problem 1419

Internal problem ID [9746]
Internal ﬁle name [OUTPUT/8688_Monday_June_06_2022_05_12_47_AM_25680586/index.tex]

Book: Diﬀerential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1419.
ODE order: 2.
ODE degree: 1.

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

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

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

In normal form the ode \begin {align*} y^{\prime \prime } x^{2} \cos \left (x \right )+x \left (x \sin \left (x \right )-2 \cos \left (x \right )\right ) y^{\prime }+\left (2 \cos \left (x \right )-x \sin \left (x \right )\right ) 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 {\tan \left (x \right ) x -2}{x}\\ q \left (x \right )&=\frac {-\tan \left (x \right ) x +2}{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 \left (\tan \left (x \right ) x -2\right )}{x^{2}}+\frac {-\tan \left (x \right ) x +2}{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 {\tan \left (x \right ) x -2}{x}\right ) v^{\prime }\left (x \right )&=0 \\ v^{\prime \prime }\left (x \right )+\tan \left (x \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 )+\tan \left (x \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)\\ &= -\tan \left (x \right ) u \end {align*}

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

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}\\ &= c_{1} \sin \left (x \right )+c_{2} \end {align*}

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

Summary

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

Veriﬁcation of solutions

\[ y = \left (c_{1} \sin \left (x \right )+c_{2} \right ) x \] 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 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 = arccos(t)] 
   Linear ODE actually solved: 
      (2*t-arccos(t)*(-t^2+1)^(1/2))*u(t)+(2*arccos(t)*(-t^2+1)^(1/2)*t-arccos(t)^2)*diff(u(t),t)+(-t^3*arccos(t)^2+t*arccos(t)^2)*d 
<- change of variables successful`

\[ y \left (x \right ) = x \left (c_{1} +\sin \left (x \right ) c_{2} \right ) \]

3.413 problem 1419

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