosh\,x\,dx$$

$$1\,+\,sinh^2\,x\,=\,cosh^2\,x$$

$$dy\prime\,=\,\frac{1}{a}\,\sqrt{cosh^2\,x}\,dx\,=\,\frac{cosh\,x}{a}\,dx$$

$$\int dy\prime\,=\,\frac{1}{a}\,\int cosh\,x\,dx$$

$$y\prime\,=\,\frac{1}{a}\,sinh\,x\,+\,C$$

C = 0 from the initial condition $y\prime(0)\,=\,0$

$$\int y\prime\,=\,\frac{1}{a}\,\int sinh\,x\,dx$$

$$y\,=\,\frac{1}{a}\,cosh\,x\,+\,C$$

$$\frac{1}{a}\,(1)\,+\,C\,=\,a$$

$$C\,=\,a\,-\,\frac{1}{a}$$

$$y\,=\,\frac{1}{a}\left(cosh\,x\,+\,a^2\,-\,1\right)$$

That doesn’t match the answer that maple gives, what did I do wrong?

Last edited: Jul 12, 2007
2. Jul 13, 2007

### sealight

If
$$y\prime\,=\,\frac{1}{a}\,sinh\,x$$
then
$$dy\prime\,=\,\sqrt{1+y\prime^2}\, dx \,=\,\sqrt{1+\frac{(sinh\,x)^2}{a^2}}\, dx \,\neq\,\frac{1}{a}\,cosh\,x\,dx$$

Last edited: Jul 13, 2007