2.1.8 Problem 8

Internal problem ID [9079]
Book : Second order enumerated odes
Section : section 1
Problem number : 8
Date solved : Friday, February 21, 2025 at 09:07:47 PM
CAS classification : [[_2nd_order, _quadrature]]

Solve

\begin{align*} {y^{\prime \prime }}^{2}&=1 \end{align*}

Factoring the ode gives these factors

\begin{align*} \tag{1} y^{\prime \prime }-1 &= 0 \\ \tag{2} y^{\prime \prime }+1 &= 0 \\ \end{align*}

Now each of the above equations is solved in turn.

Solving equation (1)

Solved as second order missing x ode

Time used: 0.580 (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} = 1 \end{align*}

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

Factoring the ode gives these factors

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

Now each of the above equations is solved in turn.

Solving equation (1)

Integrating gives

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

Solving for \(p\) gives

\begin{align*} p &= \sqrt {2 c_1 +2 y} \\ p &= -\sqrt {2 c_1 +2 y} \\ \end{align*}

Solving equation (2)

Integrating gives

\begin{align*} \int -p d p &= dy\\ -\frac {p^{2}}{2}&= y +c_2 \end{align*}

Solving for \(p\) gives

\begin{align*} p &= \sqrt {-2 c_2 -2 y} \\ p &= -\sqrt {-2 c_2 -2 y} \\ \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 } = \sqrt {2 c_1 +2 y} \end{align*}

Integrating gives

\begin{align*} \int \frac {1}{\sqrt {2 c_1 +2 y}}d y &= dx\\ \sqrt {2 c_1 +2 y}&= x +c_3 \end{align*}

Singular solutions are found by solving

\begin{align*} \sqrt {2 c_1 +2 y}&= 0 \end{align*}

for \(y\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

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

Solving for \(y\) gives

\begin{align*} y &= -c_1 \\ y &= \frac {1}{2} c_3^{2}+c_3 x +\frac {1}{2} x^{2}-c_1 \\ \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 } = \sqrt {-2 c_2 -2 y} \end{align*}

Integrating gives

\begin{align*} \int \frac {1}{\sqrt {-2 c_2 -2 y}}d y &= dx\\ -\sqrt {-2 c_2 -2 y}&= x +c_4 \end{align*}

Singular solutions are found by solving

\begin{align*} \sqrt {-2 c_2 -2 y}&= 0 \end{align*}

for \(y\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

\begin{align*} y = -c_2 \end{align*}

Solving for \(y\) gives

\begin{align*} y &= -c_2 \\ y &= -\frac {1}{2} c_4^{2}-c_4 x -\frac {1}{2} x^{2}-c_2 \\ \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 } = -\sqrt {2 c_1 +2 y} \end{align*}

Integrating gives

\begin{align*} \int -\frac {1}{\sqrt {2 c_1 +2 y}}d y &= dx\\ -\sqrt {2 c_1 +2 y}&= x +c_5 \end{align*}

Singular solutions are found by solving

\begin{align*} -\sqrt {2 c_1 +2 y}&= 0 \end{align*}

for \(y\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

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

Solving for \(y\) gives

\begin{align*} y &= -c_1 \\ y &= \frac {1}{2} c_5^{2}+c_5 x +\frac {1}{2} x^{2}-c_1 \\ \end{align*}

For solution (4) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is

\begin{align*} y^{\prime } = -\sqrt {-2 c_2 -2 y} \end{align*}

Integrating gives

\begin{align*} \int -\frac {1}{\sqrt {-2 c_2 -2 y}}d y &= dx\\ \sqrt {-2 c_2 -2 y}&= x +c_6 \end{align*}

Singular solutions are found by solving

\begin{align*} -\sqrt {-2 c_2 -2 y}&= 0 \end{align*}

for \(y\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

\begin{align*} y = -c_2 \end{align*}

Solving for \(y\) gives

\begin{align*} y &= -c_2 \\ y &= -\frac {1}{2} c_6^{2}-c_6 x -\frac {1}{2} x^{2}-c_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. The solution

\[ y = -c_2 \]

