2.8.2 Problem 2

Internal problem ID [18510]
Book : Elementary Differential Equations. By Thornton C. Fry. D Van Nostrand. NY. First Edition (1929)
Section : Chapter VII. Linear equations of order higher than the first. section 56. Problems at page 163
Problem number : 2
Date solved : Monday, March 31, 2025 at 05:40:50 PM
CAS classification : [[_2nd_order, _missing_x]]

Solved as second order linear constant coeff ode

Time used: 0.060 (sec)

Solve

$$\begin{array}{rl}{y}^{\prime \prime }+y& =0\end{array}$$

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

$$A{y}^{″}(x)+B{y}^{\prime }(x)+Cy(x)=0$$

Where in the above $A=1,B=0,C=1$. Let the solution be $y=e^{\lambda x}$. Substituting this into the ODE gives

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

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

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

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 $A=1,B=0,C=1$ into the above gives

$$\begin{array}{rl}{\lambda }_{1,2}& =\frac{0}{(2)\left(1\right)}\pm \frac{1}{(2)\left(1\right)}\sqrt{0^2-(4)\left(1\right)\left(1\right)}\\ & =\pm i\end{array}$$

Hence

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

Which simplifies to

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

Since roots are complex conjugate of each others, then let the roots be

$${\lambda }_{1,2}=\alpha \pm i\beta$$

Where $\alpha =0$ and $\beta =1$. Therefore the final solution, when using Euler relation, can be written as

$$y=e^{\alpha x}(c_1\cos (\beta x)+c_2\sin (\beta x))$$

Which becomes

$$y=e^0(c_1\cos \left(x\right)+c_2\sin \left(x\right))$$

Or

$$y=c_1\cos \left(x\right)+c_2\sin \left(x\right)$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& =c_1\cos \left(x\right)+c_2\sin \left(x\right)\end{array}$$

Figure 2.83: Slope field $y^{\prime \prime }+y=0$

Solved as second order can be made integrable

Time used: 1.052 (sec)

Solve

$$\begin{array}{rl}{y}^{\prime \prime }+y& =0\end{array}$$

Multiplying the ode by $y^{\prime }$ gives

$$y^{\prime }{y}^{\prime \prime }+y^{\prime }y=0$$

Integrating the above w.r.t $x$ gives

$$\begin{array}{rl}\int (y^{\prime }{y}^{\prime \prime }+y^{\prime }y)dx& =0\\ \frac{{y^{\prime }}^2}{2}+\frac{y^2}{2}& =c_1\end{array}$$

Which is now solved for $y$. Let $p=y^{\prime }$ the ode becomes

$$\begin{array}{r}\frac{p^2}{2}+\frac{y^2}{2}=c_1\end{array}$$

Solving for $y$ from the above results in

$$\begin{array}{}\text{(1)}& y& =\sqrt{-p^2+2c_1}\text{(2)}& y& =-\sqrt{-p^2+2c_1}\end{array}$$

This has the form

$$\begin{array}{}\text{(*)}& y=xf(p)+g(p)\end{array}$$

Where $f,g$ are functions of $p=y^{\prime }(x)$. Each of the above ode’s is dAlembert ode which is now solved.

Solving ode 1A

Taking derivative of (*) w.r.t. $x$ gives

$$\begin{array}{rl}p& =f+(x{f}^{\prime }+g^{\prime })\frac{dp}{dx}\\ \text{(2)}& p-f& =(x{f}^{\prime }+g^{\prime })\frac{dp}{dx}\end{array}$$

Comparing the form $y=xf+g$ to (1A) shows that

$$\begin{array}{rl}f& =0\\ g& =\sqrt{-p^2+2c_1}\end{array}$$

Hence (2) becomes

$$\begin{array}{}\text{(2A)}& p=-\frac{p{p}^{\prime }\left(x\right)}{\sqrt{-p^2+2c_1}}\end{array}$$

The singular solution is found by setting $\frac{dp}{dx}=0$ in the above which gives

$$\begin{array}{r}p=0\end{array}$$

Solving the above for $p$ results in

$$\begin{array}{rl}{p}_1& =0\end{array}$$

Substituting these in (1A) and keeping singular solution that verifies the ode gives

$$\begin{array}{r}y=\sqrt{2}\sqrt{c_1}\end{array}$$

The general solution is found when $\frac{\mathrm{d}p}{\mathrm{d}x}\ne 0$. From eq. (2A). This results in

$$\begin{array}{}\text{(3)}& p^{\prime }\left(x\right)=-\sqrt{-p{\left(x\right)}^2+2c_1}\end{array}$$

This ODE is now solved for $p\left(x\right)$. No inversion is needed.

