Internal problem ID [15407]
Internal file name [OUTPUT/15408_Wednesday_May_08_2024_03_58_15_PM_82899032/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 15.5 Linear equations with variable coefficients.
The Lagrange method. Exercises page 148
Problem number: 639.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "reduction_of_order", "second_order_change_of_variable_on_y_method_2", "second_order_ode_non_constant_coeff_transformation_on_B"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {x^{2} \left (\ln \left (x \right )-1\right ) y^{\prime \prime }-x y^{\prime }+y=0} \] Given that one solution of the ode is \begin {align*} y_1 &= x \end {align*}
Given one basis solution \(y_{1}\left (x \right )\), then the second basis solution is given by \[ y_{2}\left (x \right ) = y_{1} \left (\int \frac {{\mathrm e}^{-\left (\int p d x \right )}}{y_{1}^{2}}d x \right ) \] Where \(p(x)\) is the coefficient of \(y^{\prime }\) when the ode is written in the normal form \[ y^{\prime \prime }+p \left (x \right ) y^{\prime }+q \left (x \right ) y = f \left (x \right ) \] Looking at the ode to solve shows that \[ p \left (x \right ) = -\frac {1}{x \left (\ln \left (x \right )-1\right )} \] Therefore \begin{align*} y_{2}\left (x \right ) &= x \left (\int \frac {{\mathrm e}^{-\left (\int -\frac {1}{x \left (\ln \left (x \right )-1\right )}d x \right )}}{x^{2}}d x \right ) \\ y_{2}\left (x \right ) &= x \int \frac {\ln \left (x \right )-1}{x^{2}} , dx \\ y_{2}\left (x \right ) &= x \left (\int \frac {\ln \left (x \right )-1}{x^{2}}d x \right ) \\ y_{2}\left (x \right ) &= -\ln \left (x \right ) \\ \end{align*} Hence the solution is \begin{align*} y &= c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \\ &= c_{1} x -c_{2} \ln \left (x \right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x -c_{2} \ln \left (x \right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x -c_{2} \ln \left (x \right ) \] 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`
✓ Solution by Maple
Time used: 0.125 (sec). Leaf size: 12
dsolve([x^2*(ln(x)-1)*diff(y(x),x$2)-x*diff(y(x),x)+y(x)=0,x],singsol=all)
\[ y \left (x \right ) = c_{1} x +c_{2} \ln \left (x \right ) \]
✓ Solution by Mathematica
Time used: 0.104 (sec). Leaf size: 16
DSolve[x^2*(Log[x]-1)*y''[x]-x*y'[x]+y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to c_1 x-c_2 \log (x) \]