Internal problem ID [9401]
Internal file name [OUTPUT/8338_Monday_June_06_2022_02_46_35_AM_3542497/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1069.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "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]]
\[ \boxed {y^{\prime \prime }-y^{\prime } \cot \left (x \right )+y \sin \left (x \right )^{2}=0} \]
In normal form the ode \begin {align*} y^{\prime \prime }-y^{\prime } \cot \left (x \right )+y \sin \left (x \right )^{2}&=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 )&=-\cot \left (x \right )\\ q \left (x \right )&=\sin \left (x \right )^{2} \end {align*}
Applying change of variables \(\tau = g \left (x \right )\) to (2) gives \begin {align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+p_{1} \left (\frac {d}{d \tau }y \left (\tau \right )\right )+q_{1} y \left (\tau \right )&=0 \tag {3} \end {align*}
Where \(\tau \) is the new independent variable, and \begin {align*} p_{1} \left (\tau \right ) &=\frac {\tau ^{\prime \prime }\left (x \right )+p \left (x \right ) \tau ^{\prime }\left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\tag {4} \\ q_{1} \left (\tau \right ) &=\frac {q \left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\tag {5} \end {align*}
Let \(p_{1} = 0\). Eq (4) simplifies to \begin {align*} \tau ^{\prime \prime }\left (x \right )+p \left (x \right ) \tau ^{\prime }\left (x \right )&=0 \end {align*}
This ode is solved resulting in \begin {align*} \tau &= \int {\mathrm e}^{-\left (\int p \left (x \right )d x \right )}d x\\ &= \int {\mathrm e}^{-\left (\int -\cot \left (x \right )d x \right )}d x\\ &= \int e^{\ln \left (\sin \left (x \right )\right )} \,dx\\ &= \int \sin \left (x \right )d x\\ &= -\cos \left (x \right )\tag {6} \end {align*}
Using (6) to evaluate \(q_{1}\) from (5) gives \begin {align*} q_{1} \left (\tau \right ) &= \frac {q \left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\\ &= \frac {\sin \left (x \right )^{2}}{\sin \left (x \right )^{2}}\\ &= 1\tag {7} \end {align*}
Substituting the above in (3) and noting that now \(p_{1} = 0\) results in \begin {align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+q_{1} y \left (\tau \right )&=0 \\ \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+y \left (\tau \right )&=0 \end {align*}
The above ode is now solved for \(y \left (\tau \right )\).This is second order with constant coefficients homogeneous ODE. In standard form the ODE is \[ A y''(\tau ) + B y'(\tau ) + C y(\tau ) = 0 \] Where in the above \(A=1, B=0, C=1\). Let the solution be \(y \left (\tau \right )=e^{\lambda \tau }\). Substituting this into the ODE gives \[ \lambda ^{2} {\mathrm e}^{\lambda \tau }+{\mathrm e}^{\lambda \tau } = 0 \tag {1} \] Since exponential function is never zero, then dividing Eq(2) throughout by \(e^{\lambda \tau }\) gives \[ \lambda ^{2}+1 = 0 \tag {2} \] 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 \(A=1, B=0, C=1\) 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 (1\right )}\\ &= \pm i \end {align*}
Hence \begin {align*} \lambda _1 &= + i\\ \lambda _2 &= - i \end {align*}
Which simplifies to \begin{align*} \lambda _1 &= i \\ \lambda _2 &= -i \\ \end{align*} Since roots are complex conjugate of each others, then let the roots be \[ \lambda _{1,2} = \alpha \pm i \beta \] Where \(\alpha =0\) and \(\beta =1\). Therefore the final solution, when using Euler relation, can be written as \[ y \left (\tau \right ) = e^{\alpha \tau } \left ( c_{1} \cos (\beta \tau ) + c_{2} \sin (\beta \tau ) \right ) \] Which becomes \[ y \left (\tau \right ) = e^{0}\left (c_{1} \cos \left (\tau \right )+c_{2} \sin \left (\tau \right )\right ) \] Or \[ y \left (\tau \right ) = c_{1} \cos \left (\tau \right )+c_{2} \sin \left (\tau \right ) \] The above solution is now transformed back to \(y\) using (6) which results in \begin {align*} y &= c_{1} \cos \left (\cos \left (x \right )\right )-c_{2} \sin \left (\cos \left (x \right )\right ) \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \cos \left (\cos \left (x \right )\right )-c_{2} \sin \left (\cos \left (x \right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} \cos \left (\cos \left (x \right )\right )-c_{2} \sin \left (\cos \left (x \right )\right ) \] Verified OK.
