Internal problem ID [9745]
Internal file name [OUTPUT/8687_Monday_June_06_2022_05_12_26_AM_74752486/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1418.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_change_of_variable_on_y_method_2"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {y^{\prime \prime }+\frac {x \sin \left (x \right ) y^{\prime }}{\cos \left (x \right ) x -\sin \left (x \right )}-\frac {\sin \left (x \right ) y}{\cos \left (x \right ) x -\sin \left (x \right )}=0} \]
In normal form the ode \begin {align*} y^{\prime \prime } \left (-\cos \left (x \right ) x +\sin \left (x \right )\right )-x \sin \left (x \right ) y^{\prime }+y \sin \left (x \right )&=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 {x \sin \left (x \right )}{\cos \left (x \right ) x -\sin \left (x \right )}\\ q \left (x \right )&=\frac {\sin \left (x \right )}{-\cos \left (x \right ) x +\sin \left (x \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 \sin \left (x \right )}{\cos \left (x \right ) x -\sin \left (x \right )}+\frac {\sin \left (x \right )}{-\cos \left (x \right ) x +\sin \left (x \right )}&=0 \tag {5} \end {align*}
Solving (5) for \(n\) gives \begin {align*} n&=1 \tag {6} \end {align*}
Substituting this value in (3) gives \begin {align*} v^{\prime \prime }\left (x \right )+\left (\frac {2}{x}+\frac {x \sin \left (x \right )}{\cos \left (x \right ) x -\sin \left (x \right )}\right ) v^{\prime }\left (x \right )&=0 \\ v^{\prime \prime }\left (x \right )+\left (\frac {2}{x}+\frac {x \sin \left (x \right )}{\cos \left (x \right ) x -\sin \left (x \right )}\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 )+\left (\frac {2}{x}+\frac {x \sin \left (x \right )}{\cos \left (x \right ) x -\sin \left (x \right )}\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)\\ &= \frac {u \left (x^{2} \sin \left (x \right )+2 \cos \left (x \right ) x -2 \sin \left (x \right )\right )}{x \left (-\cos \left (x \right ) x +\sin \left (x \right )\right )} \end {align*}
Where \(f(x)=\frac {x^{2} \sin \left (x \right )+2 \cos \left (x \right ) x -2 \sin \left (x \right )}{x \left (-\cos \left (x \right ) x +\sin \left (x \right )\right )}\) and \(g(u)=u\). Integrating both sides gives \begin {align*} \frac {1}{u} \,du &= \frac {x^{2} \sin \left (x \right )+2 \cos \left (x \right ) x -2 \sin \left (x \right )}{x \left (-\cos \left (x \right ) x +\sin \left (x \right )\right )} \,d x\\ \int { \frac {1}{u} \,du} &= \int {\frac {x^{2} \sin \left (x \right )+2 \cos \left (x \right ) x -2 \sin \left (x \right )}{x \left (-\cos \left (x \right ) x +\sin \left (x \right )\right )} \,d x}\\ \ln \left (u \right )&=-2 \ln \left (x \right )-\ln \left (1+\tan \left (\frac {x}{2}\right )^{2}\right )+\ln \left (x \tan \left (\frac {x}{2}\right )^{2}-x +2 \tan \left (\frac {x}{2}\right )\right )+c_{1}\\ u&={\mathrm e}^{-2 \ln \left (x \right )-\ln \left (1+\tan \left (\frac {x}{2}\right )^{2}\right )+\ln \left (x \tan \left (\frac {x}{2}\right )^{2}-x +2 \tan \left (\frac {x}{2}\right )\right )+c_{1}}\\ &=c_{1} {\mathrm e}^{-2 \ln \left (x \right )-\ln \left (1+\tan \left (\frac {x}{2}\right )^{2}\right )+\ln \left (x \tan \left (\frac {x}{2}\right )^{2}-x +2 \tan \left (\frac {x}{2}\right )\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}\\ &= -\frac {c_{1} \sin \left (x \right )}{x}+c_{2} \end {align*}
Hence \begin {align*} y&= v \left (x \right ) x^{n}\\ &= \left (-\frac {c_{1} \sin \left (x \right )}{x}+c_{2} \right ) x\\ &= -c_{1} \sin \left (x \right )+c_{2} x\\ \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \left (-\frac {c_{1} \sin \left (x \right )}{x}+c_{2} \right ) x \\ \end{align*}
Verification of solutions
\[ y = \left (-\frac {c_{1} \sin \left (x \right )}{x}+c_{2} \right ) x \] 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 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 a symmetry of the form [xi=0, eta=F(x)] trying 2nd order exact linear trying symmetries linear in x and y(x) trying to convert to a linear ODE with constant coefficients 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 a symmetry of the form [xi=0, eta=F(x)] trying 2nd order exact linear trying symmetries linear in x and y(x) <- linear symmetries successful Change of variables used: [x = arcsin(t)] Linear ODE actually solved: -t*u(t)+t^2*diff(u(t),t)+((-t^2+1)^(3/2)*arcsin(t)+t^3-t)*diff(diff(u(t),t),t) = 0 <- change of variables successful`
✓ Solution by Maple
Time used: 0.36 (sec). Leaf size: 47
dsolve(diff(diff(y(x),x),x) = -x*sin(x)/(cos(x)*x-sin(x))*diff(y(x),x)+sin(x)/(cos(x)*x-sin(x))*y(x),y(x), singsol=all)
\[ y \left (x \right ) = \sin \left (x \right ) \left (c_{1} +c_{2} \left (\int {\mathrm e}^{-\left (\int \frac {2 \cos \left (x \right ) \cot \left (x \right ) x -3 \cos \left (x \right )+\sec \left (x \right )}{-\sin \left (x \right )+\cos \left (x \right ) x}d x \right )} \cos \left (x \right )d x \right )\right ) \]
✓ Solution by Mathematica
Time used: 0.172 (sec). Leaf size: 15
DSolve[y''[x] == (Sin[x]*y[x])/(x*Cos[x] - Sin[x]) - (x*Sin[x]*y'[x])/(x*Cos[x] - Sin[x]),y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to c_1 x+c_2 \sin (x) \]