Internal problem ID [15437]
Internal file name [OUTPUT/15438_Wednesday_May_08_2024_03_58_53_PM_72127052/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: 673.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_change_of_variable_on_y_method_2"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
Unable to solve or complete the solution.
\[ \boxed {x^{3} \left (-1+\ln \left (x \right )\right ) y^{\prime \prime }-x^{2} y^{\prime }+x y=2 \ln \left (x \right )} \] With initial conditions \begin {align*} [y \left (\infty \right ) = 0] \end {align*}
This is a linear ODE. In canonical form it is written as \begin {align*} y^{\prime \prime } + p(x)y^{\prime } + q(x) y &= F \end {align*}
Where here \begin {align*} p(x) &=-\frac {1}{x \left (-1+\ln \left (x \right )\right )}\\ q(x) &=\frac {1}{x^{2} \left (-1+\ln \left (x \right )\right )}\\ F &=\frac {2 \ln \left (x \right )}{x^{3} \left (-1+\ln \left (x \right )\right )} \end {align*}
Hence the ode is \begin {align*} y^{\prime \prime }-\frac {y^{\prime }}{x \left (-1+\ln \left (x \right )\right )}+\frac {y}{x^{2} \left (-1+\ln \left (x \right )\right )} = \frac {2 \ln \left (x \right )}{x^{3} \left (-1+\ln \left (x \right )\right )} \end {align*}
The domain of \(p(x)=-\frac {1}{x \left (-1+\ln \left (x \right )\right )}\) is \[
\{0
This is second order non-homogeneous ODE. In standard form the ODE is \[
A y''(x) + B y'(x) + C y(x) = f(x)
\] Where \(A=x^{3} \left (-1+\ln \left (x \right )\right ), B=-x^{2}, C=x, f(x)=2 \ln \left (x \right )\). Let the
solution be \[
y = y_h + y_p
\] Where \(y_h\) is the solution to the homogeneous ODE \( A y''(x) + B y'(x) + C y(x) = 0\), and \(y_p\) is a particular
solution to the non-homogeneous ODE \(A y''(x) + B y'(x) + C y(x) = f(x)\). Solving for \(y_h\) from \[
x^{3} \left (-1+\ln \left (x \right )\right ) y^{\prime \prime }-x^{2} y^{\prime }+x y = 0
\] In normal form the ode
\begin {align*} x^{3} \left (-1+\ln \left (x \right )\right ) y^{\prime \prime }-x^{2} 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 (-1+\ln \left (x \right )\right )}\\ q \left (x \right )&=\frac {1}{x^{2} \left (-1+\ln \left (x \right )\right )} \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 (-1+\ln \left (x \right )\right )}+\frac {1}{x^{2} \left (-1+\ln \left (x \right )\right )}&=0 \tag {5} \end {align*}
Solving (5) for \(n\) gives \begin {align*} n&=1 \tag {6} \end {align*}
Substituting this value in (3) gives \begin {align*} v^{\prime \prime }\left (x \right )+\left (\frac {2}{x}-\frac {1}{x \left (-1+\ln \left (x \right )\right )}\right ) v^{\prime }\left (x \right )&=0 \\ v^{\prime \prime }\left (x \right )+\left (\frac {2}{x}-\frac {1}{x \left (-1+\ln \left (x \right )\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 (-1+\ln \left (x \right )\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 (-3+2 \ln \left (x \right )\right )}{x \left (-1+\ln \left (x \right )\right )} \end {align*}
Where \(f(x)=-\frac {-3+2 \ln \left (x \right )}{x \left (-1+\ln \left (x \right )\right )}\) and \(g(u)=u\). Integrating both sides gives \begin {align*} \frac {1}{u} \,du &= -\frac {-3+2 \ln \left (x \right )}{x \left (-1+\ln \left (x \right )\right )} \,d x\\ \int { \frac {1}{u} \,du} &= \int {-\frac {-3+2 \ln \left (x \right )}{x \left (-1+\ln \left (x \right )\right )} \,d x}\\ \ln \left (u \right )&=-2 \ln \left (x \right )+\ln \left (-1+\ln \left (x \right )\right )+c_{1}\\ u&={\mathrm e}^{-2 \ln \left (x \right )+\ln \left (-1+\ln \left (x \right )\right )+c_{1}}\\ &=c_{1} {\mathrm e}^{-2 \ln \left (x \right )+\ln \left (-1+\ln \left (x \right )\right )} \end {align*}
Which simplifies to \[
u \left (x \right ) = c_{1} \left (-\frac {1}{x^{2}}+\frac {\ln \left (x \right )}{x^{2}}\right )
\] 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*}
Now the particular solution to this ODE is found \[
x^{3} \left (-1+\ln \left (x \right )\right ) y^{\prime \prime }-x^{2} y^{\prime }+x y = 2 \ln \left (x \right )
\] The particular solution \(y_p\) can
be found using either the method of undetermined coefficients, or the method
of variation of parameters. The method of variation of parameters will be used
as it is more general and can be used when the coefficients of the ODE depend
on \(x\) as well. Let \begin{equation}
\tag{1} y_p(x) = u_1 y_1 + u_2 y_2
\end{equation} Where \(u_1,u_2\) to be determined, and \(y_1,y_2\) are the two basis solutions (the
two linearly independent solutions of the homogeneous ODE) found earlier when
