2.3.5 Problem 7 (v)

Internal problem ID [18458]
Book : Elementary Differential Equations. By R.L.E. Schwarzenberger. Chapman and Hall. London. First Edition (1969)
Section : Chapter 5. Linear equations. Exercises at page 85
Problem number : 7 (v)
Date solved : Monday, March 31, 2025 at 05:29:51 PM
CAS classification : [[_2nd_order, _linear, _nonhomogeneous]]

Existence and uniqueness analysis

Solve

$$\begin{array}{rl}{x}^{\prime \prime }+4x^{\prime }+3x& =t\sin \left(t\right)\end{array}$$

With initial conditions

$$\begin{array}{rl}x\left(0\right)& =0\\ x^{\prime }\left(0\right)& =1\end{array}$$

This is a linear ODE. In canonical form it is written as

$$\begin{array}{rl}{x}^{\prime \prime }+p(t)x^{\prime }+q(t)x& =F\end{array}$$

Where here

$$\begin{array}{rl}p(t)& =4\\ q(t)& =3\\ F& =t\sin \left(t\right)\end{array}$$

Hence the ode is

$$\begin{array}{r}{x}^{\prime \prime }+4x^{\prime }+3x=t\sin \left(t\right)\end{array}$$

The domain of $p(t)=4$ is

$$\{-\mathrm{\infty }<t<\mathrm{\infty }\}$$

And the point $t_0=0$ is inside this domain. The domain of $q(t)=3$ is

$$\{-\mathrm{\infty }<t<\mathrm{\infty }\}$$

And the point $t_0=0$ is also inside this domain. The domain of $F=t\sin \left(t\right)$ is

$$\{-\mathrm{\infty }<t<\mathrm{\infty }\}$$

And the point $t_0=0$ is also inside this domain. Hence solution exists and is unique.

Solved as second order linear constant coeff ode

Time used: 0.211 (sec)

Solve

$$\begin{array}{rl}{x}^{\prime \prime }+4x^{\prime }+3x& =t\sin \left(t\right)\end{array}$$

With initial conditions

$$\begin{array}{rl}x\left(0\right)& =0\\ x^{\prime }\left(0\right)& =1\end{array}$$

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

$$A{x}^{″}(t)+B{x}^{\prime }(t)+Cx(t)=f(t)$$

Where $A=1,B=4,C=3,f(t)=t\sin \left(t\right)$. Let the solution be

$$x=x_h+x_p$$

Where $x_h$ is the solution to the homogeneous ODE $A{x}^{″}(t)+B{x}^{\prime }(t)+Cx(t)=0$, and $x_p$ is a particular solution to the non-homogeneous ODE $A{x}^{″}(t)+B{x}^{\prime }(t)+Cx(t)=f(t)$. $x_h$ is the solution to

$$x^{\prime \prime }+4x^{\prime }+3x=0$$

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

$$A{x}^{″}(t)+B{x}^{\prime }(t)+Cx(t)=0$$

Where in the above $A=1,B=4,C=3$. Let the solution be $x=e^{\lambda t}$. Substituting this into the ODE gives

$$\begin{array}{}\text{(1)}& {\lambda }^2{\mathrm{e}}^{t\lambda }+4\lambda {\mathrm{e}}^{t\lambda }+3{\mathrm{e}}^{t\lambda }=0\end{array}$$

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

$$\begin{array}{}\text{(2)}& {\lambda }^2+4\lambda +3=0\end{array}$$

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 $A=1,B=4,C=3$ into the above gives

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

Hence

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

Which simplifies to

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

Since roots are real and distinct, then the solution is

$$\begin{array}{rl}x& =c_1{e}^{{\lambda }_{1}t}+c_2{e}^{{\lambda }_{2}t}\\ x& =c_1{e}^{(-1)t}+c_2{e}^{(-3)t}\end{array}$$