In normal form the ode \begin {align*} y^{\prime \prime }-y^{\prime } \cot \left (x \right )+y \sin \left (x \right )^{2}&=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 )&=-\cot \left (x \right )\\ q \left (x \right )&=\sin \left (x \right )^{2} \end {align*}
Applying change of variables \(\tau = g \left (x \right )\) to (2) results \begin {align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+p_{1} \left (\frac {d}{d \tau }y \left (\tau \right )\right )+q_{1} y \left (\tau \right )&=0 \tag {3} \end {align*}
Where \(\tau \) is the new independent variable, and \begin {align*} p_{1} \left (\tau \right ) &=\frac {\tau ^{\prime \prime }\left (x \right )+p \left (x \right ) \tau ^{\prime }\left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\tag {4} \\ q_{1} \left (\tau \right ) &=\frac {q \left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\tag {5} \end {align*}
Let \(q_1=c^2\) where \(c\) is some constant. Therefore from (5) \begin {align*} \tau ' &= \frac {1}{c}\sqrt {q}\\ &= \frac {\sqrt {\frac {1}{2}-\frac {\cos \left (2 x \right )}{2}}}{c}\tag {6} \\ \tau '' &= \frac {2 \cos \left (x \right ) \sin \left (x \right )}{c \sqrt {2-2 \cos \left (2 x \right )}} \end {align*}
Substituting the above into (4) results in \begin {align*} p_{1} \left (\tau \right ) &=\frac {\tau ^{\prime \prime }\left (x \right )+p \left (x \right ) \tau ^{\prime }\left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\\ &=\frac {\frac {2 \cos \left (x \right ) \sin \left (x \right )}{c \sqrt {2-2 \cos \left (2 x \right )}}-\cot \left (x \right )\frac {\sqrt {\frac {1}{2}-\frac {\cos \left (2 x \right )}{2}}}{c}}{\left (\frac {\sqrt {\frac {1}{2}-\frac {\cos \left (2 x \right )}{2}}}{c}\right )^2} \\ &=0 \end {align*}
Therefore ode (3) now becomes \begin {align*} y \left (\tau \right )'' + p_1 y \left (\tau \right )' + q_1 y \left (\tau \right ) &= 0 \\ \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+c^{2} y \left (\tau \right ) &= 0 \tag {7} \end {align*}
The above ode is now solved for \(y \left (\tau \right )\). Since the ode is now constant coefficients, it can be easily solved to give \begin {align*} y \left (\tau \right ) &= c_{1} \cos \left (c \tau \right )+c_{2} \sin \left (c \tau \right ) \end {align*}
Now from (6) \begin {align*} \tau &= \int \frac {1}{c} \sqrt q \,dx \\ &= \frac {\int \sqrt {\frac {1}{2}-\frac {\cos \left (2 x \right )}{2}}d x}{c}\\ &= -\frac {\sqrt {\frac {1}{2}-\frac {\cos \left (2 x \right )}{2}}\, \cos \left (x \right )}{c \sin \left (x \right )} \end {align*}
Substituting the above into the solution obtained gives \[ y = c_{1} \cos \left (\cos \left (x \right )\right )-c_{2} \sin \left ({| \sin \left (x \right )|} \cot \left (x \right )\right ) \]
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \cos \left (\cos \left (x \right )\right )-c_{2} \sin \left ({| \sin \left (x \right )|} \cot \left (x \right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} \cos \left (\cos \left (x \right )\right )-c_{2} \sin \left ({| \sin \left (x \right )|} \cot \left (x \right )\right ) \] 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 checking if the LODE is of Euler type trying a symmetry of the form [xi=0, eta=F(x)] <- linear_1 successful Change of variables used: [x = arcsin(t)] Linear ODE actually solved: 2*t^3*u(t)-2*diff(u(t),t)+(-2*t^3+2*t)*diff(diff(u(t),t),t) = 0 <- change of variables successful`
✓ Solution by Maple
Time used: 0.141 (sec). Leaf size: 15
dsolve(diff(diff(y(x),x),x)-diff(y(x),x)*cot(x)+y(x)*sin(x)^2=0,y(x), singsol=all)
\[ y \left (x \right ) = c_{1} \sin \left (\cos \left (x \right )\right )+c_{2} \cos \left (\cos \left (x \right )\right ) \]
✓ Solution by Mathematica
Time used: 0.078 (sec). Leaf size: 19
DSolve[Sin[x]^2*y[x] - Cot[x]*y'[x] + y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to c_1 \cos (\cos (x))-c_2 \sin (\cos (x)) \]