1.8 problem 8

Internal problem ID [7397]
Internal file name [OUTPUT/6364_Sunday_June_05_2022_04_41_44_PM_99458875/index.tex]

Book: Second order enumerated odes
Section: section 1
Problem number: 8.
ODE order: 2.
ODE degree: 2.

The type(s) of ODE detected by this program : "second_order_ode_high_degree", "second_order_ode_missing_x", "second_order_ode_missing_y"

Maple gives the following as the ode type

[[_2nd_order, _quadrature]]

\[ \boxed {{y^{\prime \prime }}^{2}=1} \]

1.8.1 Solving as second order ode missing y ode

This is second order ode with missing dependent variable \(y\). Let \begin {align*} p(x) &= y^{\prime } \end {align*}

Then \begin {align*} p'(x) &= y^{\prime \prime } \end {align*}

Hence the ode becomes \begin {align*} {p^{\prime }\left (x \right )}^{2}-1 = 0 \end {align*}

Which is now solve for \(p(x)\) as first order ode. The ode \begin {align*} {p^{\prime }\left (x \right )}^{2} = 1 \end {align*}

is factored to \begin {align*} \left (p^{\prime }\left (x \right )-1\right ) \left (p^{\prime }\left (x \right )+1\right ) = 0 \end {align*}

Which gives the following equations \begin {align*} p^{\prime }\left (x \right )-1 = 0\tag {1} \\ p^{\prime }\left (x \right )+1 = 0\tag {2} \\ \end {align*}

Each of the above equations is now solved.

Solving ODE (1) Integrating both sides gives \begin {align*} p \left (x \right ) &= \int { 1\,\mathop {\mathrm {d}x}}\\ &= x +c_{1} \end {align*}

Solving ODE (2) Integrating both sides gives \begin {align*} p \left (x \right ) &= \int { -1\,\mathop {\mathrm {d}x}}\\ &= -x +c_{2} \end {align*}

For solution (1) found earlier, since \(p=y^{\prime }\) then the new first order ode to solve is \begin {align*} y^{\prime } = x +c_{1} \end {align*}

Integrating both sides gives \begin {align*} y &= \int { x +c_{1}\,\mathop {\mathrm {d}x}}\\ &= \frac {x \left (x +2 c_{1} \right )}{2}+c_{3} \end {align*}

Since \(p=y^{\prime }\) then the new first order ode to solve is \begin {align*} y^{\prime } = -x +c_{2} \end {align*}

Integrating both sides gives \begin {align*} y &= \int { -x +c_{2}\,\mathop {\mathrm {d}x}}\\ &= -\frac {x \left (x -2 c_{2} \right )}{2}+c_{4} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {x \left (x +2 c_{1} \right )}{2}+c_{3} \\ \tag{2} y &= -\frac {x \left (x -2 c_{2} \right )}{2}+c_{4} \\ \end{align*}

Verification of solutions

\[ y = \frac {x \left (x +2 c_{1} \right )}{2}+c_{3} \] Verified OK.

\[ y = -\frac {x \left (x -2 c_{2} \right )}{2}+c_{4} \] Verified OK.

1.8.2 Solving as second order ode missing x ode

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(y) \end {align*}

Then \begin {align*} y'' &= \frac {dp}{dx}\\ &= \frac {dy}{dx} \frac {dp}{dy}\\ &= 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 \(p(y)\). The ode \begin {align*} p \left (y \right )^{2} \left (\frac {d}{d y}p \left (y \right )\right )^{2} = 1 \end {align*}

is factored to \begin {align*} \left (p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )-1\right ) \left (p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )+1\right ) = 0 \end {align*}

Which gives the following equations \begin {align*} p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )-1 = 0\tag {1} \\ p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )+1 = 0\tag {2} \\ \end {align*}

Each of the above equations is now solved.

