76.25.15 problem 19

Internal problem ID [17828]
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 6.3 (Homogeneous Linear Systems with Constant Coefficients). Problems at page 408
Problem number : 19
Date solved : Tuesday, January 28, 2025 at 11:03:31 AM
CAS classification : system_of_ODEs

\begin{align*} \frac {d}{d t}x_{1} \left (t \right )&=2 x_{1} \left (t \right )+2 x_{2} \left (t \right )-x_{4} \left (t \right )\\ \frac {d}{d t}x_{2} \left (t \right )&=2 x_{1} \left (t \right )-x_{2} \left (t \right )+2 x_{4} \left (t \right )\\ \frac {d}{d t}x_{3} \left (t \right )&=3 x_{3} \left (t \right )\\ \frac {d}{d t}x_{4} \left (t \right )&=-x_{1} \left (t \right )+2 x_{2} \left (t \right )+2 x_{4} \left (t \right ) \end{align*}

✓ Solution by Maple

Time used: 0.151 (sec). Leaf size: 74

dsolve([diff(x__1(t),t)=2*x__1(t)+2*x__2(t)-0*x__3(t)-1*x__4(t),diff(x__2(t),t)=2*x__1(t)-1*x__2(t)+0*x__3(t)+2*x__4(t),diff(x__3(t),t)=0*x__1(t)-0*x__2(t)+3*x__3(t)+0*x__4(t),diff(x__4(t),t)=-1*x__1(t)+2*x__2(t)+0*x__3(t)+2*x__4(t)],singsol=all)
 

\begin{align*} x_{1} \left (t \right ) &= {\mathrm e}^{-3 t} c_{2} +c_{3} {\mathrm e}^{3 t} \\ x_{2} \left (t \right ) &= -2 \,{\mathrm e}^{-3 t} c_{2} -2 c_{3} {\mathrm e}^{3 t}+c_{1} {\mathrm e}^{3 t} \\ x_{3} \left (t \right ) &= c_4 \,{\mathrm e}^{3 t} \\ x_{4} \left (t \right ) &= {\mathrm e}^{-3 t} c_{2} -5 c_{3} {\mathrm e}^{3 t}+2 c_{1} {\mathrm e}^{3 t} \\ \end{align*}

✓ Solution by Mathematica

Time used: 0.025 (sec). Leaf size: 282

DSolve[{D[x1[t],t]==2*x1[t]+2*x2[t]-0*x3[t]-1*x4[t],D[x2[t],t]==2*x1[t]-1*x2[t]+0*x3[t]+2*x4[t],D[x3[t],t]==0*x1[t]-0*x2[t]+3*x3[t]+0*x4[t],D[x4[t],t]==-1*x1[t]+2*x2[t]+0*x3[t]+2*x4[t]},{x1[t],x2[t],x3[t],x4[t]},t,IncludeSingularSolutions -> True]
 

\begin{align*} \text {x1}(t)\to \frac {1}{6} e^{-3 t} \left (c_1 \left (5 e^{6 t}+1\right )+(2 c_2-c_3) \left (e^{6 t}-1\right )\right ) \\ \text {x2}(t)\to \frac {1}{3} e^{-3 t} \left (c_1 \left (e^{6 t}-1\right )+c_2 \left (e^{6 t}+2\right )+c_3 \left (e^{6 t}-1\right )\right ) \\ \text {x4}(t)\to \frac {1}{6} e^{-3 t} \left (-\left (c_1 \left (e^{6 t}-1\right )\right )+2 c_2 \left (e^{6 t}-1\right )+c_3 \left (5 e^{6 t}+1\right )\right ) \\ \text {x3}(t)\to c_4 e^{3 t} \\ \text {x1}(t)\to \frac {1}{6} e^{-3 t} \left (c_1 \left (5 e^{6 t}+1\right )+(2 c_2-c_3) \left (e^{6 t}-1\right )\right ) \\ \text {x2}(t)\to \frac {1}{3} e^{-3 t} \left (c_1 \left (e^{6 t}-1\right )+c_2 \left (e^{6 t}+2\right )+c_3 \left (e^{6 t}-1\right )\right ) \\ \text {x4}(t)\to \frac {1}{6} e^{-3 t} \left (-\left (c_1 \left (e^{6 t}-1\right )\right )+2 c_2 \left (e^{6 t}-1\right )+c_3 \left (5 e^{6 t}+1\right )\right ) \\ \text {x3}(t)\to 0 \\ \end{align*}