2.1.2 Problem 2

Internal problem ID [9073]
Book : Second order enumerated odes
Section : section 1
Problem number : 2
Date solved : Friday, February 21, 2025 at 09:07:40 PM
CAS classification : [[_2nd_order, _quadrature]]

Solve

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

Solved as second order missing x ode

Time used: 0.194 (sec)

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 \(p(y)\).

Factoring the ode gives these factors

\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 \(p = 0\)

Solving equation (2)

Solving for the derivative gives these ODE’s to solve

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

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*}

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_3 \\ y &= c_3 \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_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*}

For solution (3) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is

\begin{align*} y^{\prime } = c_2 \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_2\, dx}\\ y &= c_2 x + c_5 \end{align*}

Will add steps showing solving for IC soon.

Summary of solutions found

\begin{align*} y &= c_3 \\ y &= c_1 x +c_4 \\ y &= c_2 x +c_5 \\ \end{align*}

Solved as second order missing y ode

Time used: 0.130 (sec)

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.

Solving for the derivative gives these ODE’s to solve

\begin{align*} \tag{1} u^{\prime }\left (x \right )&=0 \\ \tag{2} u^{\prime }\left (x \right )&=0 \\ \end{align*}

Now each of the above is solved separately.

Solving Eq. (1)

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_1 \\ u \left (x \right ) &= c_1 \end{align*}

Solving Eq. (2)

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_2 \\ u \left (x \right ) &= c_2 \end{align*}

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

\begin{align*} u \left (x \right ) &= c_1 \\ u \left (x \right ) &= c_2 \\ \end{align*}

For solution \(u \left (x \right ) = c_1\), since \(u=y^{\prime }\) 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_3 \end{align*}

For solution \(u \left (x \right ) = c_2\), since \(u=y^{\prime }\) then we now have a new first order ode to solve which is

\begin{align*} y^{\prime } = c_2 \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_2\, dx}\\ y &= c_2 x + c_4 \end{align*}

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

\begin{align*} y &= c_1 x +c_3 \\ y &= c_2 x +c_4 \\ \end{align*}

Will add steps showing solving for IC soon.

Summary of solutions found

\begin{align*} y &= c_1 x +c_3 \\ y &= c_2 x +c_4 \\ \end{align*}

Maple step by step solution

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

Maple trace

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

Maple dsolve solution

Solving time : 0.022 (sec)
Leaf size : 9

dsolve(diff(diff(y(x),x),x)^2 = 0,y(x),singsol=all)
 

\[ y = c_{1} x +c_{2} \]

✓Mathematica DSolve solution

Solving time : 0.003 (sec)
Leaf size : 12

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

\[ y(x)\to c_2 x+c_1 \]

✓Sympy solution

Solving time : 0.038 (sec)
Leaf size : 7

Python version: 3.13.1 (main, Dec  4 2024, 18:05:56) [GCC 14.2.1 20240910] 
Sympy version 1.13.3
 

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)
 

Eq(y(x), C1 + C2*x)
 

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