problem 671

Internal problem ID [15435]
Internal ﬁle name [OUTPUT/15436_Wednesday_May_08_2024_03_58_47_PM_46832233/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: 671.
ODE order: 2.
ODE degree: 1.

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

\[ \boxed {2 x^{2} \left (2-\ln \left (x \right )\right ) y^{\prime \prime }+x \left (4-\ln \left (x \right )\right ) y^{\prime }-y=\frac {\left (2-\ln \left (x \right )\right )^{2}}{\sqrt {x}}} \] With initial conditions \begin {align*} [y \left (\infty \right ) = 0] \end {align*}

20.32.1 Existence and uniqueness analysis

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 {-x \ln \left (x \right )+4 x}{2 x^{2} \left (-2+\ln \left (x \right )\right )}\\ q(x) &=\frac {1}{2 x^{2} \left (-2+\ln \left (x \right )\right )}\\ F &=-\frac {-2+\ln \left (x \right )}{2 x^{\frac {5}{2}}} \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }-\frac {\left (-x \ln \left (x \right )+4 x \right ) y^{\prime }}{2 x^{2} \left (-2+\ln \left (x \right )\right )}+\frac {y}{2 x^{2} \left (-2+\ln \left (x \right )\right )} = -\frac {-2+\ln \left (x \right )}{2 x^{\frac {5}{2}}} \end {align*}

The domain of \(p(x)=-\frac {-x \ln \left (x \right )+4 x}{2 x^{2} \left (-2+\ln \left (x \right )\right )}\) is \[ \{0

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

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 \(A=-2 x^{2} \left (-2+\ln \left (x \right )\right ), B=-x \ln \left (x \right )+4 x, C=-1, f(x)=\frac {\left (-2+\ln \left (x \right )\right )^{2}}{\sqrt {x}}\). 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 \[ -2 y^{\prime \prime } x^{2} \left (-2+\ln \left (x \right )\right )+\left (-x \ln \left (x \right )+4 x \right ) y^{\prime }-y = 0 \] In normal form the ode \begin {align*} -2 y^{\prime \prime } x^{2} \left (-2+\ln \left (x \right )\right )+\left (-x \ln \left (x \right )+4 x \right ) y^{\prime }-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 {4-\ln \left (x \right )}{x \left (4-2 \ln \left (x \right )\right )}\\ q \left (x \right )&=\frac {1}{2 x^{2} \left (-2+\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 \left (4-\ln \left (x \right )\right )}{x^{2} \left (4-2 \ln \left (x \right )\right )}+\frac {1}{2 x^{2} \left (-2+\ln \left (x \right )\right )}&=0 \tag {5} \end {align*}

Solving (5) for \(n\) gives \begin {align*} n&={\frac {1}{2}} \tag {6} \end {align*}

Substituting this value in (3) gives \begin {align*} v^{\prime \prime }\left (x \right )+\left (\frac {1}{x}+\frac {4-\ln \left (x \right )}{x \left (4-2 \ln \left (x \right )\right )}\right ) v^{\prime }\left (x \right )&=0 \\ v^{\prime \prime }\left (x \right )+\frac {\left (-8+3 \ln \left (x \right )\right ) v^{\prime }\left (x \right )}{2 x \left (-2+\ln \left (x \right )\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 )+\frac {\left (-8+3 \ln \left (x \right )\right ) u \left (x \right )}{2 x \left (-2+\ln \left (x \right )\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 {\left (-8+3 \ln \left (x \right )\right ) u}{2 x \left (-2+\ln \left (x \right )\right )} \end {align*}

Where \(f(x)=-\frac {-8+3 \ln \left (x \right )}{2 x \left (-2+\ln \left (x \right )\right )}\) and \(g(u)=u\). Integrating both sides gives \begin {align*} \frac {1}{u} \,du &= -\frac {-8+3 \ln \left (x \right )}{2 x \left (-2+\ln \left (x \right )\right )} \,d x\\ \int { \frac {1}{u} \,du} &= \int {-\frac {-8+3 \ln \left (x \right )}{2 x \left (-2+\ln \left (x \right )\right )} \,d x}\\ \ln \left (u \right )&=-\frac {3 \ln \left (x \right )}{2}+\ln \left (-2+\ln \left (x \right )\right )+c_{1}\\ u&={\mathrm e}^{-\frac {3 \ln \left (x \right )}{2}+\ln \left (-2+\ln \left (x \right )\right )+c_{1}}\\ &=c_{1} {\mathrm e}^{-\frac {3 \ln \left (x \right )}{2}+\ln \left (-2+\ln \left (x \right )\right )} \end {align*}

