Internal problem ID [9748]
Internal file name [OUTPUT/8690_Monday_June_06_2022_05_13_09_AM_18212898/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1421.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_change_of_variable_on_y_method_1"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {y^{\prime \prime }+\frac {a \left (n -1\right ) \sin \left (2 a x \right ) y^{\prime }}{\cos \left (a x \right )^{2}}+\frac {n \,a^{2} \left (\left (n -1\right ) \sin \left (a x \right )^{2}+\cos \left (a x \right )^{2}\right ) y}{\cos \left (a x \right )^{2}}=0} \]
In normal form the given ode is written as \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 {a \sin \left (2 a x \right ) n -a \sin \left (2 a x \right )}{\cos \left (a x \right )^{2}}\\ q \left (x \right )&=\frac {n^{2} a^{2} \sin \left (a x \right )^{2}+2 \cos \left (a x \right )^{2} a^{2} n -n \,a^{2}}{\cos \left (a x \right )^{2}} \end {align*}
Calculating the Liouville ode invariant \(Q\) given by \begin {align*} Q &= q - \frac {p'}{2}- \frac {p^2}{4} \\ &= \frac {n^{2} a^{2} \sin \left (a x \right )^{2}+2 \cos \left (a x \right )^{2} a^{2} n -n \,a^{2}}{\cos \left (a x \right )^{2}} - \frac {\left (\frac {a \sin \left (2 a x \right ) n -a \sin \left (2 a x \right )}{\cos \left (a x \right )^{2}}\right )'}{2}- \frac {\left (\frac {a \sin \left (2 a x \right ) n -a \sin \left (2 a x \right )}{\cos \left (a x \right )^{2}}\right )^2}{4} \\ &= \frac {n^{2} a^{2} \sin \left (a x \right )^{2}+2 \cos \left (a x \right )^{2} a^{2} n -n \,a^{2}}{\cos \left (a x \right )^{2}} - \frac {\left (\frac {2 a^{2} \cos \left (2 a x \right ) n -2 a^{2} \cos \left (2 a x \right )}{\cos \left (a x \right )^{2}}+\frac {2 \left (a \sin \left (2 a x \right ) n -a \sin \left (2 a x \right )\right ) a \sin \left (a x \right )}{\cos \left (a x \right )^{3}}\right )}{2}- \frac {\left (\frac {\left (a \sin \left (2 a x \right ) n -a \sin \left (2 a x \right )\right )^{2}}{\cos \left (a x \right )^{4}}\right )}{4} \\ &= \frac {n^{2} a^{2} \sin \left (a x \right )^{2}+2 \cos \left (a x \right )^{2} a^{2} n -n \,a^{2}}{\cos \left (a x \right )^{2}} - \left (\frac {2 a^{2} \cos \left (2 a x \right ) n -2 a^{2} \cos \left (2 a x \right )}{2 \cos \left (a x \right )^{2}}+\frac {\left (a \sin \left (2 a x \right ) n -a \sin \left (2 a x \right )\right ) a \sin \left (a x \right )}{\cos \left (a x \right )^{3}}\right )-\frac {\left (a \sin \left (2 a x \right ) n -a \sin \left (2 a x \right )\right )^{2}}{4 \cos \left (a x \right )^{4}}\\ &= a^{2} \end {align*}
Since the Liouville ode invariant does not depend on the independent variable \(x\) then the transformation \begin {align*} y = v \left (x \right ) z \left (x \right )\tag {3} \end {align*}
is used to change the original ode to a constant coefficients ode in \(v\). In (3) the term \(z \left (x \right )\) is given by \begin {align*} z \left (x \right )&={\mathrm e}^{-\left (\int \frac {p \left (x \right )}{2}d x \right )}\\ &= e^{-\int \frac {\frac {a \sin \left (2 a x \right ) n -a \sin \left (2 a x \right )}{\cos \left (a x \right )^{2}}}{2} }\\ &= \left (\sec \left (a x \right )^{2}\right )^{-\frac {n}{2}+\frac {1}{2}}\tag {5} \end {align*}
Hence (3) becomes \begin {align*} y = v \left (x \right ) \left (\sec \left (a x \right )^{2}\right )^{-\frac {n}{2}+\frac {1}{2}}\tag {4} \end {align*}
Applying this change of variable to the original ode results in \begin {align*} \cos \left (a x \right )^{2} \left (v \left (x \right ) a^{2}+v^{\prime \prime }\left (x \right )\right ) \left (\sec \left (a x \right )^{2}\right )^{-\frac {n}{2}+\frac {1}{2}} = 0 \end {align*}
