Question

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?

Answer 1

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?

Answer 2

\(\displaystyle\cup{''}={1}+{\left({u}'\right)}^{{2}}\)
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 \(\displaystyle-\frac{{b}}{{d}}={K}_{{2}}\)