2.1.11 Problem 11
Internal
problem
ID
[9082]
Book
:
Second
order
enumerated
odes
Section
:
section
1
Problem
number
:
11
Date
solved
:
Sunday, March 30, 2025 at 02:06:29 PM
CAS
classification
:
[[_2nd_order, _quadrature]]
Solved as second order missing x ode
Time used: 0.280 (sec)
Solve
\begin{align*} {y^{\prime \prime }}^{3}&=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 )^{3} \left (\frac {d}{d y}p \left (y \right )\right )^{3} = 0 \end{align*}
Which is now solved as first order ode for \(p(y)\).
Solving for the derivative gives these ODE’s to solve
\begin{align*}
\tag{1} p^{\prime }&=0 \\
\tag{2} p^{\prime }&=0 \\
\tag{3} p^{\prime }&=0 \\
\end{align*}
Now each of the above is solved separately.
Solving Eq. (1)
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_1 \\ p &= c_1 \end{align*}
Solving Eq. (2)
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_2 \\ p &= c_2 \end{align*}
Solving Eq. (3)
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*}
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 } = c_1 \end{align*}
Since the ode has the form \(y^{\prime }=f(x)\), then we only need to integrate \(f(x)\).
\begin{align*} \int {dy} &= \int {c_1\, dx}\\ y &= c_1 x + c_4 \end{align*}
Will add steps showing solving for IC soon.
Summary of solutions found
\begin{align*}
y &= c_1 x +c_4 \\
\end{align*}
Solved as second order missing y ode
Time used: 0.227 (sec)
Solve
\begin{align*} {y^{\prime \prime }}^{3}&=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 )}^{3} = 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 }&=0 \\
\tag{2} u^{\prime }&=0 \\
\tag{3} u^{\prime }&=0 \\
\end{align*}
Now each of the above is solved separately.
Solving Eq. (1)
Since the ode has the form \(u^{\prime }=f(x)\), then we only need to integrate \(f(x)\).
\begin{align*} \int {du} &= \int {0\, dx} + c_1 \\ u &= c_1 \end{align*}
Solving Eq. (2)
Since the ode has the form \(u^{\prime }=f(x)\), then we only need to integrate \(f(x)\).
\begin{align*} \int {du} &= \int {0\, dx} + c_2 \\ u &= c_2 \end{align*}
Solving Eq. (3)
Since the ode has the form \(u^{\prime }=f(x)\), then we only need to integrate \(f(x)\).
\begin{align*} \int {du} &= \int {0\, dx} + c_3 \\ u &= c_3 \end{align*}
In summary, these are the solution found for \(u(x)\)
\begin{align*}
u &= c_1 \\
\end{align*}
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{align*} y^{\prime }\left (x \right ) = c_1 \end{align*}
Since the ode has the form \(y^{\prime }=f(x)\), then we only need to integrate \(f(x)\).
\begin{align*} \int {dy} &= \int {c_1\, dx}\\ y &= c_1 x + c_4 \end{align*}
In summary, these are the solution found for \((y)\)
\begin{align*}
y &= c_1 x +c_4 \\
\end{align*}
Will add steps showing solving for IC soon.
Summary of solutions found
\begin{align*}
y &= c_1 x +c_4 \\
\end{align*}
✓ Maple. Time used: 0.009 (sec). Leaf size: 9
ode:=diff(diff(y(x),x),x)^3 = 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
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}\frac {d}{d x}y \left (x \right )\right )^{3}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}\frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}\frac {d}{d x}y \left (x \right )=0 \\ \bullet & {} & \textrm {Characteristic polynomial of ODE}\hspace {3pt} \\ {} & {} & r^{2}=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {0\pm \left (\sqrt {0}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =0 \\ \bullet & {} & \textrm {1st solution of the ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (x \right )=1 \\ \bullet & {} & \textrm {Repeated root, multiply}\hspace {3pt} y_{1}\left (x \right )\hspace {3pt}\textrm {by}\hspace {3pt} x \hspace {3pt}\textrm {to ensure linear independence}\hspace {3pt} \\ {} & {} & y_{2}\left (x \right )=x \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & y \left (x \right )=\mathit {C1} y_{1}\left (x \right )+\mathit {C2} y_{2}\left (x \right ) \\ \bullet & {} & \textrm {Substitute in solutions}\hspace {3pt} \\ {} & {} & y \left (x \right )=\mathit {C2} x +\mathit {C1} \end {array} \]
✓ Mathematica. Time used: 0.002 (sec). Leaf size: 12
ode=(D[y[x],{x,2}])^3==0;
ic={};
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
\[
y(x)\to c_2 x+c_1
\]
✓ Sympy. Time used: 0.036 (sec). Leaf size: 7
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq(Derivative(y(x), (x, 2))**3,0)
ics = {}
dsolve(ode,func=y(x),ics=ics)
\[
y{\left (x \right )} = C_{1} + C_{2} x
\]