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

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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Helen Rodriguez

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 $$\displaystyle{\ln{}}$$

###### Have a similar question?
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.