This is about attempting to solve ##\left( y'\right)^2 = y^2 - 1 ##.(adsbygoogle = window.adsbygoogle || []).push({});

[tex]\int\frac{dy}{\sqrt{y^2 -1}} = \pm \int dx[/tex]

using a trig. substitution and another trick,

[tex]\int\frac{dy}{\sqrt{y^2 -1}} = \ln\left(y \pm \sqrt{y^2 - 1} \right) + C[/tex]

I'm not sure about that [itex]\pm[/itex] sign. It came in when doing [itex]\tan (\sec^{-1} y)[/itex] since I had [tex]\sec \theta = y \longrightarrow\tan^2 \theta = y^2 -1\longrightarrow \tan \theta = \pm \sqrt{y^2 -1}[/tex]

The solution is supposed to be something like [itex]y = \cosh x[/itex]. But I am not sure how I'm supposed to be getting that from separating and integrating.

[itex]\ln\left(y + \sqrt{y^2 - 1} \right) = \pm x + C [/itex]

[itex]y + \sqrt{y^2 - 1} = C_1e^{\pm x} [/itex]

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# Relationship integration math problem

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