Solving ODE (1) Integrating both sides gives \begin {align*} \int p d p &= y +c_{1}\\ \frac {p^{2}}{2}&=y +c_{1} \end {align*}

Solving for \(p\) gives these solutions \begin {align*} p_1&=\sqrt {2 y +2 c_{1}}\\ p_2&=-\sqrt {2 y +2 c_{1}} \end {align*}

Solving ODE (2) Integrating both sides gives \begin {align*} \int -p d p &= y +c_{2}\\ -\frac {p^{2}}{2}&=y +c_{2} \end {align*}

Solving for \(p\) gives these solutions \begin {align*} p_1&=\sqrt {-2 c_{2} -2 y}\\ p_2&=-\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 y+2 c_{1}} \end {align*}

Integrating both sides gives \begin{align*} \int \frac {1}{\sqrt {2 y +2 c_{1}}}d y &= \int d x \\ \sqrt {2 y+2 c_{1}}&=x +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 } = -\sqrt {2 y+2 c_{1}} \end {align*}

Integrating both sides gives \begin{align*} \int -\frac {1}{\sqrt {2 y +2 c_{1}}}d y &= \int d x \\ -\sqrt {2 y+2 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 } = \sqrt {-2 c_{2} -2 y} \end {align*}

Integrating both sides gives \begin{align*} \int \frac {1}{\sqrt {-2 c_{2} -2 y}}d y &= \int d x \\ -\sqrt {-2 c_{2} -2 y}&=x +c_{5} \\ \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 both sides gives \begin{align*} \int -\frac {1}{\sqrt {-2 c_{2} -2 y}}d y &= \int d x \\ \sqrt {-2 c_{2} -2 y}&=x +c_{6} \\ \end{align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {1}{2} c_{3}^{2}+c_{3} x +\frac {1}{2} x^{2}-c_{1} \\ \tag{2} y &= \frac {1}{2} c_{4}^{2}+c_{4} x +\frac {1}{2} x^{2}-c_{1} \\ \tag{3} y &= -\frac {1}{2} c_{5}^{2}-c_{5} x -\frac {1}{2} x^{2}-c_{2} \\ \tag{4} y &= -\frac {1}{2} c_{6}^{2}-c_{6} x -\frac {1}{2} x^{2}-c_{2} \\ \end{align*}

Verification of solutions

\[ y = \frac {1}{2} c_{3}^{2}+c_{3} x +\frac {1}{2} x^{2}-c_{1} \] Verified OK.

\[ y = \frac {1}{2} c_{4}^{2}+c_{4} x +\frac {1}{2} x^{2}-c_{1} \] Verified OK.

\[ y = -\frac {1}{2} c_{5}^{2}-c_{5} x -\frac {1}{2} x^{2}-c_{2} \] Verified OK.

\[ y = -\frac {1}{2} c_{6}^{2}-c_{6} x -\frac {1}{2} x^{2}-c_{2} \] Verified OK.

1.8.3 Solving using Kovacic algorithm

Solving for \(y^{\prime \prime }\) from the ode gives \begin{align*} \tag{1} y^{\prime \prime } &= -1 \\ \tag{2} y^{\prime \prime } &= 1 \\ \end{align*} Now each ode is solved. The ODE can be written as \[ y^{\prime \prime } = -1 \] Integrating once gives \[ y^{\prime }= -x + c_{1} \] Integrating again gives \[ y= -\frac {x^{2}}{2} + c_{1} x + c_{2} \] The ODE can be written as \[ y^{\prime \prime } = 1 \] Integrating once gives \[ y^{\prime }= x + c_{3} \] Integrating again gives \[ y= \frac {x^{2}}{2} + c_{3} x + c_{4} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {1}{2} x^{2}+c_{1} x +c_{2} \\ \tag{2} y &= \frac {1}{2} x^{2}+c_{3} x +c_{4} \\ \end{align*}

Verification of solutions

\[ y = -\frac {1}{2} x^{2}+c_{1} x +c_{2} \] Verified OK.

\[ y = \frac {1}{2} x^{2}+c_{3} x +c_{4} \] Verified OK.

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`
 

✓ Solution by Maple

Time used: 0.015 (sec). Leaf size: 27

dsolve(diff(y(x),x$2)^2=1,y(x), singsol=all)
 

\begin{align*} y \left (x \right ) &= \frac {1}{2} x^{2}+c_{1} x +c_{2} \\ y \left (x \right ) &= -\frac {1}{2} x^{2}+c_{1} x +c_{2} \\ \end{align*}

✓ Solution by Mathematica

Time used: 0.002 (sec). Leaf size: 37

DSolve[(y''[x])^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*}