2.2.25 Problem 25

Internal problem ID [10436]
Book : Second order enumerated odes
Section : section 2
Problem number : 25
Date solved : Monday, December 08, 2025 at 08:55:00 PM
CAS classification : [[_2nd_order, _linear, _nonhomogeneous]]

2.2.25.1 second order change of variable on x method 2

0.873 (sec)

\begin{align*} \cos \left (x \right ) y^{\prime \prime }+y^{\prime } \sin \left (x \right )-2 y \cos \left (x \right )^{3}&=2 \cos \left (x \right )^{5} \\ \end{align*}

Entering second order change of variable on \(x\) method 2 solverThis is second order non-homogeneous ODE. Let the solution be

\[ y = y_h + y_p \]

Where \(y_h\) is the solution to the homogeneous ODE

\begin{align*} A y''(x) + B y'(x) + C y(x) &= 0 \end{align*}

And \(y_p\) is a particular solution to the non-homogeneous ODE

\begin{align*} A y''(x) + B y'(x) + C y(x) &= f(x) \end{align*}

\(y_h\) is the solution to

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

In normal form the ode

\begin{align*} \cos \left (x \right ) y^{\prime \prime }+y^{\prime } \sin \left (x \right )-2 y \cos \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 )&=\tan \left (x \right )\\ q \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}^{-\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 {-2 \cos \left (x \right )^{2}}{\cos \left (x \right )^{2}}\\ &= -2\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 )-2 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 \(A=1, B=0, C=-2\). Let the solution be \(y \left (\tau \right )=e^{\lambda \tau }\). Substituting this into the ODE gives

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

Since exponential function is never zero, then dividing Eq(2) throughout by \(e^{\lambda \tau }\) gives

\[ \lambda ^{2}-2 = 0 \tag {2} \]

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 \(A=1, B=0, C=-2\) 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 (-2\right )}\\ &= \pm \sqrt {2} \end{align*}

Hence

\begin{align*} \lambda _1 &= + \sqrt {2} \\ \lambda _2 &= - \sqrt {2} \\ \end{align*}

Which simplifies to

\begin{align*} \lambda _1 &= \sqrt {2} \\ \lambda _2 &= -\sqrt {2} \\ \end{align*}

Since roots are 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 (\sqrt {2}\right )\tau } +c_2 e^{\left (-\sqrt {2}\right )\tau } \\ \end{align*}

Or

\[ y \left (\tau \right ) =c_1 \,{\mathrm e}^{\sqrt {2}\, \tau }+c_2 \,{\mathrm e}^{-\sqrt {2}\, \tau } \]

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

\[ y = c_1 \,{\mathrm e}^{\sin \left (x \right ) \sqrt {2}}+c_2 \,{\mathrm e}^{-\sin \left (x \right ) \sqrt {2}} \]

Therefore the homogeneous solution \(y_h\) is

\[ y_h = c_1 \,{\mathrm e}^{\sin \left (x \right ) \sqrt {2}}+c_2 \,{\mathrm e}^{-\sin \left (x \right ) \sqrt {2}} \]

The particular solution \(y_p\) can be found using either the method of undetermined coefficients, or the method of variation of parameters. The method of variation of parameters will be used as it is more general and can be used when the coefficients of the ODE depend on \(x\) as well. Let

\begin{equation} \tag{1} y_p(x) = u_1 y_1 + u_2 y_2 \end{equation}

Where \(u_1,u_2\) to be determined, and \(y_1,y_2\) are the two basis solutions (the two linearly independent solutions of the homogeneous ODE) found earlier when solving the homogeneous ODE as

\begin{align*} y_1 &= {\mathrm e}^{-\sin \left (x \right ) \sqrt {2}} \\ y_2 &= {\mathrm e}^{\sin \left (x \right ) \sqrt {2}} \\ \end{align*}

In the Variation of parameters \(u_1,u_2\) are found using

\begin{align*} \tag{2} u_1 &= -\int \frac {y_2 f(x)}{a W(x)} \\ \tag{3} u_2 &= \int \frac {y_1 f(x)}{a W(x)} \\ \end{align*}

Where \(W(x)\) is the Wronskian and \(a\) is the coefficient in front of \(y''\) in the given ODE. The Wronskian is given by \(W= \begin {vmatrix} y_1 & y_{2} \\ y_{1}^{\prime } & y_{2}^{\prime } \end {vmatrix} \). Hence

\[ W = \begin {vmatrix} {\mathrm e}^{-\sin \left (x \right ) \sqrt {2}} & {\mathrm e}^{\sin \left (x \right ) \sqrt {2}} \\ \frac {d}{dx}\left ({\mathrm e}^{-\sin \left (x \right ) \sqrt {2}}\right ) & \frac {d}{dx}\left ({\mathrm e}^{\sin \left (x \right ) \sqrt {2}}\right ) \end {vmatrix} \]

