We have the following differential equation uu''=1+(u')^2 i found that the general

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We have the following differential equation uu''=1+(u')^2 i found that the general

jazzcutie0h 2021-11-19 Answered

We have the following differential equation

\cup^{″} = 1 + {(u^{'})}^{2}

i found that the general solution of this equation is

u = d cosh ((x - \frac{b}{d})

where b and d are constats
Please how we found this general solution?

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Alicia Washington

Answered 2021-11-20 Author has 23 answers

You can separate that equation as

\frac{2 u^{'} u^{″}}{1 + u^{2}} = 2 \frac{u^{'}}{u}

where both sides are complete differentials which integrate to

\ln (1 + u^{' 2}) = \ln (u^{2}) + c \Rightarrow 1 + u^{' 2} = C u^{2}

Can you continue?

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Ourst1977

Answered 2021-11-21 Author has 21 answers

\cup^{″} = 1 + {(u^{'})}^{2}

Substitute

p = u^{'}

u \frac{d p}{d x} = 1 + p^{2}

u \frac{d p}{d u} \frac{d u}{d x} = 1 + p^{2}

u \frac{d p}{d u} p = 1 + p^{2}

Now it's separable

\int \frac{p d p}{1 + p^{2}} = \int \frac{d u}{u}

It should be easy to integrate now..
Edit

p^{2} + 1 = K u^{2} \Rightarrow \int \frac{d u}{\sqrt{K u^{2} - 1}} = \pm x + K_{2}

\frac{a r cosh (\sqrt{K} u} {\sqrt{K}} = x + K_{2}}{}

Taking

cosh

on both side

\sqrt{K} u = cosh (\sqrt{K} (x + K_{2}))

u = \frac{1}{\sqrt{K}} cosh (\sqrt{K} x + K_{2})

Which is close to your formula

u = d cosh (\frac{x - b}{d}) \Rightarrow d = \frac{1}{\sqrt{K}}

and

- \frac{b}{d} = K_{2}

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We have the following differential equation uu''=1+(u')^2 i found that the general

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