Integrating gives

$$\begin{array}{rl}\int -\frac{1}{\sqrt{-p^2+2c_1}}dp& =dx\\ -\arctan \left(\frac{p}{\sqrt{-p^2+2c_1}}\right)& =x+c_2\end{array}$$

Singular solutions are found by solving

$$\begin{array}{rl}-\sqrt{-p^2+2c_1}& =0\end{array}$$

for $p\left(x\right)$. This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

$$\begin{array}{r}p\left(x\right)=\sqrt{2}\sqrt{c_1}\\ p\left(x\right)=-\sqrt{2}\sqrt{c_1}\end{array}$$

Substituing the above solution for $p$ in (2A) gives

$$\begin{array}{rl}y& =\sqrt{-\frac{2\tan {(x+c_2)}^2{c}_1}{\tan {(x+c_2)}^2+1}+2c_1}\\ y& =0\\ y& =0\end{array}$$

Solving ode 2A

Taking derivative of (*) w.r.t. $x$ gives

$$\begin{array}{rl}p& =f+(x{f}^{\prime }+g^{\prime })\frac{dp}{dx}\\ \text{(2)}& p-f& =(x{f}^{\prime }+g^{\prime })\frac{dp}{dx}\end{array}$$

Comparing the form $y=xf+g$ to (1A) shows that

$$\begin{array}{rl}f& =0\\ g& =-\sqrt{-p^2+2c_1}\end{array}$$

Hence (2) becomes

$$\begin{array}{}\text{(2A)}& p=\frac{p{p}^{\prime }\left(x\right)}{\sqrt{-p^2+2c_1}}\end{array}$$

The singular solution is found by setting $\frac{dp}{dx}=0$ in the above which gives

$$\begin{array}{r}p=0\end{array}$$

Solving the above for $p$ results in

$$\begin{array}{rl}{p}_1& =0\end{array}$$

Substituting these in (1A) and keeping singular solution that verifies the ode gives

$$\begin{array}{r}y=-\sqrt{2}\sqrt{c_1}\end{array}$$

The general solution is found when $\frac{\mathrm{d}p}{\mathrm{d}x}\ne 0$. From eq. (2A). This results in

$$\begin{array}{}\text{(3)}& p^{\prime }\left(x\right)=\sqrt{-p{\left(x\right)}^2+2c_1}\end{array}$$

This ODE is now solved for $p\left(x\right)$. No inversion is needed.

Integrating gives

$$\begin{array}{rl}\int \frac{1}{\sqrt{-p^2+2c_1}}dp& =dx\\ \arctan \left(\frac{p}{\sqrt{-p^2+2c_1}}\right)& =x+c_3\end{array}$$

Singular solutions are found by solving

$$\begin{array}{rl}\sqrt{-p^2+2c_1}& =0\end{array}$$

for $p\left(x\right)$. This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

$$\begin{array}{r}p\left(x\right)=\sqrt{2}\sqrt{c_1}\\ p\left(x\right)=-\sqrt{2}\sqrt{c_1}\end{array}$$

Substituing the above solution for $p$ in (2A) gives

$$\begin{array}{rl}y& =-\sqrt{-\frac{2\tan {(x+c_3)}^2{c}_1}{\tan {(x+c_3)}^2+1}+2c_1}\\ y& =0\\ y& =0\end{array}$$

Will add steps showing solving for IC soon.

The solution

$$y=\sqrt{2}\sqrt{c_1}$$

was found not to satisfy the ode or the IC. Hence it is removed. The solution

$$y=-\sqrt{2}\sqrt{c_1}$$

was found not to satisfy the ode or the IC. Hence it is removed.

