14.12 problem 338

Internal problem ID [15195]
Internal file name [OUTPUT/15196_Tuesday_April_23_2024_04_53_54_PM_68807839/index.tex]

Book: A book of problems in ordinary differential equations. M.L. KRASNOV, A.L. KISELYOV, G.I. MARKARENKO. MIR, MOSCOW. 1983
Section: Chapter 2 (Higher order ODE’s). Section 14. Differential equations admitting of depression of their order. Exercises page 107
Problem number: 338.
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_poly_yn]]

\[ \boxed {2 y^{\prime \prime }-\frac {y^{\prime }}{x}-\frac {x^{2}}{y^{\prime }}=0} \] With initial conditions \begin {align*} \left [y \left (1\right ) = \frac {\sqrt {2}}{5}, y^{\prime }\left (1\right ) = \frac {\sqrt {2}}{2}\right ] \end {align*}

14.12.1 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*} 2 p^{\prime }\left (x \right ) x p \left (x \right )-x^{3}-p \left (x \right )^{2} = 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 )&=\frac {x^{3}+p^{2}}{2 x p}\\ 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 known. It is of type Bernoulli. Therefore we do not need to solve the PDE (A), and can just use the lookup table shown below to find \(\xi ,\eta \)

Table 198: Lie symmetry infinitesimal lookup table for known first order ODE’s

ODE class	Form	\(\xi \)	\(\eta \)
linear ode	\(y'=f(x) y(x) +g(x)\)	\(0\)	\(e^{\int fdx}\)
separable ode	\(y^{\prime }=f\left ( x\right ) g\left ( y\right ) \)	\(\frac {1}{f}\)	\(0\)
quadrature ode	\(y^{\prime }=f\left ( x\right ) \)	\(0\)	\(1\)
quadrature ode	\(y^{\prime }=g\left ( y\right ) \)	\(1\)	\(0\)
homogeneous ODEs of Class A	\(y^{\prime }=f\left ( \frac {y}{x}\right ) \)	\(x\)	\(y\)
homogeneous ODEs of Class C	\(y^{\prime }=\left ( a+bx+cy\right ) ^{\frac {n}{m}}\)	\(1\)	\(-\frac {b}{c}\)
homogeneous class D	\(y^{\prime }=\frac {y}{x}+g\left ( x\right ) F\left (\frac {y}{x}\right ) \)	\(x^{2}\)	\(xy\)
First order special form ID 1	\(y^{\prime }=g\left ( x\right ) e^{h\left (x\right ) +by}+f\left ( x\right ) \)	\(\frac {e^{-\int bf\left ( x\right )dx-h\left ( x\right ) }}{g\left ( x\right ) }\)	\(\frac {f\left ( x\right )e^{-\int bf\left ( x\right ) dx-h\left ( x\right ) }}{g\left ( x\right ) }\)
polynomial type ode	\(y^{\prime }=\frac {a_{1}x+b_{1}y+c_{1}}{a_{2}x+b_{2}y+c_{2}}\)	\(\frac {a_{1}b_{2}x-a_{2}b_{1}x-b_{1}c_{2}+b_{2}c_{1}}{a_{1}b_{2}-a_{2}b_{1}}\)	\(\frac {a_{1}b_{2}y-a_{2}b_{1}y-a_{1}c_{2}-a_{2}c_{1}}{a_{1}b_{2}-a_{2}b_{1}}\)
Bernoulli ode	\(y^{\prime }=f\left ( x\right ) y+g\left ( x\right ) y^{n}\)	\(0\)	\(e^{-\int \left ( n-1\right ) f\left ( x\right ) dx}y^{n}\)
Reduced Riccati	\(y^{\prime }=f_{1}\left ( x\right ) y+f_{2}\left ( x\right )y^{2}\)	\(0\)	\(e^{-\int f_{1}dx}\)

The above table shows that \begin {align*} \xi \left (x,p\right ) &=0\\ \tag {A1} \eta \left (x,p\right ) &=\frac {x}{p} \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}{\frac {x}{p}}} dy \end {align*}

Which results in \begin {align*} S&= \frac {p^{2}}{2 x} \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) &= \frac {x^{3}+p^{2}}{2 x p} \end {align*}

Evaluating all the partial derivatives gives \begin {align*} R_{x} &= 1\\ R_{p} &= 0\\ S_{x} &= -\frac {p^{2}}{2 x^{2}}\\ S_{p} &= \frac {p}{x} \end {align*}

Substituting all the above in (2) and simplifying gives the ode in canonical coordinates. \begin {align*} \frac {dS}{dR} &= \frac {x}{2}\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} &= \frac {R}{2} \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 ) = \frac {R^{2}}{4}+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 {p \left (x \right )^{2}}{2 x} = \frac {x^{2}}{4}+c_{1} \end {align*}

Which simplifies to \begin {align*} \frac {p \left (x \right )^{2}}{2 x} = \frac {x^{2}}{4}+c_{1} \end {align*}

Initial conditions are used to solve for \(c_{1}\). Substituting \(x=1\) and \(p=\frac {\sqrt {2}}{2}\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} {\frac {1}{4}} = \frac {1}{4}+c_{1} \end {align*}

The solutions are \begin {align*} c_{1} = 0 \end {align*}

Trying the constant \begin {align*} c_{1} = 0 \end {align*}

Substituting \(c_{1}\) found above in the general solution gives \begin {align*} \frac {p^{2}}{2 x} = \frac {x^{2}}{4} \end {align*}

The constant \(c_{1} = 0\) gives valid solution.