Which is now solved for \(v \left (x \right )\) This is second order with constant coefficients homogeneous ODE. In standard form the ODE is \[ A v''(x) + B v'(x) + C v(x) = 0 \] Where in the above \(A=1, B=0, C=a^{2}\). Let the solution be \(v \left (x \right )=e^{\lambda x}\). Substituting this into the ODE gives \[ \lambda ^{2} {\mathrm e}^{\lambda x}+a^{2} {\mathrm e}^{\lambda x} = 0 \tag {1} \] Since exponential function is never zero, then dividing Eq(2) throughout by \(e^{\lambda x}\) gives \[ a^{2}+\lambda ^{2} = 0 \tag {2} \] Equation (2) is the characteristic equation of the ODE. Its roots determine the general solution form.Using the quadratic formula \[ \lambda _{1,2} = \frac {-B}{2 A} \pm \frac {1}{2 A} \sqrt {B^2 - 4 A C} \] Substituting \(A=1, B=0, C=a^{2}\) into the above gives \begin {align*} \lambda _{1,2} &= \frac {0}{(2) \left (1\right )} \pm \frac {1}{(2) \left (1\right )} \sqrt {0^2 - (4) \left (1\right )\left (a^{2}\right )}\\ &= \pm \sqrt {-a^{2}} \end {align*}
Hence \begin{align*} \lambda _1 &= + \sqrt {-a^{2}} \\ \lambda _2 &= - \sqrt {-a^{2}} \\ \end{align*} Which simplifies to \begin{align*} \lambda _1 &= \sqrt {-a^{2}} \\ \lambda _2 &= -\sqrt {-a^{2}} \\ \end{align*} Since roots are real and distinct, then the solution is \begin{align*} v \left (x \right ) &= c_{1} e^{\lambda _1 x} + c_{2} e^{\lambda _2 x} \\ v \left (x \right ) &= c_{1} e^{\left (\sqrt {-a^{2}}\right )x} +c_{2} e^{\left (-\sqrt {-a^{2}}\right )x} \\ \end{align*} Or \[ v \left (x \right ) =c_{1} {\mathrm e}^{\sqrt {-a^{2}}\, x}+c_{2} {\mathrm e}^{-\sqrt {-a^{2}}\, x} \] Now that \(v \left (x \right )\) is known, then \begin {align*} y&= v \left (x \right ) z \left (x \right )\\ &= \left (c_{1} {\mathrm e}^{\sqrt {-a^{2}}\, x}+c_{2} {\mathrm e}^{-\sqrt {-a^{2}}\, x}\right ) \left (z \left (x \right )\right )\tag {7} \end {align*}
But from (5) \begin {align*} z \left (x \right )&= \left (\sec \left (a x \right )^{2}\right )^{-\frac {n}{2}+\frac {1}{2}} \end {align*}
Hence (7) becomes \begin {align*} y = \left (c_{1} {\mathrm e}^{\sqrt {-a^{2}}\, x}+c_{2} {\mathrm e}^{-\sqrt {-a^{2}}\, x}\right ) \left (\sec \left (a x \right )^{2}\right )^{-\frac {n}{2}+\frac {1}{2}} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \left (c_{1} {\mathrm e}^{\sqrt {-a^{2}}\, x}+c_{2} {\mathrm e}^{-\sqrt {-a^{2}}\, x}\right ) \left (\sec \left (a x \right )^{2}\right )^{-\frac {n}{2}+\frac {1}{2}} \\ \end{align*}
Verification of solutions
\[ y = \left (c_{1} {\mathrm e}^{\sqrt {-a^{2}}\, x}+c_{2} {\mathrm e}^{-\sqrt {-a^{2}}\, x}\right ) \left (\sec \left (a x \right )^{2}\right )^{-\frac {n}{2}+\frac {1}{2}} \] 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 Group is reducible or imprimitive <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 28
dsolve(diff(diff(y(x),x),x) = -a*(n-1)*sin(2*a*x)/cos(a*x)^2*diff(y(x),x)-n*a^2*((n-1)*sin(a*x)^2+cos(a*x)^2)/cos(a*x)^2*y(x),y(x), singsol=all)
\[ y \left (x \right ) = \sec \left (a x \right )^{-n +1} \left (c_{1} \sin \left (a x \right )+c_{2} \cos \left (a x \right )\right ) \]
✓ Solution by Mathematica
Time used: 0.244 (sec). Leaf size: 65
DSolve[y''[x] == -(a^2*n*Sec[a*x]^2*(Cos[a*x]^2 + (-1 + n)*Sin[a*x]^2)*y[x]) - a*(-1 + n)*Sec[a*x]^2*Sin[2*a*x]*y'[x],y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {2^{-n} \left (2 a c_1-i c_2 e^{2 i a x}\right ) \left (e^{-i a x}+e^{i a x}\right )^n}{a \left (1+e^{2 i a x}\right )} \]