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

Summary of solutions found

\begin{align*} y &= \frac {1}{2} c_3^{2}+c_3 x +\frac {1}{2} x^{2}-c_1 \\ y &= -\frac {1}{2} c_4^{2}-c_4 x -\frac {1}{2} x^{2}-c_2 \\ y &= \frac {1}{2} c_5^{2}+c_5 x +\frac {1}{2} x^{2}-c_1 \\ y &= -\frac {1}{2} c_6^{2}-c_6 x -\frac {1}{2} x^{2}-c_2 \\ \end{align*}

Solved as second order missing y ode

Time used: 0.078 (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}-1 = 0 \end{align*}

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

Factoring the ode gives these factors

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

Now each of the above equations is solved in turn.

Solving equation (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 {1\, dx}\\ u \left (x \right ) &= x + c_1 \end{align*}

Solving equation (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 {-1\, dx}\\ u \left (x \right ) &= -x + c_2 \end{align*}

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

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

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

\begin{align*} y^{\prime } = -x +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 {-x +c_2\, dx}\\ y &= -\frac {1}{2} x^{2}+c_2 x + c_3 \end{align*}

For solution \(u \left (x \right ) = x +c_1\), since \(u=y^{\prime }\) then we now have a new first order ode to solve which is

\begin{align*} y^{\prime } = 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_4 \end{align*}

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

\begin{align*} y &= -\frac {1}{2} x^{2}+c_2 x +c_3 \\ y &= \frac {1}{2} x^{2}+c_1 x +c_4 \\ \end{align*}

Will add steps showing solving for IC soon.

Summary of solutions found

\begin{align*} y &= -\frac {1}{2} x^{2}+c_2 x +c_3 \\ y &= \frac {1}{2} x^{2}+c_1 x +c_4 \\ \end{align*}

Solving equation (2)

Solved as second order missing x ode

Time used: 0.325 (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} = 1 \end{align*}

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

Factoring the ode gives these factors

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

Now each of the above equations is solved in turn.

Solving equation (1)

Integrating gives

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

Solving for \(p\) gives

\begin{align*} p &= \sqrt {2 c_1 +2 y} \\ p &= -\sqrt {2 c_1 +2 y} \\ \end{align*}

Solving equation (2)

Integrating gives

\begin{align*} \int -p d p &= dy\\ -\frac {p^{2}}{2}&= y +c_2 \end{align*}

Solving for \(p\) gives

\begin{align*} p &= \sqrt {-2 c_2 -2 y} \\ p &= -\sqrt {-2 c_2 -2 y} \\ \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 } = \sqrt {2 c_1 +2 y} \end{align*}

Integrating gives

\begin{align*} \int \frac {1}{\sqrt {2 c_1 +2 y}}d y &= dx\\ \sqrt {2 c_1 +2 y}&= x +c_3 \end{align*}

Singular solutions are found by solving

\begin{align*} \sqrt {2 c_1 +2 y}&= 0 \end{align*}

for \(y\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

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

Solving for \(y\) gives

\begin{align*} y &= -c_1 \\ y &= \frac {1}{2} c_3^{2}+c_3 x +\frac {1}{2} x^{2}-c_1 \\ \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 } = \sqrt {-2 c_2 -2 y} \end{align*}

Integrating gives

\begin{align*} \int \frac {1}{\sqrt {-2 c_2 -2 y}}d y &= dx\\ -\sqrt {-2 c_2 -2 y}&= x +c_4 \end{align*}

Singular solutions are found by solving

\begin{align*} \sqrt {-2 c_2 -2 y}&= 0 \end{align*}

for \(y\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

\begin{align*} y = -c_2 \end{align*}

Solving for \(y\) gives

\begin{align*} y &= -c_2 \\ y &= -\frac {1}{2} c_4^{2}-c_4 x -\frac {1}{2} x^{2}-c_2 \\ \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 } = -\sqrt {2 c_1 +2 y} \end{align*}