Which simpliﬁes to \[ u \left (x \right ) = c_{1} \left (-\frac {2}{x^{\frac {3}{2}}}+\frac {\ln \left (x \right )}{x^{\frac {3}{2}}}\right ) \] 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 {2 c_{1} \ln \left (x \right )}{\sqrt {x}}+c_{2} \end {align*}

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

Now the particular solution to this ODE is found \[ -2 y^{\prime \prime } x^{2} \left (-2+\ln \left (x \right )\right )+\left (-x \ln \left (x \right )+4 x \right ) y^{\prime }-y = \frac {\left (-2+\ln \left (x \right )\right )^{2}}{\sqrt {x}} \] 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 &= \sqrt {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 \[ W = \begin {vmatrix} \sqrt {x} & \ln \left (x \right ) \\ \frac {d}{dx}\left (\sqrt {x}\right ) & \frac {d}{dx}\left (\ln \left (x \right )\right ) \end {vmatrix} \] Which gives \[ W = \begin {vmatrix} \sqrt {x} & \ln \left (x \right ) \\ \frac {1}{2 \sqrt {x}} & \frac {1}{x} \end {vmatrix} \] Therefore \[ W = \left (\sqrt {x}\right )\left (\frac {1}{x}\right ) - \left (\ln \left (x \right )\right )\left (\frac {1}{2 \sqrt {x}}\right ) \] Which simpliﬁes to \[ W = -\frac {-2+\ln \left (x \right )}{2 \sqrt {x}} \] Which simpliﬁes to \[ W = -\frac {-2+\ln \left (x \right )}{2 \sqrt {x}} \] Therefore Eq. (2) becomes \[ u_1 = -\int \frac {\frac {\ln \left (x \right ) \left (-2+\ln \left (x \right )\right )^{2}}{\sqrt {x}}}{x^{\frac {3}{2}} \left (-2+\ln \left (x \right )\right )^{2}}\,dx \] Which simpliﬁes to \[ u_1 = - \int \frac {\ln \left (x \right )}{x^{2}}d x \] Hence \[ u_1 = \frac {\ln \left (x \right )}{x}+\frac {1}{x} \] And Eq. (3) becomes \[ u_2 = \int \frac {\left (-2+\ln \left (x \right )\right )^{2}}{x^{\frac {3}{2}} \left (-2+\ln \left (x \right )\right )^{2}}\,dx \] Which simpliﬁes to \[ u_2 = \int \frac {1}{x^{\frac {3}{2}}}d x \] Hence \[ u_2 = -\frac {2}{\sqrt {x}} \] Which simpliﬁes to \begin{align*} u_1 &= \frac {\ln \left (x \right )+1}{x} \\ u_2 &= -\frac {2}{\sqrt {x}} \\ \end{align*} Therefore the particular solution, from equation (1) is \[ y_p(x) = \frac {\ln \left (x \right )+1}{\sqrt {x}}-\frac {2 \ln \left (x \right )}{\sqrt {x}} \] Which simpliﬁes to \[ y_p(x) = -\frac {-1+\ln \left (x \right )}{\sqrt {x}} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (\left (-\frac {2 c_{1} \ln \left (x \right )}{\sqrt {x}}+c_{2} \right ) \sqrt {x}\right ) + \left (-\frac {-1+\ln \left (x \right )}{\sqrt {x}}\right ) \\ &= -\frac {-1+\ln \left (x \right )}{\sqrt {x}}+\left (-\frac {2 c_{1} \ln \left (x \right )}{\sqrt {x}}+c_{2} \right ) \sqrt {x} \\ \end{align*} Which simpliﬁes to \[ y = \frac {c_{2} x -2 \sqrt {x}\, \ln \left (x \right ) c_{1} -\ln \left (x \right )+1}{\sqrt {x}} \] Initial conditions are used to solve for the constants of integration.

Looking at the above solution \begin {align*} y = \frac {c_{2} x -2 \sqrt {x}\, \ln \left (x \right ) c_{1} -\ln \left (x \right )+1}{\sqrt {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 Kovacic algorithm successful

`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 solving first the homogeneous part of the ODE 
   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)+(2*t-4)*diff(diff(u(t),t),t) = 0 
   <- change of variables successful 
<- solving first the homogeneous part of the ODE successful`

\[ y \left (x \right ) = \frac {\sqrt {x}\, \ln \left (x \right ) c_{2} -\ln \left (x \right )+1}{\sqrt {x}} \]

20.32 problem 671

20.32.1 Existence and uniqueness analysis

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