76.27.21 problem 22

Internal problem ID [17867]
Book : Differential equations. An introduction to modern methods and applications. James Brannan, William E. Boyce. Third edition. Wiley 2015
Section : Chapter 6. Systems of First Order Linear Equations. Section 6.5 (Fundamental Matrices and the Exponential of a Matrix). Problems at page 430
Problem number : 22
Date solved : Tuesday, January 28, 2025 at 11:04:15 AM
CAS classification : system_of_ODEs

\begin{align*} \frac {d}{d t}x_{1} \left (t \right )&=-k_{1} x_{1} \left (t \right )\\ \frac {d}{d t}x_{2} \left (t \right )&=k_{1} x_{1} \left (t \right )-k_{2} x_{2} \left (t \right )\\ \frac {d}{d t}x_{3} \left (t \right )&=k_{2} x_{2} \left (t \right ) \end{align*}

With initial conditions

\begin{align*} x_{1} \left (0\right ) = m_{0}\\ x_{2} \left (0\right ) = 0\\ x_{3} \left (0\right ) = 0 \end{align*}

✓ Solution by Maple

Time used: 0.379 (sec). Leaf size: 112

dsolve([diff(x__1(t),t) = -k__1*x__1(t), diff(x__2(t),t) = k__1*x__1(t)-k__2*x__2(t), diff(x__3(t),t) = k__2*x__2(t), x__1(0) = m__0, x__2(0) = 0, x__3(0) = 0], singsol=all)
 

\begin{align*} x_{1} \left (t \right ) &= m_{0} {\mathrm e}^{-k_{1} t} \\ x_{2} \left (t \right ) &= \left (-\frac {m_{0} k_{1} {\mathrm e}^{-k_{1} t +k_{2} t}}{k_{1} -k_{2}}+\frac {k_{1} m_{0}}{k_{1} -k_{2}}\right ) {\mathrm e}^{-k_{2} t} \\ x_{3} \left (t \right ) &= \frac {{\mathrm e}^{-k_{1} t} m_{0} k_{2} -\frac {{\mathrm e}^{-k_{2} t} k_{1}^{2} m_{0}}{k_{1} -k_{2}}+\frac {{\mathrm e}^{-k_{2} t} k_{1} m_{0} k_{2}}{k_{1} -k_{2}}+k_{1} m_{0} -k_{2} m_{0}}{k_{1} -k_{2}} \\ \end{align*}

✓ Solution by Mathematica

Time used: 0.009 (sec). Leaf size: 75

DSolve[{D[x1[t],t]==-k1*x1[t]-0*x2[t]+0*x3[t],D[x2[t],t]==k1*x1[t]-k2*x2[t]-0*x3[t],D[x3[t],t]==0*x1[t]+k2*x2[t]-0*x3[t]},{x1[0]==m0,x2[0]==0,x3[0]==0},{x1[t],x2[t],x3[t]},t,IncludeSingularSolutions -> True]
 

\begin{align*} \text {x1}(t)\to \text {m0} e^{-\text {k1} t} \\ \text {x2}(t)\to -\frac {\text {k1} \text {m0} \left (e^{-\text {k1} t}-e^{-\text {k2} t}\right )}{\text {k1}-\text {k2}} \\ \text {x3}(t)\to \frac {\text {m0} \left (\text {k1} \left (-e^{-\text {k2} t}\right )+\text {k2} \left (e^{-\text {k1} t}-1\right )+\text {k1}\right )}{\text {k1}-\text {k2}} \\ \end{align*}