2.2.22 Problem 23

Internal problem ID [10433]
Book : Second order enumerated odes
Section : section 2
Problem number : 23
Date solved : Monday, December 08, 2025 at 08:54:17 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

2.2.22.1 second order change of variable on x method 2

0.352 (sec)

\begin{align*} y^{\prime \prime }+\tan \left (x \right ) y^{\prime }+y \cos \left (x \right )^{2}&=0 \\ \end{align*}

Entering second order change of variable on \(x\) method 2 solverIn normal form the ode

\begin{align*} y^{\prime \prime }+\tan \left (x \right ) y^{\prime }+y \cos \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 )&=\tan \left (x \right )\\ q \left (x \right )&=\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}^{-\int p \left (x \right )d x}d x\\ &= \int {\mathrm e}^{-\int \tan \left (x \right )d x}d x\\ &= \int e^{\ln \left (\cos \left (x \right )\right )} \,dx\\ &= \int \cos \left (x \right )d x\\ &= \sin \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 {\cos \left (x \right )^{2}}{\cos \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 \(y \left (\tau \right )\).Entering second order linear constant coefficient ode solver

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

\[ \lambda ^{2} {\mathrm e}^{\tau \lambda }+{\mathrm e}^{\tau \lambda } = 0 \tag {1} \]

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

\[ y = c_1 \cos \left (\sin \left (x \right )\right )+c_2 \sin \left (\sin \left (x \right )\right ) \]

Summary of solutions found

\begin{align*} y &= c_1 \cos \left (\sin \left (x \right )\right )+c_2 \sin \left (\sin \left (x \right )\right ) \\ \end{align*}

2.2.22.2 SSolved using second order ode arccos transformation

0.485 (sec)

\begin{align*} y^{\prime \prime }+\tan \left (x \right ) y^{\prime }+y \cos \left (x \right )^{2}&=0 \\ \end{align*}

Applying change of variable \(x = \arccos \left (\tau \right )\) to the above ode results in the following new ode

\[ \frac {\left (-\tau ^{3}+\tau \right ) \left (\frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )\right )+y \left (\tau \right ) \tau ^{3}-\frac {d}{d \tau }y \left (\tau \right )}{\tau } = 0 \]

Which is now solved for \(y \left (\tau \right )\). Entering second order change of variable on \(x\) method 2 solverIn normal form the ode

\begin{align*} \frac {\left (-\tau ^{3}+\tau \right ) \left (\frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )\right )+y \left (\tau \right ) \tau ^{3}-\frac {d}{d \tau }y \left (\tau \right )}{\tau } = 0\tag {1} \end{align*}

Becomes

\begin{align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+p \left (\tau \right ) \left (\frac {d}{d \tau }y \left (\tau \right )\right )+q \left (\tau \right ) y \left (\tau \right )&=0 \tag {2} \end{align*}

Where

\begin{align*} p \left (\tau \right )&=\frac {1}{\tau ^{3}-\tau }\\ q \left (\tau \right )&=-\frac {\tau ^{2}}{\tau ^{2}-1} \end{align*}

Applying change of variables \(\tau = g \left (\tau \right )\) 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 {\frac {d^{2}}{d \tau ^{2}}\tau \left (\tau \right )+p \left (\tau \right ) \left (\frac {d}{d \tau }\tau \left (\tau \right )\right )}{\left (\frac {d}{d \tau }\tau \left (\tau \right )\right )^{2}}\tag {4} \\ q_{1} \left (\tau \right ) &=\frac {q \left (\tau \right )}{\left (\frac {d}{d \tau }\tau \left (\tau \right )\right )^{2}}\tag {5} \end{align*}

Let \(p_{1} = 0\). Eq (4) simplifies to

\begin{align*} \frac {d^{2}}{d \tau ^{2}}\tau \left (\tau \right )+p \left (\tau \right ) \left (\frac {d}{d \tau }\tau \left (\tau \right )\right )&=0 \end{align*}

