2.1.83 Problem 82

Internal problem ID [8795]
Book : Own collection of miscellaneous problems
Section : section 1.0
Problem number : 82
Date solved : Sunday, March 30, 2025 at 01:36:47 PM
CAS classification : [[_2nd_order, _linear, _nonhomogeneous]]

Solved as second order linear constant coeff ode

Time used: 0.158 (sec)

Solve

$$\begin{array}{rl}{z}^{\prime \prime }+3z^{\prime }+2z& =24{\mathrm{e}}^{-3t}-24{\mathrm{e}}^{-4t}\end{array}$$

This is second order non-homogeneous ODE. In standard form the ODE is

$$A{z}^{″}(t)+B{z}^{\prime }(t)+Cz(t)=f(t)$$

Where $A=1,B=3,C=2,f(t)=24{\mathrm{e}}^{-3t}-24{\mathrm{e}}^{-4t}$. Let the solution be

$$z=z_h+z_p$$

Where $z_h$ is the solution to the homogeneous ODE $A{z}^{″}(t)+B{z}^{\prime }(t)+Cz(t)=0$, and $z_p$ is a particular solution to the non-homogeneous ODE $A{z}^{″}(t)+B{z}^{\prime }(t)+Cz(t)=f(t)$. $z_h$ is the solution to

$$z^{\prime \prime }+3z^{\prime }+2z=0$$

This is second order with constant coefficients homogeneous ODE. In standard form the ODE is

$$A{z}^{″}(t)+B{z}^{\prime }(t)+Cz(t)=0$$

Where in the above $A=1,B=3,C=2$. Let the solution be $z=e^{\lambda t}$. Substituting this into the ODE gives

$$\begin{array}{}\text{(1)}& {\lambda }^2{\mathrm{e}}^{t\lambda }+3\lambda {\mathrm{e}}^{t\lambda }+2{\mathrm{e}}^{t\lambda }=0\end{array}$$

Since exponential function is never zero, then dividing Eq(2) throughout by $e^{\lambda t}$ gives

$$\begin{array}{}\text{(2)}& {\lambda }^2+3\lambda +2=0\end{array}$$

Equation (2) is the characteristic equation of the ODE. Its roots determine the general solution form.Using the quadratic formula

$${\lambda }_{1,2}=\frac{-B}{2A}\pm \frac{1}{2A}\sqrt{B^2-4AC}$$

Substituting $A=1,B=3,C=2$ into the above gives

$$\begin{array}{rl}{\lambda }_{1,2}& =\frac{-3}{(2)\left(1\right)}\pm \frac{1}{(2)\left(1\right)}\sqrt{3^2-(4)\left(1\right)\left(2\right)}\\ & =-\frac{3}{2}\pm \frac{1}{2}\end{array}$$

Hence

$$\begin{array}{rl}{\lambda }_1& =-\frac{3}{2}+\frac{1}{2}\\ {\lambda }_2& =-\frac{3}{2}-\frac{1}{2}\end{array}$$

Which simplifies to

$$\begin{array}{rl}{\lambda }_1& =-1\\ {\lambda }_2& =-2\end{array}$$

Since roots are real and distinct, then the solution is

$$\begin{array}{rl}z& =c_1{e}^{{\lambda }_{1}t}+c_2{e}^{{\lambda }_{2}t}\\ z& =c_1{e}^{(-1)t}+c_2{e}^{(-2)t}\end{array}$$

Or

$$z=c_1{\mathrm{e}}^{-t}+c_2{\mathrm{e}}^{-2t}$$

Therefore the homogeneous solution $z_h$ is

$$z_h=c_1{\mathrm{e}}^{-t}+c_2{\mathrm{e}}^{-2t}$$

The particular solution is now found using the method of undetermined coefficients. Looking at the RHS of the ode, which is

$$24{\mathrm{e}}^{-3t}-24{\mathrm{e}}^{-4t}$$

Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is

$$[\{{\mathrm{e}}^{-4t}\},\{{\mathrm{e}}^{-3t}\}]$$

While the set of the basis functions for the homogeneous solution found earlier is

$$\{{\mathrm{e}}^{-2t},{\mathrm{e}}^{-t}\}$$

Since there is no duplication between the basis function in the UC_set and the basis functions of the homogeneous solution, the trial solution is a linear combination of all the basis in the UC_set.

$$z_p=A_1{\mathrm{e}}^{-4t}+A_2{\mathrm{e}}^{-3t}$$

The unknowns $\{A_1,A_2\}$ are found by substituting the above trial solution $z_p$ into the ODE and comparing coefficients. Substituting the trial solution into the ODE and simplifying gives

$$6A_1{\mathrm{e}}^{-4t}+2A_2{\mathrm{e}}^{-3t}=24{\mathrm{e}}^{-3t}-24{\mathrm{e}}^{-4t}$$

