Internal problem ID [9756]
Internal file name [OUTPUT/8698_Monday_June_06_2022_05_14_31_AM_59060654/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1429.
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], [_2nd_order, _linear, `_with_symmetry_[0,F(x)]`]]
\[ \boxed {y^{\prime \prime }+\frac {\cos \left (x \right ) y^{\prime }}{\sin \left (x \right )}-\frac {y}{\sin \left (x \right )^{2}}=0} \]
In normal form the ode \begin {align*} y^{\prime \prime } \sin \left (x \right )^{2}+\frac {\sin \left (2 x \right ) y^{\prime }}{2}-y&=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 {\sin \left (2 x \right )}{2 \sin \left (x \right )^{2}}\\ q \left (x \right )&=-\frac {1}{\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 \frac {\sin \left (2 x \right )}{2 \sin \left (x \right )^{2}}d x \right )}d x\\ &= \int e^{-\ln \left (\sin \left (x \right )\right )} \,dx\\ &= \int \csc \left (x \right )d x\\ &= -\ln \left (\cot \left (x \right )+\csc \left (x \right )\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 {-\frac {1}{\sin \left (x \right )^{2}}}{\csc \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 1 \end {align*}
Hence \begin{align*} \lambda _1 &= + 1 \\ \lambda _2 &= - 1 \\ \end{align*} Which simplifies to \begin{align*} \lambda _1 &= 1 \\ \lambda _2 &= -1 \\ \end{align*} Since roots are real and distinct, then the solution is \begin{align*} y \left (\tau \right ) &= c_{1} e^{\lambda _1 \tau } + c_{2} e^{\lambda _2 \tau } \\ y \left (\tau \right ) &= c_{1} e^{\left (1\right )\tau } +c_{2} e^{\left (-1\right )\tau } \\ \end{align*} Or \[ y \left (\tau \right ) =c_{1} {\mathrm e}^{\tau }+c_{2} {\mathrm e}^{-\tau } \] The above solution is now transformed back to \(y\) using (6) which results in \begin {align*} y &= -\csc \left (x \right ) \left (\left (c_{1} -c_{2} \right ) \cos \left (x \right )-c_{1} -c_{2} \right ) \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -\csc \left (x \right ) \left (\left (c_{1} -c_{2} \right ) \cos \left (x \right )-c_{1} -c_{2} \right ) \\ \end{align*}
Verification of solutions
\[ y = -\csc \left (x \right ) \left (\left (c_{1} -c_{2} \right ) \cos \left (x \right )-c_{1} -c_{2} \right ) \] Verified OK.
In normal form the ode \begin {align*} y^{\prime \prime } \sin \left (x \right )^{2}+\frac {\sin \left (2 x \right ) y^{\prime }}{2}-y&=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 )&=-\csc \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 {-\csc \left (x \right )^{2}}}{c}\tag {6} \\ \tau '' &= \frac {\cot \left (x \right ) \csc \left (x \right )^{2}}{c \sqrt {-\csc \left (x \right )^{2}}} \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 {\cot \left (x \right ) \csc \left (x \right )^{2}}{c \sqrt {-\csc \left (x \right )^{2}}}+\cot \left (x \right )\frac {\sqrt {-\csc \left (x \right )^{2}}}{c}}{\left (\frac {\sqrt {-\csc \left (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 {-\csc \left (x \right )^{2}}d x}{c}\\ &= \frac {\ln \left (-\cot \left (x \right )+\csc \left (x \right )\right ) \sqrt {-\csc \left (x \right )^{2}}\, \sin \left (x \right )}{c} \end {align*}
Substituting the above into the solution obtained gives \[ y = -i \cot \left (x \right ) c_{2} +c_{1} \cosh \left (\ln \left (-\cot \left (x \right )+\csc \left (x \right )\right )\right ) \]
The solution(s) found are the following \begin{align*} \tag{1} y &= -i \cot \left (x \right ) c_{2} +c_{1} \left (-\frac {\cot \left (x \right )}{2}+\frac {\csc \left (x \right )}{2}+\frac {1}{-2 \cot \left (x \right )+2 \csc \left (x \right )}\right ) \\ \end{align*}
Verification of solutions
\[ y = -i \cot \left (x \right ) c_{2} +c_{1} \left (-\frac {\cot \left (x \right )}{2}+\frac {\csc \left (x \right )}{2}+\frac {1}{-2 \cot \left (x \right )+2 \csc \left (x \right )}\right ) \] Verified OK.
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying a symmetry of the form [xi=0, eta=F(x)] <- linear_1 successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 18
dsolve(diff(diff(y(x),x),x) = -1/sin(x)*cos(x)*diff(y(x),x)+1/sin(x)^2*y(x),y(x), singsol=all)
\[ y \left (x \right ) = \csc \left (x \right ) \left (\left (c_{1} -c_{2} \right ) \cos \left (x \right )+c_{1} +c_{2} \right ) \]
✓ Solution by Mathematica
Time used: 0.053 (sec). Leaf size: 25
DSolve[y''[x] == Csc[x]^2*y[x] - Cot[x]*y'[x],y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {c_1-i c_2 \cos (x)}{\sqrt {\sin ^2(x)}} \]