Internal problem ID [9744]
Internal file name [OUTPUT/8686_Monday_June_06_2022_05_11_51_AM_31405053/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1417.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_change_of_variable_on_x_method_2"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {y^{\prime \prime }+\frac {\left (\sin \left (x \right )^{2}-\cos \left (x \right )\right ) y^{\prime }}{\sin \left (x \right )}+y \sin \left (x \right )^{2}=0} \]
In normal form the ode \begin {align*} -y^{\prime \prime } \sin \left (x \right )+\left (-\sin \left (x \right )^{2}+\cos \left (x \right )\right ) y^{\prime }-y \sin \left (x \right )^{3}&=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 (x \right )^{2}-\cos \left (x \right )}{\sin \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 \frac {\sin \left (x \right )^{2}-\cos \left (x \right )}{\sin \left (x \right )}d x \right )}d x\\ &= \int e^{\cos \left (x \right )+\ln \left (\sin \left (x \right )\right )} \,dx\\ &= \int \sin \left (x \right ) {\mathrm e}^{\cos \left (x \right )}d x\\ &= -{\mathrm e}^{\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} {\mathrm e}^{2 \cos \left (x \right )}}\\ &= {\mathrm e}^{-2 \cos \left (x \right )}\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 )+{\mathrm e}^{-2 \cos \left (x \right )} y \left (\tau \right )&=0 \\ \end {align*}
But in terms of \(\tau \) \begin {align*} {\mathrm e}^{-2 \cos \left (x \right )}&=\frac {1}{\tau ^{2}} \end {align*}
Hence the above ode becomes \begin {align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+\frac {y \left (\tau \right )}{\tau ^{2}}&=0 \end {align*}
The above ode is now solved for \(y \left (\tau \right )\). The ode can be written as \[ \left (\frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )\right ) \tau ^{2}+y \left (\tau \right ) = 0 \] Which shows it is a Euler ODE. This is Euler second order ODE. Let the solution be \(y \left (\tau \right ) = \tau ^r\), then \(y'=r \tau ^{r-1}\) and \(y''=r(r-1) \tau ^{r-2}\). Substituting these back into the given ODE gives \[ \tau ^{2}(r(r-1))\tau ^{r-2}+0 r \tau ^{r-1}+\tau ^{r} = 0 \] Simplifying gives \[ r \left (r -1\right )\tau ^{r}+0\,\tau ^{r}+\tau ^{r} = 0 \] Since \(\tau ^{r}\neq 0\) then dividing throughout by \(\tau ^{r}\) gives \[ r \left (r -1\right )+0+1 = 0 \] Or \[ r^{2}-r +1 = 0 \tag {1} \] Equation (1) is the characteristic equation. Its roots determine the form of the general solution. Using the quadratic equation the roots are \begin {align*} r_1 &= \frac {1}{2}-\frac {i \sqrt {3}}{2}\\ r_2 &= \frac {1}{2}+\frac {i \sqrt {3}}{2} \end {align*}
The roots are complex conjugate of each others. Let the roots be \begin {align*} r_1 &= \alpha + i \beta \\ r_2 &= \alpha - i \beta \\ \end {align*}
