problem 1413

Internal problem ID [9740]
Internal ﬁle name [OUTPUT/8682_Monday_June_06_2022_05_10_53_AM_68339033/index.tex]

Book: Diﬀerential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1413.
ODE order: 2.
ODE degree: 1.

The type(s) of ODE detected by this program : "second_order_change_of_variable_on_y_method_2", "second_order_ode_non_constant_coeﬀ_transformation_on_B"

\[ \boxed {y^{\prime \prime }-\frac {y^{\prime }}{x \left (\ln \left (x \right )-1\right )}+\frac {y}{x^{2} \left (\ln \left (x \right )-1\right )}=0} \]

3.407.1 Solving as second order change of variable on y method 2 ode

In normal form the ode \begin {align*} x^{2} \left (-\ln \left (x \right )+1\right ) y^{\prime \prime }+y^{\prime } x -y&=0 \tag {1} \end {align*}

Becomes \begin {align*} y^{\prime \prime }+p \left (x \right ) y^{\prime }+q \left (x \right ) y&=0 \tag {2} \end {align*}

Where \begin {align*} p \left (x \right )&=\frac {1}{x \left (-\ln \left (x \right )+1\right )}\\ q \left (x \right )&=\frac {1}{\left (\ln \left (x \right )-1\right ) x^{2}} \end {align*}

Applying change of variables on the depndent variable \(y = v \left (x \right ) x^{n}\) to (2) gives the following ode where the dependent variables is \(v \left (x \right )\) and not \(y\). \begin {align*} v^{\prime \prime }\left (x \right )+\left (\frac {2 n}{x}+p \right ) v^{\prime }\left (x \right )+\left (\frac {n \left (n -1\right )}{x^{2}}+\frac {n p}{x}+q \right ) v \left (x \right )&=0 \tag {3} \end {align*}

Let the coeﬃcient of \(v \left (x \right )\) above be zero. Hence \begin {align*} \frac {n \left (n -1\right )}{x^{2}}+\frac {n p}{x}+q&=0 \tag {4} \end {align*}

Substituting the earlier values found for \(p \left (x \right )\) and \(q \left (x \right )\) into (4) gives \begin {align*} \frac {n \left (n -1\right )}{x^{2}}+\frac {n}{x^{2} \left (-\ln \left (x \right )+1\right )}+\frac {1}{\left (\ln \left (x \right )-1\right ) x^{2}}&=0 \tag {5} \end {align*}

Substituting this value in (3) gives \begin {align*} v^{\prime \prime }\left (x \right )+\left (\frac {2}{x}+\frac {1}{x \left (-\ln \left (x \right )+1\right )}\right ) v^{\prime }\left (x \right )&=0 \\ v^{\prime \prime }\left (x \right )+\left (\frac {2}{x}-\frac {1}{x \left (\ln \left (x \right )-1\right )}\right ) v^{\prime }\left (x \right )&=0 \tag {7} \\ \end {align*}

Using the substitution \begin {align*} u \left (x \right ) = v^{\prime }\left (x \right ) \end {align*}

Then (7) becomes \begin {align*} u^{\prime }\left (x \right )+\left (\frac {2}{x}-\frac {1}{x \left (\ln \left (x \right )-1\right )}\right ) u \left (x \right ) = 0 \tag {8} \\ \end {align*}

The above is now solved for \(u \left (x \right )\). In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= -\frac {u \left (2 \ln \left (x \right )-3\right )}{x \left (\ln \left (x \right )-1\right )} \end {align*}

Where \(f(x)=-\frac {2 \ln \left (x \right )-3}{x \left (\ln \left (x \right )-1\right )}\) and \(g(u)=u\). Integrating both sides gives \begin {align*} \frac {1}{u} \,du &= -\frac {2 \ln \left (x \right )-3}{x \left (\ln \left (x \right )-1\right )} \,d x\\ \int { \frac {1}{u} \,du} &= \int {-\frac {2 \ln \left (x \right )-3}{x \left (\ln \left (x \right )-1\right )} \,d x}\\ \ln \left (u \right )&=-2 \ln \left (x \right )+\ln \left (\ln \left (x \right )-1\right )+c_{1}\\ u&={\mathrm e}^{-2 \ln \left (x \right )+\ln \left (\ln \left (x \right )-1\right )+c_{1}}\\ &=c_{1} {\mathrm e}^{-2 \ln \left (x \right )+\ln \left (\ln \left (x \right )-1\right )} \end {align*}

Which simpliﬁes to \[ u \left (x \right ) = c_{1} \left (\frac {\ln \left (x \right )}{x^{2}}-\frac {1}{x^{2}}\right ) \] Now that \(u \left (x \right )\) is known, then \begin {align*} v^{\prime }\left (x \right )&= u \left (x \right )\\ v \left (x \right )&= \int u \left (x \right )d x +c_{2}\\ &= -\frac {c_{1} \ln \left (x \right )}{x}+c_{2} \end {align*}

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

Summary

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

Veriﬁcation of solutions

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

3.407.2 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^{2} \left (-\ln \left (x \right )+1\right )\\ B &= x\\ C &= -1\\ F &= 0 \end {align*}

The above shows that for this ode \begin {align*} AB''+BB'+CB &= \left (x^{2} \left (-\ln \left (x \right )+1\right )\right ) \left (0\right ) + \left (x\right ) \left (1\right ) + \left (-1\right ) \left (x\right ) \\ &=0 \end {align*}

Hence the ode in \(v\) given in (1) now simpliﬁes to \begin {align*} x^{3} \left (-\ln \left (x \right )+1\right ) v'' +\left ( x^{2} \left (-2 \ln \left (x \right )+3\right )\right ) v' & =0 \end {align*}

Now by applying \(v'=u\) the above becomes \begin {align*} -\left (\left (\ln \left (x \right )-1\right ) x u^{\prime }\left (x \right )+2 u \left (x \right ) \ln \left (x \right )-3 u \left (x \right )\right ) x^{2} = 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 (2 \ln \left (x \right )-3\right )}{x \left (\ln \left (x \right )-1\right )} \end {align*}

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

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

Therefore the solution is \begin {align*} y(x) &= B v\\ &= \left (x\right ) \left (-\frac {c_{1} \ln \left (x \right )}{x}+c_{2}\right ) \\ &= -c_{1} \ln \left (x \right )+c_{2} x \end {align*}

Summary

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

Veriﬁcation of solutions

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

Maple trace Kovacic algorithm successful

`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 
-> 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 
   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) 
      Reducible group (found another exponential solution) 
   <- Kovacics algorithm successful 
   Change of variables used: 
      [x = exp(t)] 
   Linear ODE actually solved: 
      u(t)-t*diff(u(t),t)+(t-1)*diff(diff(u(t),t),t) = 0 
<- change of variables successful`

3.407 problem 1413

3.407.1 Solving as second order change of variable on y method 2 ode

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