- #1

squaremeplz

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## Homework Statement

[tex] \frac { d^2 \theta }{d x'^2 } = -y *exp(\theta) [/tex] eq. 1

mayb be integrated to yield

[tex] exp(\theta) = \frac {a}{cosh^2(b \frac{+}{-} \sqrt \frac{a*y}{2} * x')} [/tex]

[tex] \theta = f(y,x') [/tex]

## Homework Equations

## The Attempt at a Solution

the exponent is throwing me off, but i probably have to use the following properties

[tex] \frac { d^2 \theta }{d x'^2 } + \sqrt{y}*exp(/theta) = 0 [/tex]

[tex] g = [1/2 (e^k^x + e^-^k^x)]^2 = cosh^2(kx) [/tex]

the problem is the exp(theta), do I have to replace the variable with something like

g = ln(theta)

then find

[tex] \frac { d^2 g }{d x'^2 } [/tex]

to get

[tex] \frac { d^2 g }{d x'^2 } + \sqrt{y}*g = 0 [/tex]

then apply

m^2 + k^2 = 0?

thanks