Solving for the unknowns by comparing coefficients results in

$$[A_1=-4,A_2=12]$$

Substituting the above back in the above trial solution $z_p$, gives the particular solution

$$z_p=-4{\mathrm{e}}^{-4t}+12{\mathrm{e}}^{-3t}$$

Therefore the general solution is

$$\begin{array}{rl}z& =z_h+z_p\\ & =(c_1{\mathrm{e}}^{-t}+c_2{\mathrm{e}}^{-2t})+(-4{\mathrm{e}}^{-4t}+12{\mathrm{e}}^{-3t})\end{array}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}z& =-4{\mathrm{e}}^{-4t}+12{\mathrm{e}}^{-3t}+c_1{\mathrm{e}}^{-t}+c_2{\mathrm{e}}^{-2t}\end{array}$$

Figure 2.148: Slope field $z^{\prime \prime }+3z^{\prime }+2z=24{\mathrm{e}}^{-3t}-24{\mathrm{e}}^{-4t}$

Solved as second order ode using Kovacic algorithm

Time used: 0.118 (sec)

Solve

$$\begin{array}{rl}{z}^{\prime \prime }+3z^{\prime }+2z& =24{\mathrm{e}}^{-3t}-24{\mathrm{e}}^{-4t}\end{array}$$

Writing the ode as

$$\begin{array}{}\text{(1)}& z^{\prime \prime }+3z^{\prime }+2z& =0\text{(2)}& A{z}^{\prime \prime }+B{z}^{\prime }+Cz& =0\end{array}$$

Comparing (1) and (2) shows that

$$\begin{array}{rl}A& =1\\ \text{(3)}& B& =3C& =2\end{array}$$

Applying the Liouville transformation on the dependent variable gives

$$\begin{array}{rl}z(t)& =z{e}^{\int \frac{B}{2A}dt}\end{array}$$

Then (2) becomes

$$\begin{array}{}\text{(4)}& z^{″}(t)=rz(t)\end{array}$$

Where $r$ is given by

$$\begin{array}{rl}\text{(5)}& r& =\frac{s}{t}& =\frac{2A{B}^{\prime }-2B{A}^{\prime }+B^2-4AC}{4A^2}\end{array}$$

Substituting the values of $A,B,C$ from (3) in the above and simplifying gives

$$\begin{array}{}\text{(6)}& r& =\frac{1}{4}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =1\\ t& =4\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(t)& =\frac{z}{4}\end{array}$$

Equation (7) is now solved. After finding $z(t)$ then $z$ is found using the inverse transformation

$$\begin{array}{rl}z& =z{e}^{-\int \frac{B}{2A}dt}\end{array}$$

The first step is to determine the case of Kovacic algorithm this ode belongs to. There are 3 cases depending on the order of poles of $r$ and the order of $r$ at $\mathrm{\infty }$. The following table summarizes these cases.

Case	Allowed pole order for $r$	Allowed value for $\mathcal{O}(\mathrm{\infty })$
1	$\{0,1,2,4,6,8,\cdots \}$	$\{\cdots ,-6,-4,-2,0,2,3,4,5,6,\cdots \}$
2	Need to have at least one pole that is either order $2$ or odd order greater than $2$. Any other pole order is allowed as long as the above condition is satisfied. Hence the following set of pole orders are all allowed. $\{1,2\}$,$\{1,3\}$,$\{2\}$,$\{3\}$,$\{3,4\}$,$\{1,2,5\}$.	no condition
3	$\{1,2\}$	$\{2,3,4,5,6,7,\cdots \}$

Table 2.26: Necessary conditions for each Kovacic case

The order of $r$ at $\mathrm{\infty }$ is the degree of $t$ minus the degree of $s$. Therefore

$$\begin{array}{rl}O\left(\mathrm{\infty }\right)& =\text{deg}(t)-\text{deg}(s)\\ & =0-0\\ & =0\end{array}$$

There are no poles in $r$. Therefore the set of poles $\mathrm{\Gamma }$ is empty. Since there is no odd order pole larger than $2$ and the order at $\mathrm{\infty }$ is $0$ then the necessary conditions for case one are met. Therefore

$$\begin{array}{rl}L& =[1]\end{array}$$

Since $r=\frac{1}{4}$ is not a function of $t$, then there is no need run Kovacic algorithm to obtain a solution for transformed ode $z^{″}=rz$ as one solution is

$$z_1(t)={\mathrm{e}}^{-\frac{t}{2}}$$

Using the above, the solution for the original ode can now be found. The first solution to the original ode in $z$ is found from

$$\begin{array}{rl}{z}_1& =z_1{e}^{\int -\frac{1}{2}\frac{B}{A}dt}\\ & =z_1{e}^{-\int \frac{1}{2}\frac{3}{1}dt}\\ & =z_1{e}^{-\frac{3t}{2}}\\ & =z_1\left({\mathrm{e}}^{-\frac{3t}{2}}\right)\end{array}$$

