Internal problem ID [9758]
Internal file name [OUTPUT/8700_Monday_June_06_2022_05_14_55_AM_20792929/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1431.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_ode_non_constant_coeff_transformation_on_B"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {y^{\prime \prime }-\frac {\cos \left (2 x \right ) y^{\prime }}{\sin \left (2 x \right )}+2 y=0} \]
Given an ode of the form \begin {align*} A y^{\prime \prime } + B y^{\prime } + C y &= F(x) \end {align*}
This method reduces the order ode the ODE by one by applying the transformation \begin {align*} y&= B v \end {align*}
This results in \begin {align*} y' &=B' v+ v' B \\ y'' &=B'' v+ B' v' +v'' B + v' B' \\ &=v'' B+2 v'+ B'+B'' v \end {align*}
And now the original ode becomes \begin {align*} A\left ( v'' B+2v'B'+ B'' v\right )+B\left ( B'v+ v' B\right ) +CBv & =0\\ ABv'' +\left ( 2AB'+B^{2}\right ) v'+\left (AB''+BB'+CB\right ) v & =0 \tag {1} \end {align*}
If the term \(AB''+BB'+CB\) is zero, then this method works and can be used to solve \[ ABv''+\left ( 2AB' +B^{2}\right ) v'=0 \] By Using \(u=v'\) which reduces the order of the above ode to one. The new ode is \[ ABu'+\left ( 2AB'+B^{2}\right ) u=0 \] The above ode is first order ode which is solved for \(u\). Now a new ode \(v'=u\) is solved for \(v\) as first order ode. Then the final solution is obtain from \(y=Bv\).
This method works only if the term \(AB''+BB'+CB\) is zero. The given ODE shows that \begin {align*} A &= \sin \left (2 x \right )\\ B &= -\cos \left (2 x \right )\\ C &= 2 \sin \left (2 x \right )\\ F &= 0 \end {align*}
The above shows that for this ode \begin {align*} AB''+BB'+CB &= \left (\sin \left (2 x \right )\right ) \left (4 \cos \left (2 x \right )\right ) + \left (-\cos \left (2 x \right )\right ) \left (2 \sin \left (2 x \right )\right ) + \left (2 \sin \left (2 x \right )\right ) \left (-\cos \left (2 x \right )\right ) \\ &=0 \end {align*}
Hence the ode in \(v\) given in (1) now simplifies to \begin {align*} -\frac {\sin \left (4 x \right )}{2} v'' +\left ( -3 \cos \left (2 x \right )^{2}+4\right ) v' & =0 \end {align*}
Now by applying \(v'=u\) the above becomes \begin {align*} -\sin \left (2 x \right ) \cos \left (2 x \right ) u^{\prime }\left (x \right )-3 \cos \left (2 x \right )^{2} u \left (x \right )+4 u \left (x \right ) = 0 \end {align*}
Which is now solved for \(u\). In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= -\frac {u \left (3 \cos \left (2 x \right )^{2}-4\right )}{\sin \left (2 x \right ) \cos \left (2 x \right )} \end {align*}
Where \(f(x)=-\frac {3 \cos \left (2 x \right )^{2}-4}{\sin \left (2 x \right ) \cos \left (2 x \right )}\) and \(g(u)=u\). Integrating both sides gives \begin {align*} \frac {1}{u} \,du &= -\frac {3 \cos \left (2 x \right )^{2}-4}{\sin \left (2 x \right ) \cos \left (2 x \right )} \,d x\\ \int { \frac {1}{u} \,du} &= \int {-\frac {3 \cos \left (2 x \right )^{2}-4}{\sin \left (2 x \right ) \cos \left (2 x \right )} \,d x}\\ \ln \left (u \right )&=-\frac {3 \ln \left (\sin \left (2 x \right )\right )}{2}+2 \ln \left (\tan \left (2 x \right )\right )+c_{1}\\ u&={\mathrm e}^{-\frac {3 \ln \left (\sin \left (2 x \right )\right )}{2}+2 \ln \left (\tan \left (2 x \right )\right )+c_{1}}\\ &=c_{1} {\mathrm e}^{-\frac {3 \ln \left (\sin \left (2 x \right )\right )}{2}+2 \ln \left (\tan \left (2 x \right )\right )} \end {align*}
