problem 243

Internal problem ID [11066]
Internal ﬁle name [OUTPUT/10323_Wednesday_January_24_2024_10_07_13_PM_55005500/index.tex]

Book: Handbook of exact solutions for ordinary diﬀerential equations. By Polyanin and Zaitsev. Second edition
Section: Chapter 2, Second-Order Diﬀerential Equations. section 2.1.2-8. Other equations.
Problem number: 243.
ODE order: 2.
ODE degree: 1.

The type(s) of ODE detected by this program : "second_order_ode_non_constant_coeﬀ_transformation_on_B"

\[ \boxed {x^{n} y^{\prime \prime }+\left (x a +b \right ) y^{\prime }-a y=0} \]

33.5.1 Solving as second order ode non constant coeﬀ transformation on B ode

Given an ode of the form \begin {align*} A y^{\prime \prime } + B y^{\prime } + C y &= F(x) \end {align*}

This method reduces the order ode the ODE by one by applying the transformation \begin {align*} y&= B v \end {align*}

This results in \begin {align*} y' &=B' v+ v' B \\ y'' &=B'' v+ B' v' +v'' B + v' B' \\ &=v'' B+2 v'+ B'+B'' v \end {align*}

And now the original ode becomes \begin {align*} A\left ( v'' B+2v'B'+ B'' v\right )+B\left ( B'v+ v' B\right ) +CBv & =0\\ ABv'' +\left ( 2AB'+B^{2}\right ) v'+\left (AB''+BB'+CB\right ) v & =0 \tag {1} \end {align*}

If the term \(AB''+BB'+CB\) is zero, then this method works and can be used to solve \[ ABv''+\left ( 2AB' +B^{2}\right ) v'=0 \] By Using \(u=v'\) which reduces the order of the above ode to one. The new ode is \[ ABu'+\left ( 2AB'+B^{2}\right ) u=0 \] The above ode is ﬁrst order ode which is solved for \(u\). Now a new ode \(v'=u\) is solved for \(v\) as ﬁrst order ode. Then the ﬁnal solution is obtain from \(y=Bv\).

This method works only if the term \(AB''+BB'+CB\) is zero. The given ODE shows that \begin {align*} A &= x^{n}\\ B &= x a +b\\ C &= -a\\ F &= 0 \end {align*}

The above shows that for this ode \begin {align*} AB''+BB'+CB &= \left (x^{n}\right ) \left (0\right ) + \left (x a +b\right ) \left (a\right ) + \left (-a\right ) \left (x a +b\right ) \\ &=0 \end {align*}

Hence the ode in \(v\) given in (1) now simpliﬁes to \begin {align*} x^{n} \left (x a +b \right ) v'' +\left ( 2 a \,x^{n}+\left (x a +b \right )^{2}\right ) v' & =0 \end {align*}

Now by applying \(v'=u\) the above becomes \begin {align*} x^{n} \left (x a +b \right ) u^{\prime }\left (x \right )+\left (2 a \,x^{n}+\left (x a +b \right )^{2}\right ) u \left (x \right ) = 0 \end {align*}

Which is now solved for \(u\). In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= -\frac {u \left (a^{2} x^{2}+2 a b x +2 a \,x^{n}+b^{2}\right ) x^{-n}}{x a +b} \end {align*}

Where \(f(x)=-\frac {\left (a^{2} x^{2}+2 a b x +2 a \,x^{n}+b^{2}\right ) x^{-n}}{x a +b}\) and \(g(u)=u\). Integrating both sides gives \begin {align*} \frac {1}{u} \,du &= -\frac {\left (a^{2} x^{2}+2 a b x +2 a \,x^{n}+b^{2}\right ) x^{-n}}{x a +b} \,d x\\ \int { \frac {1}{u} \,du} &= \int {-\frac {\left (a^{2} x^{2}+2 a b x +2 a \,x^{n}+b^{2}\right ) x^{-n}}{x a +b} \,d x}\\ \ln \left (u \right )&=\left (\frac {a \,x^{2}}{n -2}+\frac {b x}{n -1}\right ) {\mathrm e}^{-n \ln \left (x \right )}-2 \ln \left (x a +b \right )+c_{1}\\ u&={\mathrm e}^{\left (\frac {a \,x^{2}}{n -2}+\frac {b x}{n -1}\right ) {\mathrm e}^{-n \ln \left (x \right )}-2 \ln \left (x a +b \right )+c_{1}}\\ &=c_{1} {\mathrm e}^{\left (\frac {a \,x^{2}}{n -2}+\frac {b x}{n -1}\right ) {\mathrm e}^{-n \ln \left (x \right )}-2 \ln \left (x a +b \right )} \end {align*}