Which gives

\[ W = \begin {vmatrix} {\mathrm e}^{-\sin \left (x \right ) \sqrt {2}} & {\mathrm e}^{\sin \left (x \right ) \sqrt {2}} \\ -{\mathrm e}^{-\sin \left (x \right ) \sqrt {2}} \cos \left (x \right ) \sqrt {2} & \cos \left (x \right ) \sqrt {2}\, {\mathrm e}^{\sin \left (x \right ) \sqrt {2}} \end {vmatrix} \]

Therefore

\[ W = \left ({\mathrm e}^{-\sin \left (x \right ) \sqrt {2}}\right )\left (\cos \left (x \right ) \sqrt {2}\, {\mathrm e}^{\sin \left (x \right ) \sqrt {2}}\right ) - \left ({\mathrm e}^{\sin \left (x \right ) \sqrt {2}}\right )\left (-{\mathrm e}^{-\sin \left (x \right ) \sqrt {2}} \cos \left (x \right ) \sqrt {2}\right ) \]

Which simplifies to

\[ W = 2 \cos \left (x \right ) \sqrt {2} \]

Which simplifies to

\[ W = 2 \cos \left (x \right ) \sqrt {2} \]

Therefore Eq. (2) becomes

\[ u_1 = -\int \frac {2 \,{\mathrm e}^{\sin \left (x \right ) \sqrt {2}} \cos \left (x \right )^{5}}{2 \cos \left (x \right )^{2} \sqrt {2}}\,dx \]

Which simplifies to

\[ u_1 = - \int \frac {{\mathrm e}^{\sin \left (x \right ) \sqrt {2}} \cos \left (x \right )^{3} \sqrt {2}}{2}d x \]

Hence

\[ u_1 = \frac {{\mathrm e}^{\sin \left (x \right ) \sqrt {2}} \sin \left (x \right )^{2}}{2}-\frac {{\mathrm e}^{\sin \left (x \right ) \sqrt {2}} \sqrt {2}\, \sin \left (x \right )}{2} \]

And Eq. (3) becomes

\[ u_2 = \int \frac {2 \,{\mathrm e}^{-\sin \left (x \right ) \sqrt {2}} \cos \left (x \right )^{5}}{2 \cos \left (x \right )^{2} \sqrt {2}}\,dx \]

Which simplifies to

\[ u_2 = \int \frac {{\mathrm e}^{-\sin \left (x \right ) \sqrt {2}} \cos \left (x \right )^{3} \sqrt {2}}{2}d x \]

Hence

\[ u_2 = \frac {{\mathrm e}^{-\sin \left (x \right ) \sqrt {2}} \sin \left (x \right )^{2}}{2}+\frac {{\mathrm e}^{-\sin \left (x \right ) \sqrt {2}} \sqrt {2}\, \sin \left (x \right )}{2} \]

Therefore the particular solution, from equation (1) is

\[ y_p(x) = \left (\frac {{\mathrm e}^{\sin \left (x \right ) \sqrt {2}} \sin \left (x \right )^{2}}{2}-\frac {{\mathrm e}^{\sin \left (x \right ) \sqrt {2}} \sqrt {2}\, \sin \left (x \right )}{2}\right ) {\mathrm e}^{-\sin \left (x \right ) \sqrt {2}}+{\mathrm e}^{\sin \left (x \right ) \sqrt {2}} \left (\frac {{\mathrm e}^{-\sin \left (x \right ) \sqrt {2}} \sin \left (x \right )^{2}}{2}+\frac {{\mathrm e}^{-\sin \left (x \right ) \sqrt {2}} \sqrt {2}\, \sin \left (x \right )}{2}\right ) \]

Which simplifies to

\[ y_p(x) = \sin \left (x \right )^{2} \]

Therefore the general solution is

\begin{align*} y &= y_h + y_p \\ &= \left (c_1 \,{\mathrm e}^{\sin \left (x \right ) \sqrt {2}}+c_2 \,{\mathrm e}^{-\sin \left (x \right ) \sqrt {2}}\right ) + \left (\sin \left (x \right )^{2}\right ) \\ \end{align*}

