Internal problem ID [6855]
Internal file name [OUTPUT/6102_Friday_July_29_2022_03_09_20_AM_12648957/index.tex]
Book: Elementary differential equations. By Earl D. Rainville, Phillip E. Bedient. Macmilliam
Publishing Co. NY. 6th edition. 1981.
Section: CHAPTER 16. Nonlinear equations. Section 101. Independent variable missing.
EXERCISES Page 324
Problem number: 38.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_ode_missing_y"
Maple gives the following as the ode type
[[_2nd_order, _missing_y], [_2nd_order, _reducible, _mu_xy]]
\[ \boxed {y^{\prime \prime }-\left (x^{2}-y^{\prime }\right )^{2}=2 x} \]
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 )+\left (2 x^{2}-p \left (x \right )\right ) p \left (x \right )-x^{4}-2 x = 0 \end {align*}
Which is now solve for \(p(x)\) as first order ode. Writing the ode as \begin {align*} p^{\prime }\left (x \right )&=x^{4}-2 p \,x^{2}+p^{2}+2 x\\ p^{\prime }\left (x \right )&= \omega \left ( x,p\right ) \end {align*}
The condition of Lie symmetry is the linearized PDE given by \begin {align*} \eta _{x}+\omega \left ( \eta _{p}-\xi _{x}\right ) -\omega ^{2}\xi _{p}-\omega _{x}\xi -\omega _{p}\eta =0\tag {A} \end {align*}
The type of this ode is not in the lookup table. To determine \(\xi ,\eta \) then (A) is solved using ansatz. Making bivariate polynomials of degree 1 to use as anstaz gives \begin{align*} \tag{1E} \xi &= p a_{3}+x a_{2}+a_{1} \\ \tag{2E} \eta &= p b_{3}+x b_{2}+b_{1} \\ \end{align*} Where the unknown coefficients are \[ \{a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}\} \] Substituting equations (1E,2E) and \(\omega \) into (A) gives \begin{equation} \tag{5E} b_{2}+\left (x^{4}-2 p \,x^{2}+p^{2}+2 x \right ) \left (b_{3}-a_{2}\right )-\left (x^{4}-2 p \,x^{2}+p^{2}+2 x \right )^{2} a_{3}-\left (4 x^{3}-4 x p +2\right ) \left (p a_{3}+x a_{2}+a_{1}\right )-\left (-2 x^{2}+2 p \right ) \left (p b_{3}+x b_{2}+b_{1}\right ) = 0 \end{equation} Putting the above in normal form gives \[ -x^{8} a_{3}+4 p \,x^{6} a_{3}-6 p^{2} x^{4} a_{3}+4 p^{3} x^{2} a_{3}-4 x^{5} a_{3}-p^{4} a_{3}+4 p \,x^{3} a_{3}-5 x^{4} a_{2}+x^{4} b_{3}+6 p \,x^{2} a_{2}-4 x^{3} a_{1}+2 x^{3} b_{2}-p^{2} a_{2}-p^{2} b_{3}+4 p x a_{1}-2 p x b_{2}-4 x^{2} a_{3}+2 x^{2} b_{1}-2 p a_{3}-2 p b_{1}-4 x a_{2}+2 x b_{3}-2 a_{1}+b_{2} = 0 \] Setting the numerator to zero gives \begin{equation} \tag{6E} -x^{8} a_{3}+4 p \,x^{6} a_{3}-6 p^{2} x^{4} a_{3}+4 p^{3} x^{2} a_{3}-4 x^{5} a_{3}-p^{4} a_{3}+4 p \,x^{3} a_{3}-5 x^{4} a_{2}+x^{4} b_{3}+6 p \,x^{2} a_{2}-4 x^{3} a_{1}+2 x^{3} b_{2}-p^{2} a_{2}-p^{2} b_{3}+4 p x a_{1}-2 p x b_{2}-4 x^{2} a_{3}+2 x^{2} b_{1}-2 p a_{3}-2 p b_{1}-4 x