Which simpliﬁes to \[ u \left (x \right ) = \frac {c_{1} {\mathrm e}^{\frac {x^{2} a \,x^{-n}}{n -2}} {\mathrm e}^{\frac {x^{-n} b x}{n -1}}}{\left (x a +b \right )^{2}} \] The ode for \(v\) now becomes \begin {align*} v' &= u\\ &=\frac {c_{1} {\mathrm e}^{\frac {x^{2} a \,x^{-n}}{n -2}} {\mathrm e}^{\frac {x^{-n} b x}{n -1}}}{\left (x a +b \right )^{2}} \end {align*}

Which is now solved for \(v\). Integrating both sides gives \begin {align*} v \left (x \right ) = \int \frac {c_{1} {\mathrm e}^{\frac {x^{2} a \,x^{-n}}{n -2}} {\mathrm e}^{\frac {x^{-n} b x}{n -1}}}{\left (x a +b \right )^{2}}d x +c_{2} \end {align*}

Therefore the solution is \begin {align*} y(x) &= B v\\ &= \left (x a +b\right ) \left (\int \frac {c_{1} {\mathrm e}^{\frac {x^{2} a \,x^{-n}}{n -2}} {\mathrm e}^{\frac {x^{-n} b x}{n -1}}}{\left (x a +b \right )^{2}}d x +c_{2}\right ) \\ &= \left (x a +b \right ) \left (c_{1} \left (\int \frac {{\mathrm e}^{\frac {\left (x a \left (n -1\right )+b \left (n -2\right )\right ) x^{1-n}}{\left (n -2\right ) \left (n -1\right )}}}{\left (x a +b \right )^{2}}d x \right )+c_{2} \right ) \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \left (x a +b \right ) \left (c_{1} \left (\int \frac {{\mathrm e}^{\frac {\left (x a \left (n -1\right )+b \left (n -2\right )\right ) x^{1-n}}{\left (n -2\right ) \left (n -1\right )}}}{\left (x a +b \right )^{2}}d x \right )+c_{2} \right ) \\ \end{align*}

Veriﬁcation of solutions

\[ y = \left (x a +b \right ) \left (c_{1} \left (\int \frac {{\mathrm e}^{\frac {\left (x a \left (n -1\right )+b \left (n -2\right )\right ) x^{1-n}}{\left (n -2\right ) \left (n -1\right )}}}{\left (x a +b \right )^{2}}d x \right )+c_{2} \right ) \] Veriﬁed OK.

Maple trace

`Methods for second order ODEs: 
--- Trying classification methods --- 
trying a symmetry of the form [xi=0, eta=F(x)] 
checking if the LODE is missing y 
-> Trying an equivalence, under non-integer power transformations, 
   to LODEs admitting Liouvillian solutions. 
-> Trying a solution in terms of special functions: 
   -> Bessel 
   -> elliptic 
   -> Legendre 
   -> Whittaker 
      -> hyper3: Equivalence to 1F1 under a power @ Moebius 
   -> hypergeometric 
      -> heuristic approach 
      -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius 
   -> Mathieu 
      -> Equivalence to the rational form of Mathieu ODE under a power @ Moebius 
-> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius 
-> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius 
-> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) 
-> Trying changes of variables to rationalize or make the ODE simpler 
<- unable to find a useful change of variables 
   trying a symmetry of the form [xi=0, eta=F(x)] 
   trying differential order: 2; exact nonlinear 
   trying symmetries linear in x and y(x) 
   <- linear symmetries successful`

\[ y \left (x \right ) = -\left (c_{1} \left (\int \frac {{\mathrm e}^{\frac {x^{-n +1} \left (a x \left (n -1\right )+b \left (n -2\right )\right )}{\left (n -2\right ) \left (n -1\right )}}}{\left (a x +b \right )^{2}}d x \right )+c_{2} \right ) \left (a x +b \right ) \]

33.5 problem 243

33.5.1 Solving as second order ode non constant coeﬀ transformation on B ode