This ode is solved resulting in

\begin{align*} \tau &= \int {\mathrm e}^{-\int p \left (\tau \right )d \tau }d \tau \\ &= \int {\mathrm e}^{-\int \frac {1}{\tau ^{3}-\tau }d \tau }d \tau \\ &= \int e^{\ln \left (\tau \right )-\frac {\ln \left (\tau -1\right )}{2}-\frac {\ln \left (\tau +1\right )}{2}} \,d\tau \\ &= \int \frac {\tau }{\sqrt {\tau -1}\, \sqrt {\tau +1}}d \tau \\ &= \sqrt {\tau -1}\, \sqrt {\tau +1}\tag {6} \end{align*}

Using (6) to evaluate \(q_{1}\) from (5) gives

\begin{align*} q_{1} \left (\tau \right ) &= \frac {q \left (\tau \right )}{\left (\frac {d}{d \tau }\tau \left (\tau \right )\right )^{2}}\\ &= \frac {-\frac {\tau ^{2}}{\tau ^{2}-1}}{\frac {\tau ^{2}}{\left (\tau -1\right ) \left (\tau +1\right )}}\\ &= -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 \(y \left (\tau \right )\).Entering second order linear constant coefficient ode solver

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

\[ \lambda ^{2} {\mathrm e}^{\tau \lambda }-{\mathrm e}^{\tau \lambda } = 0 \tag {1} \]

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 distinct, then the solution is

\begin{align*} y \left (\tau \right ) &= c_3 e^{\lambda _1 \tau } + c_4 e^{\lambda _2 \tau } \\ y \left (\tau \right ) &= c_3 e^{\left (1\right )\tau } +c_4 e^{\left (-1\right )\tau } \\ \end{align*}

Or

\[ y \left (\tau \right ) =c_3 \,{\mathrm e}^{\tau }+c_4 \,{\mathrm e}^{-\tau } \]

The above solution is now transformed back to \(y \left (\tau \right )\) using (6) which results in

\[ y \left (\tau \right ) = c_3 \,{\mathrm e}^{\sqrt {\tau -1}\, \sqrt {\tau +1}}+c_4 \,{\mathrm e}^{-\sqrt {\tau -1}\, \sqrt {\tau +1}} \]

Applying change of variable \(\tau = \cos \left (x \right )\) to the solutions above gives

\begin{align*} y &= c_3 \,{\mathrm e}^{\sqrt {\cos \left (x \right )-1}\, \sqrt {\cos \left (x \right )+1}}+c_4 \,{\mathrm e}^{-\sqrt {\cos \left (x \right )-1}\, \sqrt {\cos \left (x \right )+1}} \\ \end{align*}

2.2.22.3 ✓ Maple. Time used: 0.019 (sec). Leaf size: 15

ode:=diff(diff(y(x),x),x)+tan(x)*diff(y(x),x)+y(x)*cos(x)^2 = 0; 
dsolve(ode,y(x), singsol=all);
 

\[ y = c_1 \sin \left (\sin \left (x \right )\right )+c_2 \cos \left (\sin \left (x \right )\right ) \]

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 = arcsin(t)] 
   Linear ODE actually solved: 
      (-2*t^2+2)*u(t)+(-2*t^2+2)*diff(diff(u(t),t),t) = 0 
<- change of variables successful
 

2.2.22.4 ✓ Mathematica. Time used: 1.194 (sec). Leaf size: 37

ode=D[y[x],{x,2}]+Tan[x]*D[y[x],x]+Cos[x]^2*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

\begin{align*} y(x)&\to c_1 \cosh \left (\sqrt {-\sin ^2(x)}\right )+i c_2 \sinh \left (\sqrt {-\sin ^2(x)}\right ) \end{align*}

2.2.22.5 ✗ Sympy

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(y(x)*cos(x)**2 + tan(x)*Derivative(y(x), x) + Derivative(y(x), (x, 2)),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

False