Summary of solutions found

$$\begin{array}{rl}y& =0\\ y& =\sqrt{-\frac{2\tan {(x+c_2)}^2{c}_1}{\tan {(x+c_2)}^2+1}+2c_1}\\ y& =-\sqrt{-\frac{2\tan {(x+c_3)}^2{c}_1}{\tan {(x+c_3)}^2+1}+2c_1}\end{array}$$

Figure 2.84: Slope field $y^{\prime \prime }+y=0$

Solved as second order ode using Kovacic algorithm

Time used: 0.133 (sec)

Solve

$$\begin{array}{rl}{y}^{\prime \prime }+y& =0\end{array}$$

Writing the ode as

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

Comparing (1) and (2) shows that

$$\begin{array}{rl}A& =1\\ \text{(3)}& B& =0C& =1\end{array}$$

Applying the Liouville transformation on the dependent variable gives

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

Then (2) becomes

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

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

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

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

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

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 2.20: 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 $x$, then there is no need run Kovacic algorithm to obtain a solution for transformed ode $z^{″}=rz$ as one solution is

$$z_1(x)=\cos \left(x\right)$$

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

$$y_1=z_1{e}^{\int -\frac{1}{2}\frac{B}{A}dx}$$

Since $B=0$ then the above reduces to

$$\begin{array}{rl}{y}_1& =z_1\\ & =\cos \left(x\right)\end{array}$$

Which simplifies to

$$y_1=\cos \left(x\right)$$

The second solution $y_2$ to the original ode is found using reduction of order

$$y_2=y_1\int \frac{e^{\int -\frac{B}{A}dx}}{y_1^2}dx$$

Since $B=0$ then the above becomes

$$\begin{array}{rl}{y}_2& =y_1\int \frac{1}{y_1^2}dx\\ & =\cos \left(x\right)\int \frac{1}{\cos {\left(x\right)}^2}dx\\ & =\cos \left(x\right)(\tan \left(x\right))\end{array}$$

Therefore the solution is

$$\begin{array}{rl}y& =c_1{y}_1+c_2{y}_2\\ & =c_1(\cos \left(x\right))+c_2(\cos \left(x\right)(\tan \left(x\right)))\end{array}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& =c_1\cos \left(x\right)+c_2\sin \left(x\right)\end{array}$$

Figure 2.85: Slope field $y^{\prime \prime }+y=0$

✓ Maple. Time used: 0.001 (sec). Leaf size: 13

ode:=diff(diff(y(x),x),x)+y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 

$$y=c_1\sin \left(x\right)+c_2\cos \left(x\right)$$

Maple trace

Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
checking if the LODE has constant coefficients 
<- constant coefficients successful
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & \frac{d}{dx}\frac{d}{dx}y\left(x\right)+y\left(x\right)=0\\ \bullet & & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{d}{dx}\frac{d}{dx}y\left(x\right)\\ \bullet & & \text{Characteristic polynomial of ODE}\\ & & r^2+1=0\\ \bullet & & \text{Use quadratic formula to solve for}r\\ & & r=\frac{0\pm \left(\sqrt{-4}\right)}{2}\\ \bullet & & \text{Roots of the characteristic polynomial}\\ & & r=(-\mathrm{I},\mathrm{I})\\ \bullet & & \text{1st solution of the ODE}\\ & & y_1\left(x\right)=\cos \left(x\right)\\ \bullet & & \text{2nd solution of the ODE}\\ & & y_2\left(x\right)=\sin \left(x\right)\\ \bullet & & \text{General solution of the ODE}\\ & & y\left(x\right)=\mathit{C}\mathit{1}{y}_1\left(x\right)+\mathit{C}\mathit{2}{y}_2\left(x\right)\\ \bullet & & \text{Substitute in solutions}\\ & & y\left(x\right)=\mathit{C}\mathit{1}\cos \left(x\right)+\mathit{C}\mathit{2}\sin \left(x\right)\end{array}$$

✓ Mathematica. Time used: 0.01 (sec). Leaf size: 16

ode=D[y[x],{x,2}]+y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

$$y(x)\to c_1\cos (x)+c_2\sin (x)$$

✓ Sympy. Time used: 0.040 (sec). Leaf size: 12

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

$$y\left(x\right)=C_1\sin \left(x\right)+C_2\cos \left(x\right)$$