2.1.8 Problem 8

Internal problem ID [9079]
Book : Second order enumerated odes
Section : section 1
Problem number : 8
Date solved : Sunday, March 30, 2025 at 02:06:24 PM
CAS classification : [[_2nd_order, _quadrature]]

Solved as second order missing x ode

Time used: 0.665 (sec)

Solve

\begin{align*} {y^{\prime \prime }}^{2}&=1 \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} = 1 \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 }&=\frac {1}{p} \\ \tag{2} p^{\prime }&=-\frac {1}{p} \\ \end{align*}

Now each of the above is solved separately.

Solving Eq. (1)

Integrating gives

\begin{align*} \int p d p &= dy\\ \frac {p^{2}}{2}&= y +c_1 \end{align*}

Solving Eq. (2)

Integrating gives

\begin{align*} \int -p d p &= dy\\ -\frac {p^{2}}{2}&= y +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*} \frac {{y^{\prime }}^{2}}{2} = y+c_1 \end{align*}

Let \(p=y^{\prime }\) the ode becomes

\begin{align*} \frac {p^{2}}{2} = y +c_1 \end{align*}

Solving for \(y\) from the above results in

\begin{align*} \tag{1} y &= -c_1 +\frac {p^{2}}{2} \\ \end{align*}

This has the form

\begin{align*} y=x f(p)+g(p)\tag {*} \end{align*}

Where \(f,g\) are functions of \(p=y'(x)\). The above ode is dAlembert ode which is now solved.

Taking derivative of (*) w.r.t. \(x\) gives

\begin{align*} p &= f+(x f'+g') \frac {dp}{dx}\\ p-f &= (x f'+g') \frac {dp}{dx}\tag {2} \end{align*}

Comparing the form \(y=x f + g\) to (1A) shows that

\begin{align*} f &= 0\\ g &= -c_1 +\frac {p^{2}}{2} \end{align*}

Hence (2) becomes

\begin{equation} \tag{2A} p = p p^{\prime }\left (x \right ) \end{equation}

The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives

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

Solving the above for \(p\) results in

\begin{align*} p_{1} &=0 \end{align*}

Substituting these in (1A) and keeping singular solution that verifies the ode gives

\begin{align*} y = -c_1 \end{align*}

The general solution is found when \( \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}x}}\neq 0\). From eq. (2A). This results in

\begin{equation} \tag{3} p^{\prime }\left (x \right ) = 1 \end{equation}

This ODE is now solved for \(p \left (x \right )\). No inversion is needed.

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

\begin{align*} \int {dp} &= \int {1\, dx}\\ p \left (x \right ) &= x + c_3 \end{align*}

Substituing the above solution for \(p\) in (2A) gives

\begin{align*} y &= -c_1 +\frac {1}{2} c_3^{2}+c_3 x +\frac {1}{2} x^{2} \\ \end{align*}

Will add steps showing solving for IC soon.

The solution

\[ y = -c_1 \]

was found not to satisfy the ode or the IC. Hence it is removed.

Summary of solutions found

\begin{align*} y &= -c_1 +\frac {1}{2} c_3^{2}+c_3 x +\frac {1}{2} x^{2} \\ \end{align*}

Solved as second order missing y ode

Time used: 0.252 (sec)

Solve

\begin{align*} {y^{\prime \prime }}^{2}&=1 \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}-1 = 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 }&=1 \\ \tag{2} u^{\prime }&=-1 \\ \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 {1\, dx}\\ u &= x + 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 {-1\, dx}\\ u &= -x + c_2 \end{align*}

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

\begin{align*} u &= x +c_1 \\ \end{align*}

For solution \(u = x +c_1\), 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 ) = x +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 {x +c_1\, dx}\\ y &= \frac {1}{2} x^{2}+c_1 x + c_3 \end{align*}

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

\begin{align*} y &= \frac {1}{2} x^{2}+c_1 x +c_3 \\ \end{align*}

Will add steps showing solving for IC soon.

Summary of solutions found

\begin{align*} y &= \frac {1}{2} x^{2}+c_1 x +c_3 \\ \end{align*}

✓ Maple. Time used: 0.002 (sec). Leaf size: 27

ode:=diff(diff(y(x),x),x)^2 = 1; 
dsolve(ode,y(x), singsol=all);
 

\begin{align*} y &= \frac {1}{2} x^{2}+c_1 x +c_2 \\ y &= -\frac {1}{2} x^{2}+c_1 x +c_2 \\ \end{align*}