Where in this case \(\alpha ={\frac {1}{2}}\) and \(\beta =-\frac {\sqrt {3}}{2}\). Hence the solution becomes \begin {align*} y \left (\tau \right ) &= c_{1} \tau ^{r_1} + c_{2} \tau ^{r_2} \\ &= c_{1} \tau ^{\alpha + i \beta } + c_{2} \tau ^{\alpha - i \beta } \\ &= \tau ^{\alpha } \left ( c_{1} \tau ^{i \beta } + c_{2} \tau ^{- i \beta }\right ) \\ &= \tau ^{\alpha } \left ( c_{1} e^{\ln \left (\tau ^{i \beta }\right )} + c_{2} e^{\ln \left (\tau ^{-i \beta }\right )}\right ) \\ &= \tau ^{\alpha } \left ( c_{1} e^{i \left (\beta \ln {\tau }\right )} + c_{2} e^{-i \left (\beta \ln {\tau }\right )}\right ) \end {align*}
Using the values for \(\alpha ={\frac {1}{2}}, \beta =-\frac {\sqrt {3}}{2}\), the above becomes \begin {align*} y \left (\tau \right )&= \tau ^{{\frac {1}{2}}} \left ( c_{1} e^{-\frac {i \sqrt {3}\, \ln \left (\tau \right )}{2}} + c_{2} e^{\frac {i \sqrt {3}\, \ln \left (\tau \right )}{2}} \right ) \end {align*}
Using Euler relation, the expression \(c_{1} e^{i A}+ c_{2} e^{-i A}\) is transformed to \( c_{1} \cos A+ c_{1} \sin A\) where the constants are free to change. Applying this to the above result gives \begin {align*} y \left (\tau \right )&=\sqrt {\tau }\left (c_{1} \cos \left (\frac {\sqrt {3}\, \ln \left (\tau \right )}{2}\right )+c_{2} \sin \left (\frac {\sqrt {3}\, \ln \left (\tau \right )}{2}\right )\right ) \end {align*}
The above solution is now transformed back to \(y\) using (6) which results in \begin {align*} y &= \sqrt {-{\mathrm e}^{\cos \left (x \right )}}\, \left (c_{1} \cos \left (\frac {\sqrt {3}\, \ln \left (-{\mathrm e}^{\cos \left (x \right )}\right )}{2}\right )+c_{2} \sin \left (\frac {\sqrt {3}\, \ln \left (-{\mathrm e}^{\cos \left (x \right )}\right )}{2}\right )\right ) \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \sqrt {-{\mathrm e}^{\cos \left (x \right )}}\, \left (c_{1} \cos \left (\frac {\sqrt {3}\, \ln \left (-{\mathrm e}^{\cos \left (x \right )}\right )}{2}\right )+c_{2} \sin \left (\frac {\sqrt {3}\, \ln \left (-{\mathrm e}^{\cos \left (x \right )}\right )}{2}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = \sqrt {-{\mathrm e}^{\cos \left (x \right )}}\, \left (c_{1} \cos \left (\frac {\sqrt {3}\, \ln \left (-{\mathrm e}^{\cos \left (x \right )}\right )}{2}\right )+c_{2} \sin \left (\frac {\sqrt {3}\, \ln \left (-{\mathrm e}^{\cos \left (x \right )}\right )}{2}\right )\right ) \] 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 = arccos(t)] Linear ODE actually solved: (4*t^4-8*t^2+4)*u(t)+(-4*t^4+8*t^2-4)*diff(u(t),t)+(4*t^4-8*t^2+4)*diff(diff(u(t),t),t) = 0 <- change of variables successful`
✓ Solution by Maple
Time used: 0.125 (sec). Leaf size: 31
dsolve(diff(diff(y(x),x),x) = -(sin(x)^2-cos(x))/sin(x)*diff(y(x),x)-y(x)*sin(x)^2,y(x), singsol=all)
\[ y \left (x \right ) = {\mathrm e}^{\frac {\cos \left (x \right )}{2}} \left (c_{1} \sin \left (\frac {\sqrt {3}\, \cos \left (x \right )}{2}\right )+c_{2} \cos \left (\frac {\sqrt {3}\, \cos \left (x \right )}{2}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.221 (sec). Leaf size: 45
DSolve[y''[x] == -(Sin[x]^2*y[x]) - Csc[x]*(-Cos[x] + Sin[x]^2)*y'[x],y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to e^{\frac {\cos (x)}{2}} \left (c_1 \cos \left (\frac {1}{2} \sqrt {3} \cos (x)\right )+c_2 \sin \left (\frac {1}{2} \sqrt {3} \cos (x)\right )\right ) \]