2.1.2 Problem 2

Internal problem ID [10361]
Book : Second order enumerated odes
Section : section 1
Problem number : 2
Date solved : Monday, December 29, 2025 at 10:52:20 PM
CAS classification : [[_2nd_order, _quadrature]]

2.1.2.1 second order ode missing x

0.631 (sec)

\begin{align*} {y^{\prime \prime }}^{2}&=0 \\ \end{align*}

Entering second order ode missing \(x\) solverThis is missing independent variable second order ode. Solved by reduction of order by using substitution which makes the dependent variable \(y\) an independent variable. Using

\begin{align*} y' &= p \end{align*}

Then

\begin{align*} y'' &= \frac {dp}{dx}\\ &= \frac {dp}{dy}\frac {dy}{dx}\\ &= p \frac {dp}{dy} \end{align*}

Hence the ode becomes

\begin{align*} p \left (y \right )^{2} \left (\frac {d}{d y}p \left (y \right )\right )^{2} = 0 \end{align*}

Which is now solved as first order ode for \(p(y)\).

2.1.2.2 Solved by factoring the differential equation

Time used: 0.076 (sec)

\begin{align*} p^{2} {p^{\prime }}^{2}&=0 \\ \end{align*}

Writing the ode as

\begin{align*} \left (p^{2}\right )\left ({p^{\prime }}^{2}\right )&=0 \end{align*}

Therefore we need to solve the following equations

\begin{align*} \tag{1} p^{2} &= 0 \\ \tag{2} {p^{\prime }}^{2} &= 0 \\ \end{align*}

Now each of the above equations is solved in turn.

Solving equation (1)

Entering zero order ode solverSolving for \(p\) from

\begin{align*} p^{2} = 0 \end{align*}

Solving gives

\begin{align*} p &= 0 \\ \end{align*}

Solving equation (2)

Solving for the derivative gives these ODE’s to solve

\begin{align*} \tag{1} p^{\prime }&=0 \\ \end{align*}

Now each of the above is solved separately.

Solving Eq. (1)

Entering first order ode quadrature solverSince the ode has the form \(p^{\prime }=f(y)\), then we only need to integrate \(f(y)\).

\begin{align*} \int {dp} &= \int {0\, dy} + c_3 \\ p &= c_3 \end{align*}

Summary of solutions found

\begin{align*} p &= 0 \\ p &= c_3 \\ \end{align*}

For solution (1) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is

\begin{align*} y^{\prime } = 0 \end{align*}

Entering first order ode quadrature solverSince the ode has the form \(y^{\prime }=f(x)\), then we only need to integrate \(f(x)\).

\begin{align*} \int {dy} &= \int {0\, dx} + c_4 \\ y &= c_4 \end{align*}

For solution (2) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is

\begin{align*} y^{\prime } = c_3 \end{align*}

Entering first order ode quadrature solverSince the ode has the form \(y^{\prime }=f(x)\), then we only need to integrate \(f(x)\).

\begin{align*} \int {dy} &= \int {c_3\, dx}\\ y &= c_3 x + c_5 \end{align*}

\begin{align*} y&= c_3 x +c_5 \end{align*}

Summary of solutions found

\begin{align*} y &= c_4 \\ y &= c_3 x +c_5 \\ \end{align*}

2.1.2.3 second order ode missing y

0.473 (sec)

\begin{align*} {y^{\prime \prime }}^{2}&=0 \\ \end{align*}

Entering second order ode missing \(y\) solverThis is second order ode with missing dependent variable \(y\). Let

\begin{align*} u(x) &= y^{\prime } \end{align*}

Then

\begin{align*} u'(x) &= y^{\prime \prime } \end{align*}

Hence the ode becomes

\begin{align*} {u^{\prime }\left (x \right )}^{2} = 0 \end{align*}

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

Solving for the derivative gives these ODE’s to solve

\begin{align*} \tag{1} u^{\prime }\left (x \right )&=0 \\ \end{align*}

Now each of the above is solved separately.

Solving Eq. (1)

Entering first order ode quadrature solverSince the ode has the form \(u^{\prime }\left (x \right )=f(x)\), then we only need to integrate \(f(x)\).

\begin{align*} \int {du} &= \int {0\, dx} + c_3 \\ u \left (x \right ) &= c_3 \end{align*}

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

\begin{align*} u \left (x \right ) &= c_3 \\ \end{align*}

For solution \(u \left (x \right ) = c_3\), since \(u=y^{\prime }\) then we now have a new first order ode to solve which is

\begin{align*} y^{\prime } = c_3 \end{align*}

Entering first order ode quadrature solverSince the ode has the form \(y^{\prime }=f(x)\), then we only need to integrate \(f(x)\).

\begin{align*} \int {dy} &= \int {c_3\, dx}\\ y &= c_3 x + c_4 \end{align*}

\begin{align*} y&= c_3 x +c_4 \end{align*}

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

\begin{align*} y &= c_3 x +c_4 \\ \end{align*}

Summary of solutions found

\begin{align*} y &= c_3 x +c_4 \\ \end{align*}

2.1.2.4 ✓ Maple. Time used: 0.004 (sec). Leaf size: 9

ode:=diff(diff(y(x),x),x)^2 = 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

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )^{2}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right ) \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right )=0 \\ \bullet & {} & \textrm {Characteristic polynomial of ODE}\hspace {3pt} \\ {} & {} & r^{2}=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {0\pm \left (\sqrt {0}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =0 \\ \bullet & {} & \textrm {1st solution of the ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (x \right )=1 \\ \bullet & {} & \textrm {Repeated root, multiply}\hspace {3pt} y_{1}\left (x \right )\hspace {3pt}\textrm {by}\hspace {3pt} x \hspace {3pt}\textrm {to ensure linear independence}\hspace {3pt} \\ {} & {} & y_{2}\left (x \right )=x \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & y \left (x \right )=\mathit {C1} y_{1}\left (x \right )+\mathit {C2} y_{2}\left (x \right ) \\ \bullet & {} & \textrm {Substitute in solutions}\hspace {3pt} \\ {} & {} & y \left (x \right )=\mathit {C2} x +\mathit {C1} \end {array} \]

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

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

\begin{align*} y(x)&\to c_2 x+c_1 \end{align*}

2.1.2.6 ✓ Sympy. Time used: 0.015 (sec). Leaf size: 7

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

\[ y{\left (x \right )} = C_{1} + C_{2} x \]