 # 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
$$\displaystyle\cup{''}={1}+{\left({u}'\right)}^{{2}}$$
i found that the general solution of this equation is
$$\displaystyle{u}={d}{\text{cosh}{{\left({\left({x}-\frac{{b}}{{d}}\right)}\right.}}}$$
where b and d are constats
Please how we found this general solution?

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You can separate that equation as
$$\displaystyle{\frac{{{2}{u}'{u}{''}}}{{{1}+{u}^{{2}}}}}={2}{\frac{{{u}'}}{{{u}}}}$$
where both sides are complete differentials which integrate to
$$\displaystyle{\ln{{\left({1}+{u}'^{{2}}\right)}}}={\ln{{\left({u}^{{2}}\right)}}}+{c}\Rightarrow{1}+{u}'^{{2}}={C}{u}^{{2}}$$
Can you continue?
###### Have a similar question? Ourst1977
$$\displaystyle\cup{''}={1}+{\left({u}'\right)}^{{2}}$$
Substitute $$\displaystyle{p}={u}'$$
$$\displaystyle{u}{\frac{{{d}{p}}}{{{\left.{d}{x}\right.}}}}={1}+{p}^{{2}}$$
$$\displaystyle{u}{\frac{{{d}{p}}}{{{d}{u}}}}{\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}}={1}+{p}^{{2}}$$
$$\displaystyle{u}{\frac{{{d}{p}}}{{{d}{u}}}}{p}={1}+{p}^{{2}}$$
Now it's separable
$$\displaystyle\int{\frac{{{p}{d}{p}}}{{{1}+{p}^{{2}}}}}=\int{\frac{{{d}{u}}}{{{u}}}}$$
It should be easy to integrate now..
Edit
$$\displaystyle{p}^{{2}}+{1}={K}{u}^{{2}}\Rightarrow\int{\frac{{{d}{u}}}{{\sqrt{{{K}{u}^{{2}}-{1}}}}}}=\pm{x}+{K}_{{2}}$$
$$\displaystyle{\frac{{{a}{r}{\text{cosh}{{\left(\sqrt{{{K}}}{u}\right\rbrace}}}{\left\lbrace\sqrt{{{K}}}\right\rbrace}={x}+{K}_{{2}}}}{}}$$
Taking $$\displaystyle{\text{cosh}{}}$$ on both side
$$\displaystyle\sqrt{{{K}}}{u}={\text{cosh}{{\left(\sqrt{{{K}}}{\left({x}+{K}_{{2}}\right)}\right)}}}$$
$$\displaystyle{u}={\frac{{{1}}}{{\sqrt{{{K}}}}}}{\text{cosh}{{\left(\sqrt{{{K}}}{x}+{K}_{{2}}\right)}}}$$
Which is close to your formula
$$\displaystyle{u}={d}{\text{cosh}{{\left(\frac{{{x}-{b}}}{{d}}\right)}}}\Rightarrow{d}=\frac{{1}}{\sqrt{{{K}}}}$$ and $$\displaystyle-\frac{{b}}{{d}}={K}_{{2}}$$