Integrating gives

\begin{align*} \int -\frac {1}{\sqrt {2 c_1 +2 y}}d y &= dx\\ -\sqrt {2 c_1 +2 y}&= x +c_5 \end{align*}

Singular solutions are found by solving

\begin{align*} -\sqrt {2 c_1 +2 y}&= 0 \end{align*}

for \(y\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

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

Solving for \(y\) gives

\begin{align*} y &= -c_1 \\ y &= \frac {1}{2} c_5^{2}+c_5 x +\frac {1}{2} x^{2}-c_1 \\ \end{align*}

For solution (4) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is

\begin{align*} y^{\prime } = -\sqrt {-2 c_2 -2 y} \end{align*}

Integrating gives

\begin{align*} \int -\frac {1}{\sqrt {-2 c_2 -2 y}}d y &= dx\\ \sqrt {-2 c_2 -2 y}&= x +c_6 \end{align*}

Singular solutions are found by solving

\begin{align*} -\sqrt {-2 c_2 -2 y}&= 0 \end{align*}

for \(y\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

\begin{align*} y = -c_2 \end{align*}

Solving for \(y\) gives

\begin{align*} y &= -c_2 \\ y &= -\frac {1}{2} c_6^{2}-c_6 x -\frac {1}{2} x^{2}-c_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. The solution

\[ y = -c_2 \]

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

Summary of solutions found

\begin{align*} y &= \frac {1}{2} c_3^{2}+c_3 x +\frac {1}{2} x^{2}-c_1 \\ y &= -\frac {1}{2} c_4^{2}-c_4 x -\frac {1}{2} x^{2}-c_2 \\ y &= \frac {1}{2} c_5^{2}+c_5 x +\frac {1}{2} x^{2}-c_1 \\ y &= -\frac {1}{2} c_6^{2}-c_6 x -\frac {1}{2} x^{2}-c_2 \\ \end{align*}

Solved as second order missing y ode

Time used: 0.070 (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}-1 = 0 \end{align*}

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

Factoring the ode gives these factors

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

Now each of the above equations is solved in turn.

Solving equation (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 {1\, dx}\\ u \left (x \right ) &= x + c_1 \end{align*}

Solving equation (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 {-1\, dx}\\ u \left (x \right ) &= -x + c_2 \end{align*}

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

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

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

\begin{align*} y^{\prime } = -x +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 {-x +c_2\, dx}\\ y &= -\frac {1}{2} x^{2}+c_2 x + c_3 \end{align*}

For solution \(u \left (x \right ) = x +c_1\), since \(u=y^{\prime }\) then we now have a new first order ode to solve which is

\begin{align*} y^{\prime } = 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_4 \end{align*}

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

\begin{align*} y &= -\frac {1}{2} x^{2}+c_2 x +c_3 \\ y &= \frac {1}{2} x^{2}+c_1 x +c_4 \\ \end{align*}

Will add steps showing solving for IC soon.

Summary of solutions found

\begin{align*} y &= -\frac {1}{2} x^{2}+c_2 x +c_3 \\ y &= \frac {1}{2} x^{2}+c_1 x +c_4 \\ \end{align*}

Maple step by step solution

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 dsolve solution

Solving time : 0.004 (sec)
Leaf size : 27

dsolve(diff(diff(y(x),x),x)^2 = 1,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*}

✓Mathematica DSolve solution

Solving time : 0.003 (sec)
Leaf size : 37

DSolve[{(D[y[x],{x,2}])^2==1,{}},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 solution

Solving time : 0.151 (sec)
Leaf size : 24

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 - 1,0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

[Eq(y(x), C1 + C2*x - x**2/2), Eq(y(x), C1 + C2*x + x**2/2)]
 

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