Maple trace

Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
<- quadrature successful 
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 )^{2}=1 \\ \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 {Make substitution}\hspace {3pt} u =\frac {d}{d x}y \left (x \right )\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}u \left (x \right )\right )^{2}=1 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d x}u \left (x \right )=-1, \frac {d}{d x}u \left (x \right )=1\right ] \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} \frac {d}{d x}u \left (x \right )=-1 \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \left (\frac {d}{d x}u \left (x \right )\right )d x =\int \left (-1\right )d x +\textit {\_C1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & u \left (x \right )=-x +\textit {\_C1} \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} \frac {d}{d x}u \left (x \right )=1 \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \left (\frac {d}{d x}u \left (x \right )\right )d x =\int 1d x +\textit {\_C1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & u \left (x \right )=x +\textit {\_C1} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{u \left (x \right )=-x +\mathit {C1} , u \left (x \right )=x +\mathit {C1} \right \} \\ \bullet & {} & \textrm {Define new dependent variable}\hspace {3pt} u \\ {} & {} & u \left (x \right )=\frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Compute}\hspace {3pt} \frac {d}{d x}\frac {d}{d x}y \left (x \right ) \\ {} & {} & \frac {d}{d x}u \left (x \right )=\frac {d^{2}}{d x^{2}}y \left (x \right ) \\ \bullet & {} & \textrm {Use chain rule on the lhs}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}y \left (x \right )\right ) \left (\frac {d}{d y}u \left (y \right )\right )=\frac {d^{2}}{d x^{2}}y \left (x \right ) \\ \bullet & {} & \textrm {Substitute in the definition of}\hspace {3pt} u \\ {} & {} & u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )=\frac {d^{2}}{d x^{2}}y \left (x \right ) \\ \bullet & {} & \textrm {Make substitutions}\hspace {3pt} \frac {d}{d x}y \left (x \right )=u \left (y \right ),\frac {d}{d x}\frac {d}{d x}y \left (x \right )=u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & u \left (y \right )^{2} \left (\frac {d}{d y}u \left (y \right )\right )^{2}=1 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d y}u \left (y \right )=\frac {1}{u \left (y \right )}, \frac {d}{d y}u \left (y \right )=-\frac {1}{u \left (y \right )}\right ] \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} \frac {d}{d y}u \left (y \right )=\frac {1}{u \left (y \right )} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )=1 \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} y \\ {} & {} & \int u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )d y =\int 1d y +\textit {\_C1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {u \left (y \right )^{2}}{2}=y +\textit {\_C1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} u \left (y \right ) \\ {} & {} & \left \{u \left (y \right )=\sqrt {2 \textit {\_C1} +2 y}, u \left (y \right )=-\sqrt {2 \textit {\_C1} +2 y}\right \} \\ {} & \circ & \textrm {Redefine the integration constant(s)}\hspace {3pt} \\ {} & {} & \left \{u \left (y \right )=\sqrt {\textit {\_C1} +2 y}, u \left (y \right )=-\sqrt {\textit {\_C1} +2 y}\right \} \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} \frac {d}{d y}u \left (y \right )=-\frac {1}{u \left (y \right )} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )=-1 \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} y \\ {} & {} & \int u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )d y =\int \left (-1\right )d y +\textit {\_C1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {u \left (y \right )^{2}}{2}=-y +\textit {\_C1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} u \left (y \right ) \\ {} & {} & \left \{u \left (y \right )=\sqrt {2 \textit {\_C1} -2 y}, u \left (y \right )=-\sqrt {2 \textit {\_C1} -2 y}\right \} \\ {} & \circ & \textrm {Redefine the integration constant(s)}\hspace {3pt} \\ {} & {} & \left \{u \left (y \right )=\sqrt {\textit {\_C1} -2 y}, u \left (y \right )=-\sqrt {\textit {\_C1} -2 y}\right \} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{u \left (y \right )=\sqrt {\mathit {C1} -2 y}, u \left (y \right )=\sqrt {\mathit {C1} +2 y}, u \left (y \right )=-\sqrt {\mathit {C1} -2 y}, u \left (y \right )=-\sqrt {\mathit {C1} +2 y}\right \} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=\sqrt {\mathit {C1} -2 y} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=\frac {d}{d x}y \left (x \right ),y =y \left (x \right ) \\ {} & {} & \frac {d}{d x}y \left (x \right )=\sqrt {\mathit {C1} -2 y \left (x \right )} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\sqrt {\mathit {C1} -2 y \left (x \right )} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d x}y \left (x \right )}{\sqrt {\mathit {C1} -2 y \left (x \right )}}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {\frac {d}{d x}y \left (x \right )}{\sqrt {\mathit {C1} -2 y \left (x \right )}}d x =\int 1d x +\mathit {C2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\sqrt {\mathit {C1} -2 y \left (x \right )}=x +\mathit {C2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \left (x \right ) \\ {} & {} & y \left (x \right )=-\frac {1}{2} \mathit {C2}^{2}-\mathit {C2} x -\frac {1}{2} x^{2}+\frac {1}{2} \mathit {C1} \\ \bullet & {} & \textrm {Redefine the integration constant(s)}\hspace {3pt} \\ {} & {} & y \left (x \right )=\mathit {C2} x -\frac {1}{2} x^{2}+\mathit {C1} \\ \bullet & {} & \textrm {Solve 2nd ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=\sqrt {\mathit {C1} +2 y} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=\frac {d}{d x}y \left (x \right ),y =y \left (x \right ) \\ {} & {} & \frac {d}{d x}y \left (x \right )=\sqrt {\mathit {C1} +2 y \left (x \right )} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\sqrt {\mathit {C1} +2 y \left (x \right )} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d x}y \left (x \right )}{\sqrt {\mathit {C1} +2 y \left (x \right )}}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {\frac {d}{d x}y \left (x \right )}{\sqrt {\mathit {C1} +2 y \left (x \right )}}d x =\int 1d x +\mathit {C2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \sqrt {\mathit {C1} +2 y \left (x \right )}=x +\mathit {C2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \left (x \right ) \\ {} & {} & y \left (x \right )=\frac {1}{2} \mathit {C2}^{2}+\mathit {C2} x +\frac {1}{2} x^{2}-\frac {1}{2} \mathit {C1} \\ \bullet & {} & \textrm {Redefine the integration constant(s)}\hspace {3pt} \\ {} & {} & y \left (x \right )=\mathit {C2} x +\frac {1}{2} x^{2}+\mathit {C1} \\ \bullet & {} & \textrm {Solve 3rd ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=-\sqrt {\mathit {C1} -2 y} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=\frac {d}{d x}y \left (x \right ),y =y \left (x \right ) \\ {} & {} & \frac {d}{d x}y \left (x \right )=-\sqrt {\mathit {C1} -2 y \left (x \right )} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=-\sqrt {\mathit {C1} -2 y \left (x \right )} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d x}y \left (x \right )}{\sqrt {\mathit {C1} -2 y \left (x \right )}}=-1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {\frac {d}{d x}y \left (x \right )}{\sqrt {\mathit {C1} -2 y \left (x \right )}}d x =\int \left (-1\right )d x +\mathit {C2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\sqrt {\mathit {C1} -2 y \left (x \right )}=-x +\mathit {C2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \left (x \right ) \\ {} & {} & y \left (x \right )=-\frac {1}{2} \mathit {C2}^{2}+\mathit {C2} x -\frac {1}{2} x^{2}+\frac {1}{2} \mathit {C1} \\ \bullet & {} & \textrm {Redefine the integration constant(s)}\hspace {3pt} \\ {} & {} & y \left (x \right )=\mathit {C2} x -\frac {1}{2} x^{2}+\mathit {C1} \\ \bullet & {} & \textrm {Solve 4th ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=-\sqrt {\mathit {C1} +2 y} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=\frac {d}{d x}y \left (x \right ),y =y \left (x \right ) \\ {} & {} & \frac {d}{d x}y \left (x \right )=-\sqrt {\mathit {C1} +2 y \left (x \right )} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=-\sqrt {\mathit {C1} +2 y \left (x \right )} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d x}y \left (x \right )}{\sqrt {\mathit {C1} +2 y \left (x \right )}}=-1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {\frac {d}{d x}y \left (x \right )}{\sqrt {\mathit {C1} +2 y \left (x \right )}}d x =\int \left (-1\right )d x +\mathit {C2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \sqrt {\mathit {C1} +2 y \left (x \right )}=-x +\mathit {C2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \left (x \right ) \\ {} & {} & y \left (x \right )=\frac {1}{2} \mathit {C2}^{2}-\mathit {C2} x +\frac {1}{2} x^{2}-\frac {1}{2} \mathit {C1} \\ \bullet & {} & \textrm {Redefine the integration constant(s)}\hspace {3pt} \\ {} & {} & y \left (x \right )=\mathit {C2} x +\frac {1}{2} x^{2}+\mathit {C1} \end {array} \]

✓ Mathematica. Time used: 0.003 (sec). Leaf size: 37

ode=(D[y[x],{x,2}])^2==1; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

\begin{align*} y(x)\to -\frac {x^2}{2}+c_2 x+c_1 \\ y(x)\to \frac {x^2}{2}+c_2 x+c_1 \\ \end{align*}

✓ Sympy. Time used: 0.151 (sec). Leaf size: 24

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(Derivative(y(x), (x, 2))**2 - 1,0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

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