52.10.15 problem 14

Internal problem ID [8409]
Book : DIFFERENTIAL EQUATIONS with Boundary Value Problems. DENNIS G. ZILL, WARREN S. WRIGHT, MICHAEL R. CULLEN. Brooks/Cole. Boston, MA. 2013. 8th edition.
Section : CHAPTER 8 SYSTEMS OF LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS. EXERCISES 8.2. Page 346
Problem number : 14
Date solved : Wednesday, March 05, 2025 at 05:44:48 AM
CAS classification : system_of_ODEs

\begin{align*} \frac {d}{d t}x \left (t \right )&=x \left (t \right )+y \left (t \right )+4 z \left (t \right )\\ \frac {d}{d t}y \left (t \right )&=2 y \left (t \right )\\ \frac {d}{d t}z \left (t \right )&=x \left (t \right )+y \left (t \right )+z \left (t \right ) \end{align*}

With initial conditions

\begin{align*} x \left (0\right ) = 1\\ y \left (0\right ) = 3\\ z \left (0\right ) = 0 \end{align*}

✓ Maple. Time used: 0.055 (sec). Leaf size: 52

ode:=[diff(x(t),t) = x(t)+y(t)+4*z(t), diff(y(t),t) = 2*y(t), diff(z(t),t) = x(t)+y(t)+z(t)]; 
ic:=x(0) = 1y(0) = 3z(0) = 0; 
dsolve([ode,ic]);
 

\begin{align*} x \left (t \right ) &= 5 \,{\mathrm e}^{3 t}+{\mathrm e}^{-t}-5 \,{\mathrm e}^{2 t} \\ y &= 3 \,{\mathrm e}^{2 t} \\ z \left (t \right ) &= \frac {5 \,{\mathrm e}^{3 t}}{2}-\frac {{\mathrm e}^{-t}}{2}-2 \,{\mathrm e}^{2 t} \\ \end{align*}

✓ Mathematica. Time used: 0.005 (sec). Leaf size: 63

ode={D[x[t],t]==x[t]+y[t]+4*z[t],D[y[t],t]==2*y[t],D[z[t],t]==x[t]+y[t]+z[t]}; 
ic={x[0]==1,y[0]==3,z[0]==0}; 
DSolve[{ode,ic},{x[t],y[t],z[t]},t,IncludeSingularSolutions->True]
 

\begin{align*} x(t)\to e^{-t}-5 e^{2 t}+5 e^{3 t} \\ y(t)\to 3 e^{2 t} \\ z(t)\to \frac {1}{2} e^{-t} \left (-4 e^{3 t}+5 e^{4 t}-1\right ) \\ \end{align*}

✓ Sympy. Time used: 0.123 (sec). Leaf size: 61

from sympy import * 
t = symbols("t") 
x = Function("x") 
y = Function("y") 
z = Function("z") 
ode=[Eq(-x(t) - y(t) - 4*z(t) + Derivative(x(t), t),0),Eq(-2*y(t) + Derivative(y(t), t),0),Eq(-x(t) - y(t) - z(t) + Derivative(z(t), t),0)] 
ics = {} 
dsolve(ode,func=[x(t),y(t),z(t)],ics=ics)
 

\[ \left [ x{\left (t \right )} = - 2 C_{1} e^{- t} + \frac {5 C_{2} e^{2 t}}{2} + 2 C_{3} e^{3 t}, \ y{\left (t \right )} = - \frac {3 C_{2} e^{2 t}}{2}, \ z{\left (t \right )} = C_{1} e^{- t} + C_{2} e^{2 t} + C_{3} e^{3 t}\right ] \]