solving the homogeneous ODE as \begin{align*}
y_1 &= x \\
y_2 &= \ln \left (x \right ) \\
\end{align*} In the Variation of parameters \(u_1,u_2\) are found using \begin{align*}
\tag{2} u_1 &= -\int \frac {y_2 f(x)}{a W(x)} \\
\tag{3} u_2 &= \int \frac {y_1 f(x)}{a W(x)} \\
\end{align*}
Where \(W(x)\) is the Wronskian and \(a\) is the coefficient in front of \(y''\) in the given ODE. The
Wronskian is given by \(W= \begin {vmatrix} y_1 & y_{2} \\ y_{1}^{\prime } & y_{2}^{\prime } \end {vmatrix} \). Hence \[ W = \begin {vmatrix} x & \ln \left (x \right ) \\ \frac {d}{dx}\left (x\right ) & \frac {d}{dx}\left (\ln \left (x \right )\right ) \end {vmatrix} \] Which gives \[ W = \begin {vmatrix} x & \ln \left (x \right ) \\ 1 & \frac {1}{x} \end {vmatrix} \] Therefore \[
W = \left (x\right )\left (\frac {1}{x}\right ) - \left (\ln \left (x \right )\right )\left (1\right )
\] Which simplifies to \[
W = 1-\ln \left (x \right )
\] Which
simplifies to \[
W = 1-\ln \left (x \right )
\] Therefore Eq. (2) becomes \[
u_1 = -\int \frac {2 \ln \left (x \right )^{2}}{x^{3} \left (-1+\ln \left (x \right )\right ) \left (1-\ln \left (x \right )\right )}\,dx
\] Which simplifies to \[
u_1 = - \int -\frac {2 \ln \left (x \right )^{2}}{x^{3} \left (-1+\ln \left (x \right )\right )^{2}}d x
\] Hence \[
u_1 = -\frac {\ln \left (x \right )+1}{\left (-1+\ln \left (x \right )\right ) x^{2}}
\] And Eq. (3)
becomes \[
u_2 = \int \frac {2 x \ln \left (x \right )}{x^{3} \left (-1+\ln \left (x \right )\right ) \left (1-\ln \left (x \right )\right )}\,dx
\] Which simplifies to \[
u_2 = \int -\frac {2 \ln \left (x \right )}{x^{2} \left (-1+\ln \left (x \right )\right )^{2}}d x
\] Hence \[
u_2 = \frac {2}{x \left (-1+\ln \left (x \right )\right )}
\] Which simplifies to \begin{align*}
u_1 &= \frac {-\ln \left (x \right )-1}{x^{2} \left (-1+\ln \left (x \right )\right )} \\
u_2 &= \frac {2}{x \left (-1+\ln \left (x \right )\right )} \\
\end{align*} Therefore the particular
solution, from equation (1) is \[
y_p(x) = \frac {-\ln \left (x \right )-1}{x \left (-1+\ln \left (x \right )\right )}+\frac {2 \ln \left (x \right )}{x \left (-1+\ln \left (x \right )\right )}
\] Which simplifies to \[
y_p(x) = \frac {1}{x}
\] Therefore the general solution
is \begin{align*}
y &= y_h + y_p \\
&= \left (\left (-\frac {c_{1} \ln \left (x \right )}{x}+c_{2} \right ) x\right ) + \left (\frac {1}{x}\right ) \\
&= \frac {1}{x}+\left (-\frac {c_{1} \ln \left (x \right )}{x}+c_{2} \right ) x \\
\end{align*} Which simplifies to \[
y = \frac {1}{x}+\left (-\frac {c_{1} \ln \left (x \right )}{x}+c_{2} \right ) x
\] Initial conditions are used to solve for the constants of
integration.
Looking at the above solution \begin {align*} y = \frac {1}{x}+\left (-\frac {c_{1} \ln \left (x \right )}{x}+c_{2} \right ) x \tag {1} \end {align*}
Initial conditions are now substituted in the above solution. This will generate the required
equations to solve for the integration constants. substituting \(y = 0\) and \(x = \infty \) in the above gives
\begin {align*} 0 = \operatorname {signum}\left (c_{2} \right ) \infty \tag {1A} \end {align*}
Equations {1A} are now solved for \(\{c_{1}, c_{2}\}\). There is no solution for the constants of integrations.
This solution is removed.
Verification of solutions N/A Maple trace
✓ Solution by Maple
Time used: 0.047 (sec). Leaf size: 16
\[
y \left (x \right ) = \frac {-c_{1} \ln \left (x \right ) x +1}{x}
\]
✓ Solution by Mathematica
Time used: 0.116 (sec). Leaf size: 8
\[
y(x)\to \frac {1}{x}
\]
20.34.2 Solving as second order change of variable on y method 2 ode
`Methods for second order ODEs:
--- Trying classification methods ---
trying a quadrature
trying high order exact linear fully integrable
trying differential order: 2; linear nonhomogeneous with symmetry [0,1]
trying a double symmetry of the form [xi=0, eta=F(x)]
trying symmetries linear in x and y(x)
Try integration with the canonical coordinates of the symmetry [0, x]
-> Calling odsolve with the ODE`, diff(_b(_a), _a) = -(2*_a^3*_b(_a)*ln(_a)-3*_b(_a)*_a^3-2*ln(_a))/(_a^4*(ln(_a)-1)), _b(_a), expli
Methods for first order ODEs:
--- Trying classification methods ---
trying a quadrature
trying 1st order linear
<- 1st order linear successful
<- differential order: 2; canonical coordinates successful`
dsolve([x^3*(ln(x)-1)*diff(y(x),x$2)-x^2*diff(y(x),x)+x*y(x)=2*ln(x),y(infinity) = 0],y(x), singsol=all)
DSolve[{x^3*(Log[x]-1)*y''[x]-x^2*y'[x]+x*y[x]==2*Log[x],{y[Infinity]==0}},y[x],x,IncludeSingularSolutions -> True]