Which simplifies to

$$z_1={\mathrm{e}}^{-2t}$$

The second solution $z_2$ to the original ode is found using reduction of order

$$z_2=z_1\int \frac{e^{\int -\frac{B}{A}dt}}{z_1^2}dt$$

Substituting gives

$$\begin{array}{rl}{z}_2& =z_1\int \frac{e^{\int -\frac{3}{1}dt}}{{\left(z_1\right)}^2}dt\\ & =z_1\int \frac{e^{-3t}}{{\left(z_1\right)}^2}dt\\ & =z_1\left({\mathrm{e}}^{-3t}{\mathrm{e}}^{4t}\right)\end{array}$$

Therefore the solution is

$$\begin{array}{rl}z& =c_1{z}_1+c_2{z}_2\\ & =c_1\left({\mathrm{e}}^{-2t}\right)+c_2\left({\mathrm{e}}^{-2t}\left({\mathrm{e}}^{-3t}{\mathrm{e}}^{4t}\right)\right)\end{array}$$

This is second order nonhomogeneous ODE. Let the solution be

$$z=z_h+z_p$$

Where $z_h$ is the solution to the homogeneous ODE $A{z}^{″}(t)+B{z}^{\prime }(t)+Cz(t)=0$, and $z_p$ is a particular solution to the nonhomogeneous ODE $A{z}^{″}(t)+B{z}^{\prime }(t)+Cz(t)=f(t)$. $z_h$ is the solution to

$$z^{\prime \prime }+3z^{\prime }+2z=0$$

The homogeneous solution is found using the Kovacic algorithm which results in

$$z_h=c_1{\mathrm{e}}^{-2t}+c_2{\mathrm{e}}^{-t}$$

The particular solution is now found using the method of undetermined coefficients. Looking at the RHS of the ode, which is

$$24{\mathrm{e}}^{-3t}-24{\mathrm{e}}^{-4t}$$

Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is

$$[\{{\mathrm{e}}^{-4t}\},\{{\mathrm{e}}^{-3t}\}]$$

While the set of the basis functions for the homogeneous solution found earlier is

$$\{{\mathrm{e}}^{-2t},{\mathrm{e}}^{-t}\}$$

Since there is no duplication between the basis function in the UC_set and the basis functions of the homogeneous solution, the trial solution is a linear combination of all the basis in the UC_set.

$$z_p=A_1{\mathrm{e}}^{-4t}+A_2{\mathrm{e}}^{-3t}$$

The unknowns $\{A_1,A_2\}$ are found by substituting the above trial solution $z_p$ into the ODE and comparing coefficients. Substituting the trial solution into the ODE and simplifying gives

$$6A_1{\mathrm{e}}^{-4t}+2A_2{\mathrm{e}}^{-3t}=24{\mathrm{e}}^{-3t}-24{\mathrm{e}}^{-4t}$$

Solving for the unknowns by comparing coefficients results in

$$[A_1=-4,A_2=12]$$

Substituting the above back in the above trial solution $z_p$, gives the particular solution

$$z_p=-4{\mathrm{e}}^{-4t}+12{\mathrm{e}}^{-3t}$$

Therefore the general solution is

$$\begin{array}{rl}z& =z_h+z_p\\ & =(c_1{\mathrm{e}}^{-2t}+c_2{\mathrm{e}}^{-t})+(-4{\mathrm{e}}^{-4t}+12{\mathrm{e}}^{-3t})\end{array}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}z& =-4{\mathrm{e}}^{-4t}+12{\mathrm{e}}^{-3t}+c_1{\mathrm{e}}^{-2t}+c_2{\mathrm{e}}^{-t}\end{array}$$

Figure 2.149: Slope field $z^{\prime \prime }+3z^{\prime }+2z=24{\mathrm{e}}^{-3t}-24{\mathrm{e}}^{-4t}$

✓ Maple. Time used: 0.003 (sec). Leaf size: 30

ode:=diff(diff(z(t),t),t)+3*diff(z(t),t)+2*z(t) = 24*exp(-3*t)-24*exp(-4*t); 
dsolve(ode,z(t), singsol=all);
 

$$z=-{\mathrm{e}}^{-2t}{c}_1-4{\mathrm{e}}^{-4t}+12{\mathrm{e}}^{-3t}+{\mathrm{e}}^{-t}{c}_2$$