Summary of solutions found

\begin{align*} y &= c_1 \,{\mathrm e}^{\sin \left (x \right ) \sqrt {2}}+c_2 \,{\mathrm e}^{-\sin \left (x \right ) \sqrt {2}}+\sin \left (x \right )^{2} \\ \end{align*}

2.2.25.2 SSolved using second order ode arccos transformation

0.638 (sec)

\begin{align*} \cos \left (x \right ) y^{\prime \prime }+y^{\prime } \sin \left (x \right )-2 y \cos \left (x \right )^{3}&=2 \cos \left (x \right )^{5} \\ \end{align*}

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

\[ \left (-\tau ^{3}+\tau \right ) \left (\frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )\right )-2 y \left (\tau \right ) \tau ^{3}-\frac {d}{d \tau }y \left (\tau \right ) = 2 \tau ^{5} \]

Which is now solved for \(y \left (\tau \right )\). Entering second order change of variable on \(x\) method 2 solverThis is second order non-homogeneous ODE. Let the solution be

\[ y \left (\tau \right ) = y_h + y_p \]

Where \(y_h\) is the solution to the homogeneous ODE

\begin{align*} A y''(\tau ) + B y'(\tau ) + C y(\tau ) &= 0 \end{align*}

And \(y_p\) is a particular solution to the non-homogeneous ODE

\begin{align*} A y''(\tau ) + B y'(\tau ) + C y(\tau ) &= f(\tau ) \end{align*}

\(y_h\) is the solution to

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

In normal form the ode

\begin{align*} \left (-\tau ^{3}+\tau \right ) \left (\frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )\right )-2 y \left (\tau \right ) \tau ^{3}-\frac {d}{d \tau }y \left (\tau \right ) = 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 {2 \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 {2 \tau ^{2}}{\tau ^{2}-1}}{\frac {\tau ^{2}}{\left (\tau -1\right ) \left (\tau +1\right )}}\\ &= 2\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 )+2 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 \(A=1, B=0, C=2\). Let the solution be \(y \left (\tau \right )=e^{\lambda \tau }\). Substituting this into the ODE gives

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

Since exponential function is never zero, then dividing Eq(2) throughout by \(e^{\lambda \tau }\) gives

\[ \lambda ^{2}+2 = 0 \tag {2} \]

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 \(A=1, B=0, C=2\) 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 (2\right )}\\ &= \pm i \sqrt {2} \end{align*}

Hence

\begin{align*} \lambda _1 &= + i \sqrt {2}\\ \lambda _2 &= - i \sqrt {2} \end{align*}

Which simplifies to

\begin{align*} \lambda _1 &= i \sqrt {2} \\ \lambda _2 &= -i \sqrt {2} \\ \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 \(\beta =\sqrt {2}\). Therefore the final solution, when using Euler relation, can be written as

\[ y \left (\tau \right ) = e^{\alpha \tau } \left ( c_3 \cos (\beta \tau ) + c_4 \sin (\beta \tau ) \right ) \]

Which becomes

\[ y \left (\tau \right ) = e^{0}\left (c_3 \cos \left (\sqrt {2}\, \tau \right )+c_4 \sin \left (\sqrt {2}\, \tau \right )\right ) \]

Or

\[ y \left (\tau \right ) = c_3 \cos \left (\sqrt {2}\, \tau \right )+c_4 \sin \left (\sqrt {2}\, \tau \right ) \]

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

\[ y \left (\tau \right ) = c_3 \cos \left (\sqrt {2}\, \sqrt {\tau -1}\, \sqrt {\tau +1}\right )+c_4 \sin \left (\sqrt {2}\, \sqrt {\tau -1}\, \sqrt {\tau +1}\right ) \]

Therefore the homogeneous solution \(y_h\) is

\[ y_h = c_3 \cos \left (\sqrt {2}\, \sqrt {\tau -1}\, \sqrt {\tau +1}\right )+c_4 \sin \left (\sqrt {2}\, \sqrt {\tau -1}\, \sqrt {\tau +1}\right ) \]

The particular solution is now found using the method of undetermined coefficients. Looking at the RHS of the ode, which is

\[ \tau ^{5} \]

Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is

\[ [\{1, \tau , \tau ^{2}, \tau ^{3}, \tau ^{4}, \tau ^{5}\}] \]

While the set of the basis functions for the homogeneous solution found earlier is

\[ \left \{\cos \left (\sqrt {2}\, \sqrt {\tau -1}\, \sqrt {\tau +1}\right ), \sin \left (\sqrt {2}\, \sqrt {\tau -1}\, \sqrt {\tau +1}\right )\right \} \]

