I'm currently working on some problems concerning the calculus of

Schwelliney 2021-11-23 Answered
I'm currently working on some problems concerning the calculus of variations and I have come up with the following differential equation that I now want to solve:
\(\displaystyle{1}+{y}'{\left({x}\right)}^{{2}}-{y}{''}{\left({x}\right)}{\left({y}{\left({x}\right)}-\lambda\right)}={0}\)
I'm only used to solving linear differential equations so I would appreciate some advice on where to start.

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Expert Answer

Helen Rodriguez
Answered 2021-11-24 Author has 1343 answers

Let \(\displaystyle{u}={y}-\lambda,\ {v}={u}'={y}'\). then we can rewrite the differential equation as
\(\displaystyle{1}+{v}^{{2}}-{v}'{u}={0}\to{u}{\frac{{{d}{v}}}{{{\left.{d}{x}\right.}}}}={v}'{u}={1}+{v}^{{2}},\ {\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}}={u}'={v}\)
which implies
\(\displaystyle{u}{\frac{{{d}{v}}}{{{d}{u}}}}={\frac{{{1}+{v}^{{2}}}}{{{v}}}}\to{\frac{{{2}{v}{d}{v}}}{{{1}+{v}^{{2}}}}}={\frac{{{2}{d}{u}}}{{{u}}}}\)
on integration, we get
\(\displaystyle{\left({1}+{v}^{{2}}\right)}={c}{u}^{{2}}\to{v}=\pm\sqrt{{{1}-{c}{u}^{{2}}}}\to{\frac{{{d}{u}}}{{\sqrt{{{1}-{c}{u}^{{2}}}}}}}=\pm{\left.{d}{x}\right.}\)
now the integration will depend on the sign of c: if \(\displaystyle{c}{>}{0}\), you get an inverse sine, \(\displaystyle{{\sin}^{{-{1}}}}\) and if \(\displaystyle{c}{<}{0}\), then \(\displaystyle{\ln{}}\)

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Forneadil
Answered 2021-11-25 Author has 1175 answers
As abel noted, with \(\displaystyle{u}={y}-\lambda\) we get \(\displaystyle{1}+{\left({u}'\right)}^{{2}}-{u}{''}{u}={0}\). Now if you recall that \(\displaystyle{{\text{cosh}}^{{2}}-}{{\text{sinh}}^{{2}}=}{1}\) and \(\displaystyle{\text{cosh}{{''}}}={\text{cosh}{}}\), you are led to consider something of the form \(\displaystyle{u}={a}{\text{cosh}{{\left({b}{x}+{c}\right)}}}\) and we find that this works if \(\displaystyle{b}=\frac{{1}}{{a}}\). Thus we get the solution
\(\displaystyle{y}=\lambda+{a}{\text{cosh}{{\left(\frac{{x}}{{a}}+{c}\right)}}}\)
where \(\displaystyle{a}\ne{0}\) and c are arbitrary parameters.
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