Types of singularities in ODEy⋅d2ydx2−(dydx)2+1=0with y:R→RAs seen, it does not satisfy Picard–Lindelöf theorem because of singularity at y=0. Nevertheless it possesses smooth solutions, for example y(x)=1acos⁡ax for a≠0. On the other hand, all the solutions ofy⋅d2ydx2+(dydx)2+1=0explode when y approaches 0.The first singularity leads to violation of uniqueness, while the second breaks the solutions.Are there special terms for these types of singularities? A study? The term "regular singular point" seems to have a different meaning.

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Types of singularities in ODEy⋅d2ydx2−(dydx)2+1=0with y:R→RAs seen, it does not satisfy Picard–Lindelöf theorem because of singularity at y=0. Nevertheless it possesses smooth solutions, for example y(x)=1acos⁡ax for a≠0. On the other hand, all the solutions ofy⋅d2ydx2+(dydx)2+1=0explode when y approaches 0.The first singularity leads to violation of uniqueness, while the second breaks the solutions.Are there special terms for these types of singularities? A study? The term &quot;regular singular point&quot; seems to have a different meaning.

Rodach42hj · Accepted Answer

That might be a coordinate or removable singularity. With some change of variables, there seems to be a meaningful value at y=0.Step 1yd2ydx2−(dydx)2+1=0ddx(dydxy)=−1y2=d2dx2(ln⁡y)=−1y2u=ln⁡{y}d2dx2u=−e−2ududxd2dx2u=−e−2ududx(dudx)2=c1+e−2uStep 2duc1+e−2u=dx=dyyc1+1y2=dyc1y2+1c1y=tan⁡θ⇒dy=sec2{θ}dθc11c1y2+1=cos⁡{θ}∫dy…=∫(1c1)sec⁡{θ}dθ=1c1ln|1+c1y1+c1y2|+c2=xPlug y=0 into that, you get x=c2.1+c1y1+c1y2=c3ec1x[(1+c1y2)c32e2c1x−1]2=c1y2This will give you a fourth order polynomial in y with coefficients consisting of an exponential function of x.

Types of singularities in ODE \(\displaystyle{y}\cdot{\frac{{{d}^{{{2}}}{y}}}{{{\left.{d}{x}\right.}^{{{2}}}}}}-{\left({\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}\right)}^{{{2}}}+{1}={0}\) with \(\displaystyle{y}:{\mathbb{{{R}}}}\to{\mathbb{{{R}}}}\) As

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