Or

$$x=c_1{\mathrm{e}}^{-t}+c_2{\mathrm{e}}^{-3t}$$

Therefore the homogeneous solution $x_h$ is

$$x_h=c_1{\mathrm{e}}^{-t}+c_2{\mathrm{e}}^{-3t}$$

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

$$t\sin \left(t\right)$$

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

$$[\{t\cos \left(t\right),t\sin \left(t\right),\cos \left(t\right),\sin \left(t\right)\}]$$

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

$$\{{\mathrm{e}}^{-3t},{\mathrm{e}}^{-t}\}$$

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.

$$x_p=A_{1}t\cos \left(t\right)+A_{2}t\sin \left(t\right)+A_3\cos \left(t\right)+A_4\sin \left(t\right)$$

The unknowns $\{A_1,A_2,A_3,A_4\}$ are found by substituting the above trial solution $x_p$ into the ODE and comparing coefficients. Substituting the trial solution into the ODE and simplifying gives

$$-2A_1\sin \left(t\right)+2A_{1}t\cos \left(t\right)+2A_2\cos \left(t\right)+2A_{2}t\sin \left(t\right)+2A_3\cos \left(t\right)+2A_4\sin \left(t\right)+4A_1\cos \left(t\right)-4A_{1}t\sin \left(t\right)+4A_2\sin \left(t\right)+4A_{2}t\cos \left(t\right)-4A_3\sin \left(t\right)+4A_4\cos \left(t\right)=t\sin \left(t\right)$$

Solving for the unknowns by comparing coefficients results in

$$[A_1=-\frac{1}{5},A_2=\frac{1}{10},A_3=\frac{11}{50},A_4=\frac{1}{25}]$$

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

$$x_p=-\frac{t\cos \left(t\right)}{5}+\frac{t\sin \left(t\right)}{10}+\frac{11\cos \left(t\right)}{50}+\frac{\sin \left(t\right)}{25}$$

Therefore the general solution is

$$\begin{array}{rl}x& =x_h+x_p\\ & =(c_1{\mathrm{e}}^{-t}+c_2{\mathrm{e}}^{-3t})+(-\frac{t\cos \left(t\right)}{5}+\frac{t\sin \left(t\right)}{10}+\frac{11\cos \left(t\right)}{50}+\frac{\sin \left(t\right)}{25})\end{array}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}x& =-\frac{t\cos \left(t\right)}{5}+\frac{t\sin \left(t\right)}{10}+\frac{11\cos \left(t\right)}{50}+\frac{\sin \left(t\right)}{25}+\frac{{\mathrm{e}}^{-t}}{4}-\frac{47{\mathrm{e}}^{-3t}}{100}\end{array}$$

Solution plot	Slope field $x^{\prime \prime }+4x^{\prime }+3x=t\sin \left(t\right)$

Solved as second order ode using Kovacic algorithm

Time used: 0.128 (sec)

Solve

$$\begin{array}{rl}{x}^{\prime \prime }+4x^{\prime }+3x& =t\sin \left(t\right)\end{array}$$

With initial conditions

$$\begin{array}{rl}x\left(0\right)& =0\\ x^{\prime }\left(0\right)& =1\end{array}$$

Writing the ode as

$$\begin{array}{}\text{(1)}& x^{\prime \prime }+4x^{\prime }+3x& =0\text{(2)}& A{x}^{\prime \prime }+B{x}^{\prime }+Cx& =0\end{array}$$

Comparing (1) and (2) shows that

$$\begin{array}{rl}A& =1\\ \text{(3)}& B& =4C& =3\end{array}$$

Applying the Liouville transformation on the dependent variable gives

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

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

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

Comparing the above to (5) shows that

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

Therefore eq. (4) becomes

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

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

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

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 2.22: 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 $r=1$ 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

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

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

$$\begin{array}{rl}{x}_1& =z_1{e}^{\int -\frac{1}{2}\frac{B}{A}dt}\\ & =z_1{e}^{-\int \frac{1}{2}\frac{4}{1}dt}\\ & =z_1{e}^{-2t}\\ & =z_1\left({\mathrm{e}}^{-2t}\right)\end{array}$$

