2.1.1 Problem 1

Internal problem ID [9072]
Book : Second order enumerated odes
Section : section 1
Problem number : 1
Date solved : Sunday, March 30, 2025 at 02:06:13 PM
CAS classification : [[_2nd_order, _quadrature]]

Solved as second order ode quadrature

Time used: 0.030 (sec)

Solve

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

Integrating twice gives the solution

$$y=c_{1}x+c_2$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& =c_{1}x+c_2\end{array}$$

Figure 2.1: Slope field $y^{\prime \prime }=0$

Solved as second order linear constant coeff ode

Time used: 0.024 (sec)

Solve

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

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

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

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

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

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

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

Hence this is the case of a double root ${\lambda }_{1,2}=0$. Therefore the solution is

$$\begin{array}{}\text{(1)}& y=c_{1}1+c_{2}x\end{array}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& =c_{2}x+c_1\end{array}$$

Figure 2.2: Slope field $y^{\prime \prime }=0$

Solved as second order linear exact ode

Time used: 0.088 (sec)

Solve

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

An ode of the form

$$\begin{array}{rl}p\left(x\right)y^{\prime \prime }+q\left(x\right)y^{\prime }+r\left(x\right)y& =s\left(x\right)\end{array}$$

is exact if

$$\begin{array}{}\text{(1)}& p^{″}(x)-q^{\prime }(x)+r(x)& =0\end{array}$$

For the given ode we have

$$\begin{array}{rl}p(x)& =1\\ q(x)& =0\\ r(x)& =0\\ s(x)& =0\end{array}$$

Hence

$$\begin{array}{rl}{p}^{″}(x)& =0\\ q^{\prime }(x)& =0\end{array}$$

Therefore (1) becomes

$$\begin{array}{rl}0-\left(0\right)+\left(0\right)& =0\end{array}$$

Hence the ode is exact. Since we now know the ode is exact, it can be written as

$$\begin{array}{rl}{(p\left(x\right)y^{\prime }+(q\left(x\right)-p^{\prime }\left(x\right))y)}^{\prime }& =s(x)\end{array}$$

Integrating gives

$$\begin{array}{rl}p\left(x\right)y^{\prime }+(q\left(x\right)-p^{\prime }\left(x\right))y& =\int s\left(x\right)dx\end{array}$$

Substituting the above values for $p,q,r,s$ gives

$$\begin{array}{rl}{y}^{\prime }& =c_1\end{array}$$

We now have a first order ode to solve which is

$$\begin{array}{r}{y}^{\prime }=c_1\end{array}$$

Since the ode has the form $y^{\prime }=f(x)$, then we only need to integrate $f(x)$.

$$\begin{array}{rl}\int dy& =\int c_{1}dx\\ y& =c_{1}x+c_2\end{array}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& =c_{1}x+c_2\end{array}$$

Figure 2.3: Slope field $y^{\prime \prime }=0$

Solved as second order missing y ode

Time used: 0.069 (sec)

Solve

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

This is second order ode with missing dependent variable $y$. Let

$$\begin{array}{rl}u(x)& =y^{\prime }\end{array}$$

Then

$$\begin{array}{rl}{u}^{\prime }(x)& =y^{\prime \prime }\end{array}$$

Hence the ode becomes

$$\begin{array}{r}{u}^{\prime }\left(x\right)=0\end{array}$$

Which is now solved for $u(x)$ as first order ode.

Since the ode has the form $u^{\prime }=f(x)$, then we only need to integrate $f(x)$.

$$\begin{array}{rl}\int du& =\int 0dx+c_1\\ u& =c_1\end{array}$$

In summary, these are the solution found for $u(x)$

$$\begin{array}{rl}u& =c_1\end{array}$$

For solution $u=c_1$, since $u=y^{\prime }\left(x\right)$ then we now have a new first order ode to solve which is

$$\begin{array}{r}{y}^{\prime }\left(x\right)=c_1\end{array}$$

Since the ode has the form $y^{\prime }=f(x)$, then we only need to integrate $f(x)$.

$$\begin{array}{rl}\int dy& =\int c_{1}dx\\ y& =c_{1}x+c_2\end{array}$$

In summary, these are the solution found for $(y)$

$$\begin{array}{rl}y& =c_{1}x+c_2\end{array}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& =c_{1}x+c_2\end{array}$$

Figure 2.4: Slope field $y^{\prime \prime }=0$

Solved as second order integrable as is ode

Time used: 0.099 (sec)

Solve

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

Integrating both sides of the ODE w.r.t $x$ gives

$$\begin{array}{rl}\int y^{\prime \prime }dx& =0\\ y^{\prime }=c_1\end{array}$$

Which is now solved for $y$. Since the ode has the form $y^{\prime }=f(x)$, then we only need to integrate $f(x)$.

$$\begin{array}{rl}\int dy& =\int c_{1}dx\\ y& =c_{1}x+c_2\end{array}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& =c_{1}x+c_2\end{array}$$

Figure 2.5: Slope field $y^{\prime \prime }=0$

Solved as second order integrable as is ode (ABC method)

Time used: 0.040 (sec)

Solve

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

Writing the ode as

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

Integrating both sides of the ODE w.r.t $x$ gives

$$\begin{array}{rl}\int y^{\prime \prime }dx& =0\\ y^{\prime }=c_1\end{array}$$

Which is now solved for $y$. Since the ode has the form $y^{\prime }=f(x)$, then we only need to integrate $f(x)$.

$$\begin{array}{rl}\int dy& =\int c_{1}dx\\ y& =c_{1}x+c_2\end{array}$$

Since the ode has the form $y^{\prime }=f(x)$, then we only need to integrate $f(x)$.

$$\begin{array}{rl}\int dy& =\int c_{1}dx\\ y& =c_{1}x+c_2\end{array}$$

Will add steps showing solving for IC soon.

Figure 2.6: Slope field $y^{\prime \prime }=0$

Solved as second order ode using Kovacic algorithm

Time used: 0.042 (sec)

Solve

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

Writing the ode as

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

Comparing (1) and (2) shows that

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

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

$$\begin{array}{}\text{(6)}& r& =\frac{0}{1}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =0\\ t& =1\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(x)& =0\end{array}$$

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 2.1: Necessary conditions for each Kovacic case

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

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

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 $infinity$ then the necessary conditions for case one are met. Therefore

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

Since $r=0$ 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

$$z_1(x)=1$$

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

$$\begin{array}{rl}{y}_1& =z_1\\ & =1\end{array}$$

Which simplifies to

$$y_1=1$$

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

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

Therefore the solution is

$$\begin{array}{rl}y& =c_1{y}_1+c_2{y}_2\\ & =c_1\left(1\right)+c_2\left(1\left(x\right)\right)\end{array}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& =c_{2}x+c_1\end{array}$$

Figure 2.7: Slope field $y^{\prime \prime }=0$

✓ Maple. Time used: 0.001 (sec). Leaf size: 9

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

$$y=c_{1}x+c_2$$

Maple trace

Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
<- quadrature successful
 

Maple step by step

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

✓ Mathematica. Time used: 0.002 (sec). Leaf size: 12

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

$$y(x)\to c_{2}x+c_1$$

✓ Sympy. Time used: 0.033 (sec). Leaf size: 7

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

$$y\left(x\right)=C_1+C_{2}x$$