a_{2}+2 x b_{3}-2 a_{1}+b_{2} = 0 \end{equation} Looking at the above PDE shows the following are all the terms with \(\{p, x\}\) in them. \[ \{p, x\} \] The following substitution is now made to be able to collect on all terms with \(\{p, x\}\) in them \[ \{p = v_{1}, x = v_{2}\} \] The above PDE (6E) now becomes \begin{equation} \tag{7E} -a_{3} v_{2}^{8}+4 a_{3} v_{1} v_{2}^{6}-6 a_{3} v_{1}^{2} v_{2}^{4}+4 a_{3} v_{1}^{3} v_{2}^{2}-4 a_{3} v_{2}^{5}-5 a_{2} v_{2}^{4}-a_{3} v_{1}^{4}+4 a_{3} v_{1} v_{2}^{3}+b_{3} v_{2}^{4}-4 a_{1} v_{2}^{3}+6 a_{2} v_{1} v_{2}^{2}+2 b_{2} v_{2}^{3}+4 a_{1} v_{1} v_{2}-a_{2} v_{1}^{2}-4 a_{3} v_{2}^{2}+2 b_{1} v_{2}^{2}-2 b_{2} v_{1} v_{2}-b_{3} v_{1}^{2}-4 a_{2} v_{2}-2 a_{3} v_{1}-2 b_{1} v_{1}+2 b_{3} v_{2}-2 a_{1}+b_{2} = 0 \end{equation} Collecting the above on the terms \(v_i\) introduced, and these are \[ \{v_{1}, v_{2}\} \] Equation (7E) now becomes \begin{equation} \tag{8E} -a_{3} v_{1}^{4}+4 a_{3} v_{1}^{3} v_{2}^{2}-6 a_{3} v_{1}^{2} v_{2}^{4}+\left (-a_{2}-b_{3}\right ) v_{1}^{2}+4 a_{3} v_{1} v_{2}^{6}+4 a_{3} v_{1} v_{2}^{3}+6 a_{2} v_{1} v_{2}^{2}+\left (4 a_{1}-2 b_{2}\right ) v_{1} v_{2}+\left (-2 a_{3}-2 b_{1}\right ) v_{1}-a_{3} v_{2}^{8}-4 a_{3} v_{2}^{5}+\left (-5 a_{2}+b_{3}\right ) v_{2}^{4}+\left (-4 a_{1}+2 b_{2}\right ) v_{2}^{3}+\left (-4 a_{3}+2 b_{1}\right ) v_{2}^{2}+\left (-4 a_{2}+2 b_{3}\right ) v_{2}-2 a_{1}+b_{2} = 0 \end{equation} Setting each coefficients in (8E) to zero gives the following equations to solve \begin {align*} 6 a_{2}&=0\\ -6 a_{3}&=0\\ -4 a_{3}&=0\\ -a_{3}&=0\\ 4 a_{3}&=0\\ -4 a_{1}+2 b_{2}&=0\\ -2 a_{1}+b_{2}&=0\\ 4 a_{1}-2 b_{2}&=0\\ -5 a_{2}+b_{3}&=0\\ -4 a_{2}+2 b_{3}&=0\\ -a_{2}-b_{3}&=0\\ -4 a_{3}+2 b_{1}&=0\\ -2 a_{3}-2 b_{1}&=0 \end {align*}
Solving the above equations for the unknowns gives \begin {align*} a_{1}&=a_{1}\\ a_{2}&=0\\ a_{3}&=0\\ b_{1}&=0\\ b_{2}&=2 a_{1}\\ b_{3}&=0 \end {align*}
Substituting the above solution in the anstaz (1E,2E) (using \(1\) as arbitrary value for any unknown in the RHS) gives \begin{align*} \xi &= 1 \\ \eta &= 2 x \\ \end{align*} Shifting is now applied to make \(\xi =0\) in order to simplify the rest of the computation \begin {align*} \eta &= \eta - \omega \left (x,p\right ) \xi \\ &= 2 x - \left (x^{4}-2 p \,x^{2}+p^{2}+2 x\right ) \left (1\right ) \\ &= -x^{4}+2 p \,x^{2}-p^{2}\\ \xi &= 0 \end {align*}
The next step is to determine the canonical coordinates \(R,S\). The canonical coordinates map \(\left ( x,p\right ) \to \left ( R,S \right )\) where \(\left ( R,S \right )\) are the canonical coordinates which make the original ode become a quadrature and hence solved by integration.