Which simplifies to

$$x_1={\mathrm{e}}^{-3t}$$

The second solution $x_2$ to the original ode is found using reduction of order

$$x_2=x_1\int \frac{e^{\int -\frac{B}{A}dt}}{x_1^2}dt$$

Substituting gives

$$\begin{array}{rl}{x}_2& =x_1\int \frac{e^{\int -\frac{4}{1}dt}}{{\left(x_1\right)}^2}dt\\ & =x_1\int \frac{e^{-4t}}{{\left(x_1\right)}^2}dt\\ & =x_1\left(\frac{{\mathrm{e}}^{-4t}{\mathrm{e}}^{6t}}{2}\right)\end{array}$$

Therefore the solution is

$$\begin{array}{rl}x& =c_1{x}_1+c_2{x}_2\\ & =c_1\left({\mathrm{e}}^{-3t}\right)+c_2\left({\mathrm{e}}^{-3t}\left(\frac{{\mathrm{e}}^{-4t}{\mathrm{e}}^{6t}}{2}\right)\right)\end{array}$$

This is second order nonhomogeneous ODE. Let the solution be

$$x=x_h+x_p$$

Where $x_h$ is the solution to the homogeneous ODE $A{x}^{″}(t)+B{x}^{\prime }(t)+Cx(t)=0$, and $x_p$ is a particular solution to the nonhomogeneous ODE $A{x}^{″}(t)+B{x}^{\prime }(t)+Cx(t)=f(t)$. $x_h$ is the solution to

$$x^{\prime \prime }+4x^{\prime }+3x=0$$

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

$$x_h=c_1{\mathrm{e}}^{-3t}+\frac{c_2{\mathrm{e}}^{-t}}{2}$$

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

$$t\sin \left(t\right)$$

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

$$[\{t\cos \left(t\right),t\sin \left(t\right),\cos \left(t\right),\sin \left(t\right)\}]$$

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

$$\{\frac{{\mathrm{e}}^{-t}}{2},{\mathrm{e}}^{-3t}\}$$

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.

$$x_p=A_{1}t\cos \left(t\right)+A_{2}t\sin \left(t\right)+A_3\cos \left(t\right)+A_4\sin \left(t\right)$$

The unknowns $\{A_1,A_2,A_3,A_4\}$ are found by substituting the above trial solution $x_p$ into the ODE and comparing coefficients. Substituting the trial solution into the ODE and simplifying gives

$$-2A_1\sin \left(t\right)+2A_{1}t\cos \left(t\right)+2A_2\cos \left(t\right)+2A_{2}t\sin \left(t\right)+2A_3\cos \left(t\right)+2A_4\sin \left(t\right)+4A_1\cos \left(t\right)-4A_{1}t\sin \left(t\right)+4A_2\sin \left(t\right)+4A_{2}t\cos \left(t\right)-4A_3\sin \left(t\right)+4A_4\cos \left(t\right)=t\sin \left(t\right)$$

Solving for the unknowns by comparing coefficients results in

$$[A_1=-\frac{1}{5},A_2=\frac{1}{10},A_3=\frac{11}{50},A_4=\frac{1}{25}]$$

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

$$x_p=-\frac{t\cos \left(t\right)}{5}+\frac{t\sin \left(t\right)}{10}+\frac{11\cos \left(t\right)}{50}+\frac{\sin \left(t\right)}{25}$$

Therefore the general solution is

$$\begin{array}{rl}x& =x_h+x_p\\ & =(c_1{\mathrm{e}}^{-3t}+\frac{c_2{\mathrm{e}}^{-t}}{2})+(-\frac{t\cos \left(t\right)}{5}+\frac{t\sin \left(t\right)}{10}+\frac{11\cos \left(t\right)}{50}+\frac{\sin \left(t\right)}{25})\end{array}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}x& =-\frac{t\cos \left(t\right)}{5}+\frac{t\sin \left(t\right)}{10}+\frac{11\cos \left(t\right)}{50}+\frac{\sin \left(t\right)}{25}+\frac{{\mathrm{e}}^{-t}}{4}-\frac{47{\mathrm{e}}^{-3t}}{100}\end{array}$$