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 x} = \frac {x^{2}}{4} \end {align*}

Solving the given ode for \(y^{\prime }\) results in \(2\) differential equations to solve. Each one of these will generate a solution. The equations generated are \begin {align*} y^{\prime }&=\frac {\sqrt {2}\, x^{\frac {3}{2}}}{2} \tag {1} \\ y^{\prime }&=-\frac {\sqrt {2}\, x^{\frac {3}{2}}}{2} \tag {2} \end {align*}

Now each one of the above ODE is solved.

Solving equation (1)

Integrating both sides gives \begin {align*} y &= \int { \frac {\sqrt {2}\, x^{\frac {3}{2}}}{2}\,\mathop {\mathrm {d}x}}\\ &= \frac {x^{\frac {5}{2}} \sqrt {2}}{5}+c_{2} \end {align*}

Initial conditions are used to solve for \(c_{2}\). Substituting \(x=1\) and \(y=\frac {\sqrt {2}}{5}\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} \frac {\sqrt {2}}{5} = \frac {\sqrt {2}}{5}+c_{2} \end {align*}

The solutions are \begin {align*} c_{2} = 0 \end {align*}

Trying the constant \begin {align*} c_{2} = 0 \end {align*}

Substituting this in the general solution gives \begin {align*} y&=\frac {x^{\frac {5}{2}} \sqrt {2}}{5} \end {align*}

The constant \(c_{2} = 0\) gives valid solution.

Solving equation (2)

Integrating both sides gives \begin {align*} y &= \int { -\frac {\sqrt {2}\, x^{\frac {3}{2}}}{2}\,\mathop {\mathrm {d}x}}\\ &= -\frac {x^{\frac {5}{2}} \sqrt {2}}{5}+c_{3} \end {align*}

Initial conditions are used to solve for \(c_{3}\). Substituting \(x=1\) and \(y=\frac {\sqrt {2}}{5}\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} \frac {\sqrt {2}}{5} = -\frac {\sqrt {2}}{5}+c_{3} \end {align*}

The solutions are \begin {align*} c_{3} = \frac {2 \sqrt {2}}{5} \end {align*}

Trying the constant \begin {align*} c_{3} = \frac {2 \sqrt {2}}{5} \end {align*}

Substituting this in the general solution gives \begin {align*} y&=-\frac {x^{\frac {5}{2}} \sqrt {2}}{5}+\frac {2 \sqrt {2}}{5} \end {align*}

The constant \(c_{3} = \frac {2 \sqrt {2}}{5}\) gives valid solution.

Initial conditions are used to solve for the constants of integration.

Summary

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

Figure 403: Solution plot

Verification of solutions

\[ y = \frac {x^{\frac {5}{2}} \sqrt {2}}{5} \] Verified OK.

\[ y = -\frac {x^{\frac {5}{2}} \sqrt {2}}{5}+\frac {2 \sqrt {2}}{5} \] Warning, solution could not be verified

Maple trace

`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 differential order: 2; missing variables 
`, `-> Computing symmetries using: way = 3 
-> Calling odsolve with the ODE`, diff(_b(_a), _a) = (1/2)*(_a^3+_b(_a)^2)/(_a*_b(_a)), _b(_a), HINT = [[_a, (3/2)*_b]]`   *** Suble 
   symmetry methods on request 
`, `1st order, trying reduction of order with given symmetries:`[_a, 3/2*_b]
 

✓ Solution by Maple

Time used: 0.094 (sec). Leaf size: 12

dsolve([2*diff(y(x),x$2)=diff(y(x),x)/x+x^2/diff(y(x),x),y(1) = 1/5*sqrt(2), D(y)(1) = 1/2*sqrt(2)],y(x), singsol=all)
 

\[ y = \frac {\sqrt {2}\, x^{\frac {5}{2}}}{5} \]

✓ Solution by Mathematica

Time used: 0.12 (sec). Leaf size: 26

DSolve[{2*y''[x]==y'[x]/x+x^2/y'[x],{y[1]==Sqrt[2]/5,y'[1]==Sqrt[2]/2}},y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to \frac {1}{5} \sqrt {2} x^{3/2} \sqrt {x^2} \]