Maple trace

Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying high order exact linear fully integrable 
trying differential order: 2; linear nonhomogeneous with symmetry [0,1] 
trying a double symmetry of the form [xi=0, eta=F(x)] 
<- double symmetry of the form [xi=0, eta=F(x)] successful
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & \frac{d}{dt}\frac{d}{dt}z\left(t\right)+3\frac{d}{dt}z\left(t\right)+2z\left(t\right)=24{\mathrm{e}}^{-3t}-24{\mathrm{e}}^{-4t}\\ \bullet & & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{d}{dt}\frac{d}{dt}z\left(t\right)\\ \bullet & & \text{Characteristic polynomial of homogeneous ODE}\\ & & r^2+3r+2=0\\ \bullet & & \text{Factor the characteristic polynomial}\\ & & (r+2)(r+1)=0\\ \bullet & & \text{Roots of the characteristic polynomial}\\ & & r=(-2,-1)\\ \bullet & & \text{1st solution of the homogeneous ODE}\\ & & z_1\left(t\right)={\mathrm{e}}^{-2t}\\ \bullet & & \text{2nd solution of the homogeneous ODE}\\ & & z_2\left(t\right)={\mathrm{e}}^{-t}\\ \bullet & & \text{General solution of the ODE}\\ & & z\left(t\right)=\mathit{C}\mathit{1}{z}_1\left(t\right)+\mathit{C}\mathit{2}{z}_2\left(t\right)+z_p\left(t\right)\\ \bullet & & \text{Substitute in solutions of the homogeneous ODE}\\ & & z\left(t\right)=\mathit{C}\mathit{1}{\mathrm{e}}^{-2t}+\mathit{C}\mathit{2}{\mathrm{e}}^{-t}+z_p\left(t\right)\\ ◻& & \text{Find a particular solution}{z}_p\left(t\right)\text{of the ODE}\\ & \circ & \text{Use variation of parameters to find}{z}_p\text{here}f\left(t\right)\text{is the forcing function}\\ & & [z_p\left(t\right)=-z_1\left(t\right)\int \frac{z_2\left(t\right)f\left(t\right)}{W(z_1\left(t\right),z_2\left(t\right))}dt+z_2\left(t\right)\int \frac{z_1\left(t\right)f\left(t\right)}{W(z_1\left(t\right),z_2\left(t\right))}dt,f\left(t\right)=24{\mathrm{e}}^{-3t}-24{\mathrm{e}}^{-4t}]\\ & \circ & \text{Wronskian of solutions of the homogeneous equation}\\ & & W(z_1\left(t\right),z_2\left(t\right))=\left[\begin{array}{cc}{\mathrm{e}}^{-2t}& {\mathrm{e}}^{-t}\\ -2{\mathrm{e}}^{-2t}& -{\mathrm{e}}^{-t}\end{array}\right]\\ & \circ & \text{Compute Wronskian}\\ & & W(z_1\left(t\right),z_2\left(t\right))={\mathrm{e}}^{-3t}\\ & \circ & \text{Substitute functions into equation for}{z}_p\left(t\right)\\ & & z_p\left(t\right)=-24{\mathrm{e}}^{-2t}\int (-{\mathrm{e}}^{-2t}+{\mathrm{e}}^{-t})dt+24{\mathrm{e}}^{-t}\int (-{\mathrm{e}}^{-3t}+{\mathrm{e}}^{-2t})dt\\ & \circ & \text{Compute integrals}\\ & & z_p\left(t\right)=-4{\mathrm{e}}^{-4t}+12{\mathrm{e}}^{-3t}\\ \bullet & & \text{Substitute particular solution into general solution to ODE}\\ & & z\left(t\right)=\mathit{C}\mathit{1}{\mathrm{e}}^{-2t}+\mathit{C}\mathit{2}{\mathrm{e}}^{-t}-4{\mathrm{e}}^{-4t}+12{\mathrm{e}}^{-3t}\end{array}$$

✓ Mathematica. Time used: 0.19 (sec). Leaf size: 34

ode=D[z[t],{t,2}]+3*D[z[t],t]+2*z[t]==24*(Exp[-3*t]-Exp[-4*t]); 
ic={}; 
DSolve[{ode,ic},z[t],t,IncludeSingularSolutions->True]
 

$$z(t)\to e^{-4t}(12e^t+c_1{e}^{2t}+c_2{e}^{3t}-4)$$

✓ Sympy. Time used: 0.286 (sec). Leaf size: 26

from sympy import * 
t = symbols("t") 
z = Function("z") 
ode = Eq(2*z(t) + 3*Derivative(z(t), t) + Derivative(z(t), (t, 2)) - 24*exp(-3*t) + 24*exp(-4*t),0) 
ics = {} 
dsolve(ode,func=z(t),ics=ics)
 

$$z\left(t\right)=(C_1+C_2{e}^{-t}+12e^{-2t}-4e^{-3t})e^{-t}$$