Solution plot	Slope field $x^{\prime \prime }+4x^{\prime }+3x=t\sin \left(t\right)$

✓ Maple. Time used: 0.035 (sec). Leaf size: 35

ode:=diff(diff(x(t),t),t)+4*diff(x(t),t)+3*x(t) = t*sin(t); 
ic:=x(0) = 0, D(x)(0) = 1; 
dsolve([ode,ic],x(t), singsol=all);
 

$$x=-\frac{47{\mathrm{e}}^{-3t}}{100}+\frac{{\mathrm{e}}^{-t}}{4}+\frac{(-10t+11)\cos \left(t\right)}{50}+\frac{\sin \left(t\right)(5t+2)}{50}$$

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)] 
-> Try solving first the homogeneous part of the ODE 
   checking if the LODE has constant coefficients 
   <- constant coefficients successful 
<- solving first the homogeneous part of the ODE successful
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & [\frac{d}{dt}\frac{d}{dt}x\left(t\right)+4\frac{d}{dt}x\left(t\right)+3x\left(t\right)=t\sin \left(t\right),x\left(0\right)=0,\left(\frac{d}{dt}x\left(t\right)\right)|{}_{\left\{t=0\right\}}=1]\\ \bullet & & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{d}{dt}\frac{d}{dt}x\left(t\right)\\ \bullet & & \text{Characteristic polynomial of homogeneous ODE}\\ & & r^2+4r+3=0\\ \bullet & & \text{Factor the characteristic polynomial}\\ & & (r+3)(r+1)=0\\ \bullet & & \text{Roots of the characteristic polynomial}\\ & & r=(-3,-1)\\ \bullet & & \text{1st solution of the homogeneous ODE}\\ & & x_1\left(t\right)={\mathrm{e}}^{-3t}\\ \bullet & & \text{2nd solution of the homogeneous ODE}\\ & & x_2\left(t\right)={\mathrm{e}}^{-t}\\ \bullet & & \text{General solution of the ODE}\\ & & x\left(t\right)=\mathit{C}\mathit{1}{x}_1\left(t\right)+\mathit{C}\mathit{2}{x}_2\left(t\right)+x_p\left(t\right)\\ \bullet & & \text{Substitute in solutions of the homogeneous ODE}\\ & & x\left(t\right)=\mathit{C}\mathit{1}{\mathrm{e}}^{-3t}+\mathit{C}\mathit{2}{\mathrm{e}}^{-t}+x_p\left(t\right)\\ ◻& & \text{Find a particular solution}{x}_p\left(t\right)\text{of the ODE}\\ & \circ & \text{Use variation of parameters to find}{x}_p\text{here}f\left(t\right)\text{is the forcing function}\\ & & [x_p\left(t\right)=-x_1\left(t\right)\int \frac{x_2\left(t\right)f\left(t\right)}{W(x_1\left(t\right),x_2\left(t\right))}dt+x_2\left(t\right)\int \frac{x_1\left(t\right)f\left(t\right)}{W(x_1\left(t\right),x_2\left(t\right))}dt,f\left(t\right)=t\sin \left(t\right)]\\ & \circ & \text{Wronskian of solutions of the homogeneous equation}\\ & & W(x_1\left(t\right),x_2\left(t\right))=\left[\begin{array}{cc}{\mathrm{e}}^{-3t}& {\mathrm{e}}^{-t}\\ -3{\mathrm{e}}^{-3t}& -{\mathrm{e}}^{-t}\end{array}\right]\\ & \circ & \text{Compute Wronskian}\\ & & W(x_1\left(t\right),x_2\left(t\right))=2{\mathrm{e}}^{-4t}\\ & \circ & \text{Substitute functions into equation