77.1.86 problem 113 (page 172)

Internal problem ID [17976]
Book : V.V. Stepanov, A course of differential equations (in Russian), GIFML. Moscow (1958)
Section : All content
Problem number : 113 (page 172)
Date solved : Tuesday, January 28, 2025 at 11:19:14 AM
CAS classification : [[_2nd_order, _missing_x]]

\begin{align*} y y^{\prime \prime }+{y^{\prime }}^{2}&=y^{2} \ln \left (y\right ) \end{align*}

Solution by Maple

Time used: 0.118 (sec). Leaf size: 73

dsolve(y(x)*diff(y(x),x$2)+diff(y(x),x)^2=y(x)^2*ln(y(x)),y(x), singsol=all)
 
\begin{align*} -4 \left (\int _{}^{y}\frac {\textit {\_a}}{\sqrt {8 \textit {\_a}^{4} \ln \left (\textit {\_a} \right )-2 \textit {\_a}^{4}+16 c_{1}}}d \textit {\_a} \right )-x -c_{2} &= 0 \\ 4 \left (\int _{}^{y}\frac {\textit {\_a}}{\sqrt {8 \textit {\_a}^{4} \ln \left (\textit {\_a} \right )-2 \textit {\_a}^{4}+16 c_{1}}}d \textit {\_a} \right )-x -c_{2} &= 0 \\ \end{align*}

Solution by Mathematica

Time used: 1.627 (sec). Leaf size: 311

DSolve[y[x]*D[y[x],{x,2}]+D[y[x],x]^2==y[x]^2*Log[y[x]],y[x],x,IncludeSingularSolutions -> True]
 
\begin{align*} y(x)\to \text {InverseFunction}\left [\int _1^{\text {$\#$1}}-\frac {2 \sqrt {2} K[1]}{\sqrt {4 \log (K[1]) K[1]^4-K[1]^4+8 c_1}}dK[1]\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\int _1^{\text {$\#$1}}\frac {2 \sqrt {2} K[2]}{\sqrt {4 \log (K[2]) K[2]^4-K[2]^4+8 c_1}}dK[2]\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\int _1^{\text {$\#$1}}-\frac {2 \sqrt {2} K[1]}{\sqrt {4 \log (K[1]) K[1]^4-K[1]^4+8 (-1) c_1}}dK[1]\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\int _1^{\text {$\#$1}}-\frac {2 \sqrt {2} K[1]}{\sqrt {4 \log (K[1]) K[1]^4-K[1]^4+8 c_1}}dK[1]\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\int _1^{\text {$\#$1}}\frac {2 \sqrt {2} K[2]}{\sqrt {4 \log (K[2]) K[2]^4-K[2]^4+8 (-1) c_1}}dK[2]\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\int _1^{\text {$\#$1}}\frac {2 \sqrt {2} K[2]}{\sqrt {4 \log (K[2]) K[2]^4-K[2]^4+8 c_1}}dK[2]\&\right ][x+c_2] \\ \end{align*}