2.1.2 Problem 2
Internal
problem
ID
[10361]
Book
:
Second
order
enumerated
odes
Section
:
section
1
Problem
number
:
2
Date
solved
:
Thursday, November 27, 2025 at 10:35:15 AM
CAS
classification
:
[[_2nd_order, _quadrature]]
Solved as second order missing x ode
Time used: 0.461 (sec)
Solve
\begin{align*}
{y^{\prime \prime }}^{2}&=0 \\
\end{align*}
This 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 \(p(y)\).
Solved by factoring the differential equation
Time used: 0.052 (sec)
Solve
\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)
Solving for \(p\) from
\begin{align*} p^{2} = 0 \end{align*}
Solving gives
\begin{align*}
p &= 0 \\
\end{align*}
Solving equation (2)
Solve 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)
Solve Since 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*}
Solve Since 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*}
Solve Since 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*}
Summary of solutions found
\begin{align*}
y &= c_4 \\
y &= c_3 x +c_5 \\
\end{align*}
Solved as second order missing y ode
Time used: 0.206 (sec)
Solve
\begin{align*}
{y^{\prime \prime }}^{2}&=0 \\
\end{align*}
This 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.
Solve 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)
Solve Since 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*}
Solve Since 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*}
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*}
✓ 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
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )^{2}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right ) \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right )=0 \\ \bullet & {} & \textrm {Characteristic polynomial of ODE}\hspace {3pt} \\ {} & {} & r^{2}=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {0\pm \left (\sqrt {0}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =0 \\ \bullet & {} & \textrm {1st solution of the ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (x \right )=1 \\ \bullet & {} & \textrm {Repeated root, multiply}\hspace {3pt} y_{1}\left (x \right )\hspace {3pt}\textrm {by}\hspace {3pt} x \hspace {3pt}\textrm {to ensure linear independence}\hspace {3pt} \\ {} & {} & y_{2}\left (x \right )=x \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & y \left (x \right )=\mathit {C1} y_{1}\left (x \right )+\mathit {C2} y_{2}\left (x \right ) \\ \bullet & {} & \textrm {Substitute in solutions}\hspace {3pt} \\ {} & {} & y \left (x \right )=\mathit {C2} x +\mathit {C1} \end {array} \]
✓ 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*}
✓ 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)
\[
y{\left (x \right )} = C_{1} + C_{2} x
\]