Since there is no duplication between the basis function in the UC_set and the basis functions of the homogeneous solution, the trial solution is a linear combination of all the basis in the UC_set.

\[ y_p = A_{6} \tau ^{5}+A_{5} \tau ^{4}+A_{4} \tau ^{3}+A_{3} \tau ^{2}+A_{2} \tau +A_{1} \]

The unknowns \(\{A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}\}\) are found by substituting the above trial solution \(y_p\) into the ODE and comparing coefficients. Substituting the trial solution into the ODE and simplifying gives

\[ \left (-\tau ^{3}+\tau \right ) \left (20 \tau ^{3} A_{6}+12 \tau ^{2} A_{5}+6 \tau A_{4}+2 A_{3}\right )-2 \left (A_{6} \tau ^{5}+A_{5} \tau ^{4}+A_{4} \tau ^{3}+A_{3} \tau ^{2}+A_{2} \tau +A_{1}\right ) \tau ^{3}-5 \tau ^{4} A_{6}-4 \tau ^{3} A_{5}-3 \tau ^{2} A_{4}-2 \tau A_{3}-A_{2} = 2 \tau ^{5} \]

Solving for the unknowns by comparing coefficients results in

\[ [A_{1} = 1, A_{2} = 0, A_{3} = -1, A_{4} = 0, A_{5} = 0, A_{6} = 0] \]

Substituting the above back in the above trial solution \(y_p\), gives the particular solution

\[ y_p = -\tau ^{2}+1 \]

Therefore the general solution is

\begin{align*} y &= y_h + y_p \\ &= \left (c_3 \cos \left (\sqrt {2}\, \sqrt {\tau -1}\, \sqrt {\tau +1}\right )+c_4 \sin \left (\sqrt {2}\, \sqrt {\tau -1}\, \sqrt {\tau +1}\right )\right ) + \left (-\tau ^{2}+1\right ) \\ \end{align*}

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

\begin{align*} y &= c_3 \cos \left (\sqrt {2}\, \sqrt {\cos \left (x \right )-1}\, \sqrt {\cos \left (x \right )+1}\right )+c_4 \sin \left (\sqrt {2}\, \sqrt {\cos \left (x \right )-1}\, \sqrt {\cos \left (x \right )+1}\right )-\cos \left (x \right )^{2}+1 \\ \end{align*}

2.2.25.3 ✓ Maple. Time used: 0.007 (sec). Leaf size: 30

ode:=cos(x)*diff(diff(y(x),x),x)+sin(x)*diff(y(x),x)-2*cos(x)^3*y(x) = 2*cos(x)^5; 
dsolve(ode,y(x), singsol=all);
 

\[ y = \sinh \left (\sin \left (x \right ) \sqrt {2}\right ) c_2 +\cosh \left (\sin \left (x \right ) \sqrt {2}\right ) c_1 +\frac {1}{2}-\frac {\cos \left (2 x \right )}{2} \]

Maple trace

Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying high order exact linear fully integrable 
trying differential order: 2; linear nonhomogeneous with symmetry [0,1] 
trying a double symmetry of the form [xi=0, eta=F(x)] 
trying symmetries linear in x and y(x) 
-> Try solving first the homogeneous part of the ODE 
   trying a symmetry of the form [xi=0, eta=F(x)] 
   <- linear_1 successful 
<- solving first the homogeneous part of the ODE successful
 

2.2.25.4 ✓ Mathematica. Time used: 8.198 (sec). Leaf size: 90

ode=Cos[x]*D[y[x],{x,2}]+Sin[x]*D[y[x],x]-2*y[x]*Cos[x]^3==2*Cos[x]^5; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

\begin{align*} y(x)&\to \frac {1}{2} e^{-i \sqrt {\cos (2 x)-1}} \left (\cos \left (\sqrt {\cos (2 x)-1}\right )+i \sin \left (\sqrt {\cos (2 x)-1}\right )\right ) \left (-\cos (2 x)+2 c_1 \cos \left (\sqrt {\cos (2 x)-1}\right )+2 c_2 \sin \left (\sqrt {\cos (2 x)-1}\right )+1\right ) \end{align*}

2.2.25.5 ✗ Sympy

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(-2*y(x)*cos(x)**3 + sin(x)*Derivative(y(x), x) - 2*cos(x)**5 + cos(x)*Derivative(y(x), (x, 2)),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

NotImplementedError : The given ODE -(2*y(x)*cos(x)**2 + 2*cos(x)**4 - Derivative(y(x), (x, 2)))/tan(x) + Derivative(y(x), x) cannot be solved by the factorable group method