problem 1437

Internal problem ID [9764]
Internal ﬁle name [OUTPUT/8706_Monday_June_06_2022_05_16_17_AM_28682212/index.tex]

Book: Diﬀerential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1437.
ODE order: 2.
ODE degree: 1.

The type(s) of ODE detected by this program : "second_order_change_of_variable_on_x_method_2"

\[ \boxed {y^{\prime \prime }-\frac {\left (3 \sin \left (x \right )^{2}+1\right ) y^{\prime }}{\cos \left (x \right ) \sin \left (x \right )}-\frac {y \sin \left (x \right )^{2}}{\cos \left (x \right )^{2}}=0} \]

3.431.1 Solving as second order change of variable on x method 2 ode

In normal form the ode \begin {align*} y^{\prime \prime } \cos \left (x \right )^{2} \sin \left (x \right )+\left (3 \cos \left (x \right )^{3}-4 \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 {3 \cos \left (x \right )^{2}-4}{\cos \left (x \right ) \sin \left (x \right )}\\ q \left (x \right )&=-\frac {\sin \left (x \right )^{2}}{\cos \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 {3 \cos \left (x \right )^{2}-4}{\cos \left (x \right ) \sin \left (x \right )}d x \right )}d x\\ &= \int e^{-3 \ln \left (\sin \left (x \right )\right )+4 \ln \left (\tan \left (x \right )\right )} \,dx\\ &= \int \frac {\tan \left (x \right )^{4}}{\sin \left (x \right )^{3}}d x\\ &= \frac {1}{3 \cos \left (x \right )^{3}}\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 {\sin \left (x \right )^{2}}{\cos \left (x \right )^{2}}}{\frac {\tan \left (x \right )^{8}}{\sin \left (x \right )^{6}}}\\ &= -\cos \left (x \right )^{6}\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 )-\cos \left (x \right )^{6} y \left (\tau \right )&=0 \\ \end {align*}

But in terms of \(\tau \) \begin {align*} -\cos \left (x \right )^{6}&=-\frac {1}{9 \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 )}{9 \tau ^{2}}&=0 \end {align*}

The above ode is now solved for \(y \left (\tau \right )\). The ode can be written as \[ 9 \left (\frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )\right ) \tau ^{2}-y \left (\tau \right ) = 0 \] 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 \[ 9 \tau ^{2}(r(r-1))\tau ^{r-2}+0 r \tau ^{r-1}-\tau ^{r} = 0 \] Simplifying gives \[ 9 r \left (r -1\right )\tau ^{r}+0\,\tau ^{r}-\tau ^{r} = 0 \] Since \(\tau ^{r}\neq 0\) then dividing throughout by \(\tau ^{r}\) gives \[ 9 r \left (r -1\right )+0-1 = 0 \] Or \[ 9 r^{2}-9 r -1 = 0 \tag {1} \] 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 {\sqrt {13}}{6}\\ r_2 &= \frac {1}{2}+\frac {\sqrt {13}}{6} \end {align*}

Since the roots are real and distinct, then the general solution is \[ y \left (\tau \right )= c_{1} y_1 + c_{2} y_2 \] Where \(y_1 = \tau ^{r_1}\) and \(y_2 = \tau ^{r_2} \). Hence \[ y \left (\tau \right ) = c_{1} \tau ^{\frac {1}{2}-\frac {\sqrt {13}}{6}}+c_{2} \tau ^{\frac {1}{2}+\frac {\sqrt {13}}{6}} \] The above solution is now transformed back to \(y\) using (6) which results in \begin {align*} y &= c_{1} 3^{-\frac {1}{2}+\frac {\sqrt {13}}{6}} \left (\sec \left (x \right )^{3}\right )^{\frac {1}{2}-\frac {\sqrt {13}}{6}}+c_{2} 3^{-\frac {1}{2}-\frac {\sqrt {13}}{6}} \left (\sec \left (x \right )^{3}\right )^{\frac {1}{2}+\frac {\sqrt {13}}{6}} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} 3^{-\frac {1}{2}+\frac {\sqrt {13}}{6}} \left (\sec \left (x \right )^{3}\right )^{\frac {1}{2}-\frac {\sqrt {13}}{6}}+c_{2} 3^{-\frac {1}{2}-\frac {\sqrt {13}}{6}} \left (\sec \left (x \right )^{3}\right )^{\frac {1}{2}+\frac {\sqrt {13}}{6}} \\ \end{align*}

Veriﬁcation of solutions

\[ y = c_{1} 3^{-\frac {1}{2}+\frac {\sqrt {13}}{6}} \left (\sec \left (x \right )^{3}\right )^{\frac {1}{2}-\frac {\sqrt {13}}{6}}+c_{2} 3^{-\frac {1}{2}-\frac {\sqrt {13}}{6}} \left (\sec \left (x \right )^{3}\right )^{\frac {1}{2}+\frac {\sqrt {13}}{6}} \] 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 
   <- LODE of Euler type successful 
   Change of variables used: 
      [x = arccos(t)] 
   Linear ODE actually solved: 
      -4*(-t^2+1)^(3/2)*u(t)+16*(-t^2+1)^(3/2)*t*diff(u(t),t)+4*t^2*(-t^2+1)^(3/2)*diff(diff(u(t),t),t) = 0 
<- change of variables successful`

\[ y \left (x \right ) = c_{1} \cos \left (x \right )^{-\frac {3}{2}+\frac {\sqrt {13}}{2}}+c_{2} \cos \left (x \right )^{-\frac {3}{2}-\frac {\sqrt {13}}{2}} \]

\[ y(x)\to \cos ^{-\frac {3}{2}-\frac {\sqrt {13}}{2}}(x) \left (c_2 \cos ^{\sqrt {13}}(x)+c_1\right ) \]

3.431 problem 1437

3.431.1 Solving as second order change of variable on x method 2 ode