Internal problem ID [9755]
Book: Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev. Second
edition
Section: Chapter 1, section 1.2. Riccati Equation. subsection 1.2.5-2
Problem number: 23.
ODE order: 1.
ODE degree: 1.
CAS Maple gives this as type [_Riccati]
Solve \begin {gather*} \boxed {\left (a \ln \relax (x )+b \right ) y^{\prime }-\ln \relax (x )^{n} y^{2}-c y+\lambda ^{2} \ln \relax (x )^{n}-c \lambda =0} \end {gather*}
✓ Solution by Maple
Time used: 0.004 (sec). Leaf size: 168
dsolve((a*ln(x)+b)*diff(y(x),x)=(ln(x))^n*y(x)^2+c*y(x)-lambda^2*(ln(x))^n+c*lambda,y(x), singsol=all)
\[ y \relax (x ) = -\frac {\left (\left (\int \frac {\ln \relax (x )^{n} {\mathrm e}^{\int -\frac {2 \ln \relax (x )^{n} \lambda -c}{a \ln \relax (x )+b}d x}}{a \ln \relax (x )+b}d x \right ) {\mathrm e}^{\int \frac {2 \ln \relax (x )^{n} \lambda -c}{a \ln \relax (x )+b}d x} \lambda +c_{1} {\mathrm e}^{\int \frac {2 \ln \relax (x )^{n} \lambda -c}{a \ln \relax (x )+b}d x} \lambda +1\right ) {\mathrm e}^{\int -\frac {2 \ln \relax (x )^{n} \lambda -c}{a \ln \relax (x )+b}d x}}{c_{1}+\int \frac {\ln \relax (x )^{n} {\mathrm e}^{\int -\frac {2 \ln \relax (x )^{n} \lambda -c}{a \ln \relax (x )+b}d x}}{a \ln \relax (x )+b}d x} \]
✓ Solution by Mathematica
Time used: 3.093 (sec). Leaf size: 286
DSolve[(a*Log[x]+b)*y'[x]==(Log[x])^n*y[x]^2+c*y[x]-\[Lambda]^2*(Log[x])^n+c*\[Lambda],y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [\int _1^x\frac {\exp \left (-\int _1^{K[2]}-\frac {c-2 \lambda \log ^n(K[1])}{b+a \log (K[1])}dK[1]\right ) \left (-\lambda \log ^n(K[2])+y(x) \log ^n(K[2])+c\right )}{c n (b+a \log (K[2])) (\lambda +y(x))}dK[2]+\int _1^{y(x)}\left (-\int _1^x\left (\frac {\exp \left (-\int _1^{K[2]}-\frac {c-2 \lambda \log ^n(K[1])}{b+a \log (K[1])}dK[1]\right ) \log ^n(K[2])}{c n (\lambda +K[3]) (b+a \log (K[2]))}-\frac {\exp \left (-\int _1^{K[2]}-\frac {c-2 \lambda \log ^n(K[1])}{b+a \log (K[1])}dK[1]\right ) \left (-\lambda \log ^n(K[2])+K[3] \log ^n(K[2])+c\right )}{c n (\lambda +K[3])^2 (b+a \log (K[2]))}\right )dK[2]-\frac {\exp \left (-\int _1^x-\frac {c-2 \lambda \log ^n(K[1])}{b+a \log (K[1])}dK[1]\right )}{c n (\lambda +K[3])^2}\right )dK[3]=c_1,y(x)\right ] \]