Please suggest a substitution for solving or any method of

Donald Johnson 2021-12-08 Answered
Please suggest a substitution for solving or any method of solving:
\(\displaystyle{y}{''}{\left({x}\right)}{\cot{{\left({y}{\left({x}\right)}\right)}}}={y}'{\left({x}\right)}^{{2}}+{c}\)

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Expert Answer

esfloravaou
Answered 2021-12-09 Author has 947 answers
Substituting \(\displaystyle{u}={\sin{{\left({y}\right)}}}\) one gets
\(\displaystyle{u}{''}={\left({\sin{{\left({y}\right)}}}\right)}{''}={\left({\cos{{\left({y}\right)}}}{y}'\right)}'={\cos{{\left({y}\right)}}}{y}{''}-{\sin{{\left({y}\right)}}}{y}'^{{2}}={c}{\sin{{\left({y}\right)}}}={c}{u}\)
which is, depending on the sign of c, an oscillation equation or an exponential function which is easily solvable for u and thus for y.
\[u=\begin{cases}a_1\cos(\sqrt{|c|}x)+a_2\sin(\sqrt{|c|}x)&for\ c<0\\b_1+b_2x&for\ c=0\\c_1e^{\sqrt{c}x}+c_2e^{\sqrt{c}x}&for\ c>0\end{cases}\]
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Serita Dewitt
Answered 2021-12-10 Author has 5497 answers
\(\displaystyle{\cot{{y}}}{y}{''}={y}'^{{2}}+{c}\)
Substitute \(\displaystyle{y}'={p}\Rightarrow{y}{''}={p}{\frac{{{d}{p}}}{{{\left.{d}{y}\right.}}}}\)
\(\displaystyle{p}{p}'{\cot{{y}}}={p}^{{2}}+{c}\)
\(\displaystyle\int{\frac{{{2}{p}}}{{{p}^{{2}}+{c}}}}{d}{p}={2}\int{\tan{{\left({y}\right)}}}{\left.{d}{y}\right.}\)
\(\displaystyle{\ln}{\left|{p}^{{2}}+{c}\right|}=-{2}{\ln}{\left|{\cos{{\left({y}\right)}}}\right|}+{K}_{{1}}\)
\(\displaystyle{p}^{{2}}={\frac{{{K}_{{1}}}}{{{{\cos}^{{2}}{\left({y}\right)}}}}}-{c}\)
\(\displaystyle{y}'=\sqrt{{{\frac{{{K}_{{1}}}}{{{{\cos}^{{2}}{\left({y}\right)}}}}}-{c}}}\)
\(\displaystyle\int{\frac{{{\left.{d}{y}\right.}}}{{\sqrt{{{\frac{{{K}_{{1}}}}{{{{\cos}^{{2}}{\left({y}\right)}}}}}-{c}}}}}}={x}+{K}_{{2}}\)
with \(\displaystyle{u}={\sin{{\left({y}\right)}}}\)
\(\displaystyle\int{\frac{{{d}{u}}}{{\sqrt{{{K}_{{1}}+{c}{u}^{{2}}}}}}}={x}+{K}_{{2}}\)
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