Internal problem ID [1126]
Internal file name [OUTPUT/1127_Sunday_June_05_2022_02_03_11_AM_7309673/index.tex]
Book: Elementary differential equations with boundary value problems. William F. Trench.
Brooks/Cole 2001
Section: Chapter 5 linear second order equations. Section 5.6 Reduction or order. Page
253
Problem number: 20.
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_1", "second_order_ode_non_constant_coeff_transformation_on_B"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {x^{2} \ln \left (x \right )^{2} y^{\prime \prime }-2 x \ln \left (x \right ) y^{\prime }+\left (2+\ln \left (x \right )\right ) y=0} \] Given that one solution of the ode is \begin {align*} y_1 &= \ln \left (x \right ) \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 \[ p \left (x \right ) = -\frac {2}{x \ln \left (x \right )} \] Therefore \begin{align*} y_{2}\left (x \right ) &= \ln \left (x \right ) \left (\int \frac {{\mathrm e}^{-\left (\int -\frac {2}{x \ln \left (x \right )}d x \right )}}{\ln \left (x \right )^{2}}d x \right ) \\ y_{2}\left (x \right ) &= \ln \left (x \right ) \int \frac {\ln \left (x \right )^{2}}{\ln \left (x \right )^{2}} , dx \\ y_{2}\left (x \right ) &= \ln \left (x \right ) \left (\int 1d x \right ) \\ y_{2}\left (x \right ) &= x \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} \ln \left (x \right )+\ln \left (x \right ) c_{2} x \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \ln \left (x \right )+\ln \left (x \right ) c_{2} x \\ \end{align*}
Verification of solutions
\[ y = c_{1} \ln \left (x \right )+\ln \left (x \right ) c_{2} x \] 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 -> Trying a Liouvillian solution using Kovacics algorithm A Liouvillian solution exists Reducible group (found an exponential solution) <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 12
dsolve([x^2*(ln(x))^2*diff(y(x),x$2)-2*x*ln(x)*diff(y(x),x)+(2+ln(x))*y(x)=0,ln(x)],singsol=all)
\[ y \left (x \right ) = \ln \left (x \right ) \left (c_{2} x +c_{1} \right ) \]
✓ Solution by Mathematica
Time used: 0.023 (sec). Leaf size: 15
DSolve[x^2*Log[x]^2*y''[x]-2*x*Log[x]*y'[x]+(2+Log[x])*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to (c_2 x+c_1) \log (x) \]