for}{x}_p\left(t\right)\\ & & x_p\left(t\right)=\frac{(\int {\mathrm{e}}^{t}t\sin \left(t\right)dt{\mathrm{e}}^{2t}-\int t\sin \left(t\right){\mathrm{e}}^{3t}dt){\mathrm{e}}^{-3t}}{2}\\ & \circ & \text{Compute integrals}\\ & & x_p\left(t\right)=\frac{(-10t+11)\cos \left(t\right)}{50}+\frac{\sin \left(t\right)(5t+2)}{50}\\ \bullet & & \text{Substitute particular solution into general solution to ODE}\\ & & x\left(t\right)=\mathit{C}\mathit{1}{\mathrm{e}}^{-3t}+\mathit{C}\mathit{2}{\mathrm{e}}^{-t}+\frac{(-10t+11)\cos \left(t\right)}{50}+\frac{\sin \left(t\right)(5t+2)}{50}\\ ◻& & \text{Check validity of solution}x\left(t\right)=\mathit{\text{\_C1}}{\mathrm{e}}^{-3t}+\mathit{\text{\_C2}}{\mathrm{e}}^{-t}+\frac{(-10t+11)\cos \left(t\right)}{50}+\frac{\sin \left(t\right)(5t+2)}{50}\\ & \circ & \text{Use initial condition}x\left(0\right)=0\\ & & 0=\mathit{\text{\_C1}}+\mathit{\text{\_C2}}+\frac{11}{50}\\ & \circ & \text{Compute derivative of the solution}\\ & & \frac{d}{dt}x\left(t\right)=-3\mathit{\text{\_C1}}{\mathrm{e}}^{-3t}-\mathit{\text{\_C2}}{\mathrm{e}}^{-t}-\frac{\cos \left(t\right)}{5}-\frac{(-10t+11)\sin \left(t\right)}{50}+\frac{\cos \left(t\right)(5t+2)}{50}+\frac{\sin \left(t\right)}{10}\\ & \circ & \text{Use the initial condition}\left(\frac{d}{dt}x\left(t\right)\right)|{}_{\left\{t=0\right\}}=1\\ & & 1=-3\mathit{\text{\_C1}}-\mathit{\text{\_C2}}-\frac{4}{25}\\ & \circ & \text{Solve for}\mathit{\text{\_C1}}\text{and}\mathit{\text{\_C2}}\\ & & \{\mathit{\text{\_C1}}=-\frac{47}{100},\mathit{\text{\_C2}}=\frac{1}{4}\}\\ & \circ & \text{Substitute constant values into general solution and simplify}\\ & & x\left(t\right)=-\frac{47{\mathrm{e}}^{-3t}}{100}+\frac{{\mathrm{e}}^{-t}}{4}+\frac{(-10t+11)\cos \left(t\right)}{50}+\frac{\sin \left(t\right)(5t+2)}{50}\\ \bullet & & \text{Solution to the IVP}\\ & & x\left(t\right)=-\frac{47{\mathrm{e}}^{-3t}}{100}+\frac{{\mathrm{e}}^{-t}}{4}+\frac{(-10t+11)\cos \left(t\right)}{50}+\frac{\sin \left(t\right)(5t+2)}{50}\end{array}$$

✓ Mathematica. Time used: 0.022 (sec). Leaf size: 42

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

$$x(t)\to \frac{1}{100}(e^{-3t}(25e^{2t}-47)+2(5t+2)\sin (t)+(22-20t)\cos (t))$$

✓ Sympy. Time used: 0.290 (sec). Leaf size: 42

from sympy import * 
t = symbols("t") 
x = Function("x") 
ode = Eq(-t*sin(t) + 3*x(t) + 4*Derivative(x(t), t) + Derivative(x(t), (t, 2)),0) 
ics = {x(0): 0, Subs(Derivative(x(t), t), t, 0): 1} 
dsolve(ode,func=x(t),ics=ics)
 

$$x\left(t\right)=\frac{t\sin \left(t\right)}{10}-\frac{t\cos \left(t\right)}{5}+\frac{\sin \left(t\right)}{25}+\frac{11\cos \left(t\right)}{50}+\frac{e^{-t}}{4}-\frac{47e^{-3t}}{100}$$