The ode for \(v\) now becomes \begin {align*} v' &= u\\ &=c_{1} {\mathrm e}^{-\frac {3 \ln \left (\sin \left (2 x \right )\right )}{2}+2 \ln \left (\tan \left (2 x \right )\right )} \end {align*}
Which is now solved for \(v\). Integrating both sides gives \begin {align*} v \left (x \right ) &= \int { c_{1} {\mathrm e}^{-\frac {3 \ln \left (\sin \left (2 x \right )\right )}{2}+2 \ln \left (\tan \left (2 x \right )\right )}\,\mathop {\mathrm {d}x}}\\ &= \frac {c_{1} \sqrt {\cos \left (2 x \right )^{2} \sin \left (2 x \right )}\, \left (2 \sqrt {\sin \left (2 x \right )+1}\, \sqrt {-2 \sin \left (2 x \right )+2}\, \sqrt {-\sin \left (2 x \right )}\, \operatorname {EllipticE}\left (\sqrt {\sin \left (2 x \right )+1}, \frac {\sqrt {2}}{2}\right )-\sqrt {\sin \left (2 x \right )+1}\, \sqrt {-2 \sin \left (2 x \right )+2}\, \sqrt {-\sin \left (2 x \right )}\, \operatorname {EllipticF}\left (\sqrt {\sin \left (2 x \right )+1}, \frac {\sqrt {2}}{2}\right )-2 \cos \left (2 x \right )^{2}+2\right )}{4 \sqrt {-\sin \left (2 x \right ) \left (\sin \left (2 x \right )-1\right ) \left (\sin \left (2 x \right )+1\right )}\, \cos \left (2 x \right ) \sqrt {\sin \left (2 x \right )}}+c_{2} \end {align*}
Therefore the solution is \begin {align*} y(x) &= B v\\ &= \left (-\cos \left (2 x \right )\right ) \left (\frac {c_{1} \sqrt {\cos \left (2 x \right )^{2} \sin \left (2 x \right )}\, \left (2 \sqrt {\sin \left (2 x \right )+1}\, \sqrt {-2 \sin \left (2 x \right )+2}\, \sqrt {-\sin \left (2 x \right )}\, \operatorname {EllipticE}\left (\sqrt {\sin \left (2 x \right )+1}, \frac {\sqrt {2}}{2}\right )-\sqrt {\sin \left (2 x \right )+1}\, \sqrt {-2 \sin \left (2 x \right )+2}\, \sqrt {-\sin \left (2 x \right )}\, \operatorname {EllipticF}\left (\sqrt {\sin \left (2 x \right )+1}, \frac {\sqrt {2}}{2}\right )-2 \cos \left (2 x \right )^{2}+2\right )}{4 \sqrt {-\sin \left (2 x \right ) \left (\sin \left (2 x \right )-1\right ) \left (\sin \left (2 x \right )+1\right )}\, \cos \left (2 x \right ) \sqrt {\sin \left (2 x \right )}}+c_{2}\right ) \\ &= \frac {-\sqrt {2}\, \cos \left (2 x \right ) \operatorname {csgn}\left (\cos \left (\frac {\pi }{4}+x \right )\right ) \sqrt {-\sin \left (2 x \right )}\, \operatorname {EllipticE}\left (\sin \left (\frac {\pi }{4}+x \right ) \sqrt {2}, \frac {\sqrt {2}}{2}\right ) c_{1} +\cos \left (2 x \right ) \operatorname {csgn}\left (\cos \left (\frac {\pi }{4}+x \right )\right ) \sqrt {-\sin \left (2 x \right )}\, \operatorname {EllipticF}\left (\sin \left (\frac {\pi }{4}+x \right ), \sqrt {2}\right ) c_{1} -2 c_{2} \sqrt {\sin \left (2 x \right )}\, \cos \left (2 x \right )-\sin \left (2 x \right )^{2} c_{1}}{2 \sqrt {\sin \left (2 x \right )}} \end {align*}
Simplifying the solution \(y = \frac {-\sqrt {2}\, \cos \left (2 x \right ) \operatorname {csgn}\left (\cos \left (\frac {\pi }{4}+x \right )\right ) \sqrt {-\sin \left (2 x \right )}\, \operatorname {EllipticE}\left (\sin \left (\frac {\pi }{4}+x \right ) \sqrt {2}, \frac {\sqrt {2}}{2}\right ) c_{1} +\cos \left (2 x \right ) \operatorname {csgn}\left (\cos \left (\frac {\pi }{4}+x \right )\right ) \sqrt {-\sin \left (2 x \right )}\, \operatorname {EllipticF}\left (\sin \left (\frac {\pi }{4}+x \right ), \sqrt {2}\right ) c_{1} -2 c_{2} \sqrt {\sin \left (2 x \right )}\, \cos \left (2 x \right )-\sin \left (2 x \right )^{2} c_{1}}{2 \sqrt {\sin \left (2 x \right )}}\) to \(y = \frac {-\sqrt {2}\, \cos \left (2 x \right ) \sqrt {-\sin \left (2 x \right )}\, \operatorname {EllipticE}\left (\sin \left (\frac {\pi }{4}+x \right ) \sqrt {2}, \frac {\sqrt {2}}{2}\right ) c_{1} +\cos \left (2 x \right ) \sqrt {-\sin \left (2 x \right )}\, \operatorname {EllipticF}\left (\sin \left (\frac {\pi }{4}+x \right ), \sqrt {2}\right ) c_{1} -2 c_{2} \sqrt {\sin \left (2 x \right )}\, \cos \left (2 x \right )-\sin \left (2 x \right )^{2} c_{1}}{2 \sqrt {\sin \left (2 x \right )}}\)
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {-\sqrt {2}\, \cos \left (2 x \right ) \sqrt {-\sin \left (2 x \right )}\, \operatorname {EllipticE}\left (\sin \left (\frac {\pi }{4}+x \right ) \sqrt {2}, \frac {\sqrt {2}}{2}\right ) c_{1} +\cos \left (2 x \right ) \sqrt {-\sin \left (2 x \right )}\, \operatorname {EllipticF}\left (\sin \left (\frac {\pi }{4}+x \right ), \sqrt {2}\right ) c_{1} -2 c_{2} \sqrt {\sin \left (2 x \right )}\, \cos \left (2 x \right )-\sin \left (2 x \right )^{2} c_{1}}{2 \sqrt {\sin \left (2 x \right )}} \\ \end{align*}
Verification of solutions
\[ y = \frac {-\sqrt {2}\, \cos \left (2 x \right ) \sqrt {-\sin \left (2 x \right )}\, \operatorname {EllipticE}\left (\sin \left (\frac {\pi }{4}+x \right ) \sqrt {2}, \frac {\sqrt {2}}{2}\right ) c_{1} +\cos \left (2 x \right ) \sqrt {-\sin \left (2 x \right )}\, \operatorname {EllipticF}\left (\sin \left (\frac {\pi }{4}+x \right ), \sqrt {2}\right ) c_{1} -2 c_{2} \sqrt {\sin \left (2 x \right )}\, \cos \left (2 x \right )-\sin \left (2 x \right )^{2} c_{1}}{2 \sqrt {\sin \left (2 x \right )}} \] 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) Group is reducible, not completely reducible Solution has integrals. Trying a special function solution free of integrals... -> Trying a solution in terms of special functions: -> Bessel -> elliptic -> Legendre <- Legendre successful <- special function solution successful -> Trying to convert hypergeometric functions to elementary form... <- elementary form is not straightforward to achieve - returning special function solution free of uncomputed integrals <- Kovacics algorithm successful Change of variables used: [x = 1/2*arccos(t)] Linear ODE actually solved: (-t^2+1)*u(t)+(t^3-t)*diff(u(t),t)+(2*t^4-4*t^2+2)*diff(diff(u(t),t),t) = 0 <- change of variables successful`
✓ Solution by Maple
Time used: 0.25 (sec). Leaf size: 30
dsolve(diff(diff(y(x),x),x) = cos(2*x)/sin(2*x)*diff(y(x),x)-2*y(x),y(x), singsol=all)
\[ y \left (x \right ) = \sin \left (2 x \right )^{\frac {3}{4}} \left (c_{1} \operatorname {LegendreP}\left (\frac {1}{4}, \frac {3}{4}, \cos \left (2 x \right )\right )+c_{2} \operatorname {LegendreQ}\left (\frac {1}{4}, \frac {3}{4}, \cos \left (2 x \right )\right )\right ) \]
✓ Solution by Mathematica
Time used: 20.33 (sec). Leaf size: 64
DSolve[y''[x] == -2*y[x] + Cot[2*x]*y'[x],y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to -\frac {2}{3} c_2 \cos (2 x) \cos ^{\frac {3}{2}}(x) \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {3}{4},\frac {7}{4},\cos ^2(x)\right )+\frac {1}{2} c_1 \cos (2 x)-2 c_2 \sin ^2(x)^{3/4} \cos ^{\frac {3}{2}}(x) \]