14.27.4 problem 4
Internal
problem
ID
[2792]
Book
:
Differential
equations
and
their
applications,
4th
ed.,
M.
Braun
Section
:
Chapter
4.
Qualitative
theory
of
differential
equations.
Section
4.1
(Introduction).
Page
3770
Problem
number
:
4
Date
solved
:
Tuesday, January 28, 2025 at 02:38:52 PM
CAS
classification
:
system_of_ODEs
\begin{align*} x^{\prime }\left (t \right )&=-x \left (t \right )-x \left (t \right ) y^{2}\\ y^{\prime }&=-y-y x \left (t \right )^{2}\\ z^{\prime }\left (t \right )&=1-z \left (t \right )+x \left (t \right )^{2} \end{align*}
✓ Solution by Maple
Time used: 0.800 (sec). Leaf size: 214
dsolve([diff(x(t),t)=-x(t)-x(t)*y(t)^2,diff(y(t),t)=-y(t)-y(t)*x(t)^2,diff(z(t),t)=1-z(t)+x(t)^2],singsol=all)
\begin{align*}
\\
\left [\left \{x \left (t \right ) &= \frac {\sqrt {\left ({\mathrm e}^{2 t}-{\mathrm e}^{2 c_3 c_2} {\mathrm e}^{2 t} {\mathrm e}^{{\mathrm e}^{-2 t} c_2}\right ) c_2}}{{\mathrm e}^{2 t}-{\mathrm e}^{2 c_3 c_2} {\mathrm e}^{2 t} {\mathrm e}^{{\mathrm e}^{-2 t} c_2}}, x \left (t \right ) = -\frac {\sqrt {\left ({\mathrm e}^{2 t}-{\mathrm e}^{2 c_3 c_2} {\mathrm e}^{2 t} {\mathrm e}^{{\mathrm e}^{-2 t} c_2}\right ) c_2}}{{\mathrm e}^{2 t}-{\mathrm e}^{2 c_3 c_2} {\mathrm e}^{2 t} {\mathrm e}^{{\mathrm e}^{-2 t} c_2}}\right \}, \left \{y = \frac {\sqrt {-x \left (t \right ) \left (x^{\prime }\left (t \right )+x \left (t \right )\right )}}{x \left (t \right )}, y = -\frac {\sqrt {-x \left (t \right ) \left (x^{\prime }\left (t \right )+x \left (t \right )\right )}}{x \left (t \right )}\right \}, \{z \left (t \right ) = \left (\int {\mathrm e}^{t} \left (x \left (t \right )^{2}+1\right )d t +c_1 \right ) {\mathrm e}^{-t}\}\right ] \\
\end{align*}
✓ Solution by Mathematica
Time used: 0.211 (sec). Leaf size: 445
DSolve[{D[x[t],t]==-x[t]-x[t]*y[t]^2,D[y[t],t]==-y[t]-y[t]*x[t]^2,D[z[t],t]==1-z[t]+x[t]^2},{x[t],y[t],z[t]},t,IncludeSingularSolutions -> True]
\begin{align*}
y(t)\to -\sqrt {W\left (\text {InverseFunction}\left [\int _1^{\text {$\#$1}}\frac {1}{K[1] \left (W\left (e^{K[1]^2+2 c_1} K[1]^2\right )+1\right )}dK[1]\&\right ][-t+c_2]{}^2 \exp \left (\text {InverseFunction}\left [\int _1^{\text {$\#$1}}\frac {1}{K[1] \left (W\left (e^{K[1]^2+2 c_1} K[1]^2\right )+1\right )}dK[1]\&\right ][-t+c_2]{}^2+2 c_1\right )\right )} \\
x(t)\to \text {InverseFunction}\left [\int _1^{\text {$\#$1}}\frac {1}{K[1] \left (W\left (e^{K[1]^2+2 c_1} K[1]^2\right )+1\right )}dK[1]\&\right ][-t+c_2] \\
z(t)\to e^{-t} \left (\int _1^te^{K[3]} \left (\text {InverseFunction}\left [\int _1^{\text {$\#$1}}\frac {1}{K[1] \left (W\left (e^{K[1]^2+2 c_1} K[1]^2\right )+1\right )}dK[1]\&\right ][c_2-K[3]]{}^2+1\right )dK[3]+c_3\right ) \\
y(t)\to \sqrt {W\left (\text {InverseFunction}\left [\int _1^{\text {$\#$1}}\frac {1}{K[2] \left (W\left (e^{K[2]^2+2 c_1} K[2]^2\right )+1\right )}dK[2]\&\right ][-t+c_2]{}^2 \exp \left (\text {InverseFunction}\left [\int _1^{\text {$\#$1}}\frac {1}{K[2] \left (W\left (e^{K[2]^2+2 c_1} K[2]^2\right )+1\right )}dK[2]\&\right ][-t+c_2]{}^2+2 c_1\right )\right )} \\
x(t)\to \text {InverseFunction}\left [\int _1^{\text {$\#$1}}\frac {1}{K[2] \left (W\left (e^{K[2]^2+2 c_1} K[2]^2\right )+1\right )}dK[2]\&\right ][-t+c_2] \\
z(t)\to e^{-t} \left (\int _1^te^{K[4]} \left (\text {InverseFunction}\left [\int _1^{\text {$\#$1}}\frac {1}{K[2] \left (W\left (e^{K[2]^2+2 c_1} K[2]^2\right )+1\right )}dK[2]\&\right ][c_2-K[4]]{}^2+1\right )dK[4]+c_3\right ) \\
\end{align*}