61.3.20 problem 20
Internal
problem
ID
[12104]
Book
:
Handbook
of
exact
solutions
for
ordinary
differential
equations.
By
Polyanin
and
Zaitsev.
Second
edition
Section
:
Chapter
1,
section
1.2.
Riccati
Equation.
subsection
1.2.3.
Equations
Containing
Exponential
Functions
Problem
number
:
20
Date
solved
:
Tuesday, January 28, 2025 at 07:08:49 PM
CAS
classification
:
[_Riccati]
\begin{align*} \left (a \,{\mathrm e}^{\lambda x}+b \,{\mathrm e}^{\mu x}+c \right ) y^{\prime }&=y^{2}+k \,{\mathrm e}^{\nu x} y-m^{2}+k m \,{\mathrm e}^{\nu x} \end{align*}
✓ Solution by Maple
Time used: 0.006 (sec). Leaf size: 202
dsolve((a*exp(lambda*x)+b*exp(mu*x)+c)*diff(y(x),x)=y(x)^2+k*exp(nu*x)*y(x)-m^2+k*m*exp(nu*x),y(x), singsol=all)
\[
y = \frac {-m \left (\int \frac {{\mathrm e}^{k \left (\int \frac {{\mathrm e}^{\nu x}}{{\mathrm e}^{\lambda x} a +b \,{\mathrm e}^{\mu x}+c}d x \right )-2 m \left (\int \frac {1}{{\mathrm e}^{\lambda x} a +b \,{\mathrm e}^{\mu x}+c}d x \right )}}{{\mathrm e}^{\lambda x} a +b \,{\mathrm e}^{\mu x}+c}d x \right )+c_{1} m -{\mathrm e}^{k \left (\int \frac {{\mathrm e}^{\nu x}}{{\mathrm e}^{\lambda x} a +b \,{\mathrm e}^{\mu x}+c}d x \right )-2 m \left (\int \frac {1}{{\mathrm e}^{\lambda x} a +b \,{\mathrm e}^{\mu x}+c}d x \right )}}{\int \frac {{\mathrm e}^{k \left (\int \frac {{\mathrm e}^{\nu x}}{{\mathrm e}^{\lambda x} a +b \,{\mathrm e}^{\mu x}+c}d x \right )-2 m \left (\int \frac {1}{{\mathrm e}^{\lambda x} a +b \,{\mathrm e}^{\mu x}+c}d x \right )}}{{\mathrm e}^{\lambda x} a +b \,{\mathrm e}^{\mu x}+c}d x -c_{1}}
\]
✓ Solution by Mathematica
Time used: 5.560 (sec). Leaf size: 358
DSolve[(a*Exp[\[Lambda]*x]+b*Exp[\[Mu]*x]+c)*D[y[x],x]==y[x]^2+k*Exp[\[Nu]*x]*y[x]-m^2+k*m*Exp[\[Nu]*x],y[x],x,IncludeSingularSolutions -> True]
\[
\text {Solve}\left [\int _1^x-\frac {\exp \left (-\int _1^{K[2]}-\frac {e^{\nu K[1]} k-2 m}{e^{\lambda K[1]} a+b e^{\mu K[1]}+c}dK[1]\right ) \left (e^{\nu K[2]} k-m+y(x)\right )}{\left (e^{\lambda K[2]} a+b e^{\mu K[2]}+c\right ) k \nu (m+y(x))}dK[2]+\int _1^{y(x)}\left (\frac {\exp \left (-\int _1^x-\frac {e^{\nu K[1]} k-2 m}{e^{\lambda K[1]} a+b e^{\mu K[1]}+c}dK[1]\right )}{k \nu (m+K[3])^2}-\int _1^x\left (\frac {\exp \left (-\int _1^{K[2]}-\frac {e^{\nu K[1]} k-2 m}{e^{\lambda K[1]} a+b e^{\mu K[1]}+c}dK[1]\right ) \left (e^{\nu K[2]} k-m+K[3]\right )}{\left (e^{\lambda K[2]} a+b e^{\mu K[2]}+c\right ) k \nu (m+K[3])^2}-\frac {\exp \left (-\int _1^{K[2]}-\frac {e^{\nu K[1]} k-2 m}{e^{\lambda K[1]} a+b e^{\mu K[1]}+c}dK[1]\right )}{\left (e^{\lambda K[2]} a+b e^{\mu K[2]}+c\right ) k \nu (m+K[3])}\right )dK[2]\right )dK[3]=c_1,y(x)\right ]
\]