The characteristic pde which is used to find the canonical coordinates is \begin {align*} \frac {d x}{\xi } &= \frac {d p}{\eta } = dS \tag {1} \end {align*}
The above comes from the requirements that \(\left ( \xi \frac {\partial }{\partial x} + \eta \frac {\partial }{\partial p}\right ) S(x,p) = 1\). Starting with the first pair of ode’s in (1) gives an ode to solve for the independent variable \(R\) in the canonical coordinates, where \(S(R)\). Since \(\xi =0\) then in this special case \begin {align*} R = x \end {align*}
\(S\) is found from \begin {align*} S &= \int { \frac {1}{\eta }} dy\\ &= \int { \frac {1}{-x^{4}+2 p \,x^{2}-p^{2}}} dy \end {align*}
Which results in \begin {align*} S&= \frac {1}{-x^{2}+p} \end {align*}
Now that \(R,S\) are found, we need to setup the ode in these coordinates. This is done by evaluating \begin {align*} \frac {dS}{dR} &= \frac { S_{x} + \omega (x,p) S_{p} }{ R_{x} + \omega (x,p) R_{p} }\tag {2} \end {align*}
Where in the above \(R_{x},R_{p},S_{x},S_{p}\) are all partial derivatives and \(\omega (x,p)\) is the right hand side of the original ode given by \begin {align*} \omega (x,p) &= x^{4}-2 p \,x^{2}+p^{2}+2 x \end {align*}
Evaluating all the partial derivatives gives \begin {align*} R_{x} &= 1\\ R_{p} &= 0\\ S_{x} &= \frac {2 x}{\left (-x^{2}+p \right )^{2}}\\ S_{p} &= -\frac {1}{\left (-x^{2}+p \right )^{2}} \end {align*}
Substituting all the above in (2) and simplifying gives the ode in canonical coordinates. \begin {align*} \frac {dS}{dR} &= -1\tag {2A} \end {align*}
We now need to express the RHS as function of \(R\) only. This is done by solving for \(x,p\) in terms of \(R,S\) from the result obtained earlier and simplifying. This gives \begin {align*} \frac {dS}{dR} &= -1 \end {align*}
The above is a quadrature ode. This is the whole point of Lie symmetry method. It converts an ode, no matter how complicated it is, to one that can be solved by integration when the ode is in the canonical coordiates \(R,S\). Integrating the above gives \begin {align*} S \left (R \right ) = -R +c_{1}\tag {4} \end {align*}
To complete the solution, we just need to transform (4) back to \(x,p\) coordinates. This results in \begin {align*} \frac {1}{-x^{2}+p \left (x \right )} = -x +c_{1} \end {align*}
Which simplifies to \begin {align*} \frac {1}{-x^{2}+p \left (x \right )} = -x +c_{1} \end {align*}
Which gives \begin {align*} p \left (x \right ) = \frac {c_{1} x^{2}-x^{3}+1}{-x +c_{1}} \end {align*}
Since \(p=y^{\prime }\) then the new first order ode to solve is \begin {align*} y^{\prime } = \frac {c_{1} x^{2}-x^{3}+1}{-x +c_{1}} \end {align*}
Integrating both sides gives \begin {align*} y &= \int { \frac {c_{1} x^{2}-x^{3}+1}{-x +c_{1}}\,\mathop {\mathrm {d}x}}\\ &= \frac {x^{3}}{3}-\ln \left (-c_{1} +x \right )+c_{2} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {x^{3}}{3}-\ln \left (-c_{1} +x \right )+c_{2} \\ \end{align*}
Verification of solutions
\[ y = \frac {x^{3}}{3}-\ln \left (-c_{1} +x \right )+c_{2} \] Verified OK.
Maple trace Kovacic algorithm successful
`Methods for second order ODEs: --- Trying classification methods --- trying 2nd order Liouville trying 2nd order WeierstrassP trying 2nd order JacobiSN differential order: 2; trying a linearization to 3rd order trying 2nd order ODE linearizable_by_differentiation trying 2nd order, 2 integrating factors of the form mu(x,y) trying a quadrature checking if the LODE has constant coefficients checking if the LODE is of Euler type trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Trying a Liouvillian solution using Kovacics algorithm A Liouvillian solution exists Reducible group (found an exponential solution) <- Kovacics algorithm successful <- 2nd order, 2 integrating factors of the form mu(x,y) successful`
✓ Solution by Maple
Time used: 0.047 (sec). Leaf size: 20
dsolve(diff(y(x),x$2)=2*x+(x^2-diff(y(x),x))^2,y(x), singsol=all)
\[ y \left (x \right ) = \frac {x^{3}}{3}-\ln \left (c_{2} x -c_{1} \right ) \]
✓ Solution by Mathematica
Time used: 0.298 (sec). Leaf size: 24
DSolve[y''[x]==2*x+(x^2-y'[x])^2,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {x^3}{3}-\log (-x+c_1)+c_2 \]