Internal problem ID [15409]
Internal file name [OUTPUT/15410_Wednesday_May_08_2024_03_58_16_PM_27062736/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: 641.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "reduction_of_order", "second_order_change_of_variable_on_x_method_1", "second_order_change_of_variable_on_x_method_2"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {y^{\prime \prime }+y^{\prime } \tan \left (x \right )+y \cos \left (x \right )^{2}=0} \] Given that one solution of the ode is \begin {align*} y_1 &= \cos \left (\sin \left (x \right )\right ) \end {align*}
Given one basis solution \(y_{1}\left (x \right )\), then the second basis solution is given by \[ y_{2}\left (x \right ) = y_{1} \left (\int \frac {{\mathrm e}^{-\left (\int p d x \right )}}{y_{1}^{2}}d x \right ) \] Where \(p(x)\) is the coefficient of \(y^{\prime }\) when the ode is written in the normal form \[ y^{\prime \prime }+p \left (x \right ) y^{\prime }+q \left (x \right ) y = f \left (x \right ) \] Looking at the ode to solve shows that \[ p \left (x \right ) = \tan \left (x \right ) \] Therefore \begin{align*} y_{2}\left (x \right ) &= \cos \left (\sin \left (x \right )\right ) \left (\int \frac {{\mathrm e}^{-\left (\int \tan \left (x \right )d x \right )}}{\cos \left (\sin \left (x \right )\right )^{2}}d x \right ) \\ y_{2}\left (x \right ) &= \cos \left (\sin \left (x \right )\right ) \int \frac {\cos \left (x \right )}{\cos \left (\sin \left (x \right )\right )^{2}} , dx \\ y_{2}\left (x \right ) &= \cos \left (\sin \left (x \right )\right ) \left (\int \cos \left (x \right ) \sec \left (\sin \left (x \right )\right )^{2}d x \right ) \\ y_{2}\left (x \right ) &= \cos \left (\sin \left (x \right )\right ) \tan \left (\sin \left (x \right )\right ) \\ \end{align*} Hence the solution is \begin{align*} y &= c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \\ &= \cos \left (\sin \left (x \right )\right ) c_{1} +c_{2} \cos \left (\sin \left (x \right )\right ) \tan \left (\sin \left (x \right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= \cos \left (\sin \left (x \right )\right ) c_{1} +c_{2} \cos \left (\sin \left (x \right )\right ) \tan \left (\sin \left (x \right )\right ) \\ \end{align*}
Verification of solutions
\[ y = \cos \left (\sin \left (x \right )\right ) c_{1} +c_{2} \cos \left (\sin \left (x \right )\right ) \tan \left (\sin \left (x \right )\right ) \] Verified OK.
Maple trace
`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 <- constant coefficients successful Change of variables used: [x = arcsin(t)] Linear ODE actually solved: (-2*t^2+2)*u(t)+(-2*t^2+2)*diff(diff(u(t),t),t) = 0 <- change of variables successful`
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 15
dsolve([diff(y(x),x$2)+tan(x)*diff(y(x),x)+cos(x)^2*y(x)=0,cos(sin(x))],singsol=all)
\[ y \left (x \right ) = c_{1} \sin \left (\sin \left (x \right )\right )+c_{2} \cos \left (\sin \left (x \right )\right ) \]
✓ Solution by Mathematica
Time used: 0.066 (sec). Leaf size: 18
DSolve[y''[x]+Tan[x]*y'[x]+Cos[x]^2*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to c_2 \sin (\sin (x))+c_1 \cos (\sin (x)) \]