2.2.18 Problem 19

Internal problem ID [9141]
Book : Second order enumerated odes
Section : section 2
Problem number : 19
Date solved : Friday, February 21, 2025 at 09:20:45 PM
CAS classification : [[_2nd_order, _linear, _nonhomogeneous]]

Solve

\begin{align*} x^{6} y^{\prime \prime }+3 x^{5} y^{\prime }+a^{2} y&=\frac {1}{x^{2}} \end{align*}

Solved as second order ode using change of variable on x method 2

Time used: 0.526 (sec)

This 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 \( A y''(x) + B y'(x) + C y(x) = 0\), and \(y_p\) is a particular solution to the non-homogeneous ODE \(A y''(x) + B y'(x) + C y(x) = f(x)\). \(y_h\) is the solution to

\[ x^{6} y^{\prime \prime }+3 x^{5} y^{\prime }+a^{2} y = 0 \]

In normal form the ode

\begin{align*} x^{6} y^{\prime \prime }+3 x^{5} y^{\prime }+a^{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 {3}{x}\\ q \left (x \right )&=\frac {a^{2}}{x^{6}} \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 \frac {3}{x}d x}d x\\ &= \int e^{-3 \ln \left (x \right )} \,dx\\ &= \int \frac {1}{x^{3}}d x\\ &= -\frac {1}{2 x^{2}}\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 {a^{2}}{x^{6}}}{\frac {1}{x^{6}}}\\ &= a^{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 )+a^{2} 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 \(A=1, B=0, C=a^{2}\). Let the solution be \(y \left (\tau \right )=e^{\lambda \tau }\). Substituting this into the ODE gives

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

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

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

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

Hence

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

Which simplifies to

\begin{align*} \lambda _1 &= i a \\ \lambda _2 &= -i a \\ \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 \(\beta =a\). 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

\[ y \left (\tau \right ) = e^{0}\left (c_1 \cos \left (a \tau \right )+c_2 \sin \left (a \tau \right )\right ) \]

Or

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

Will add steps showing solving for IC soon.

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

\[ y = c_1 \cos \left (\frac {a}{2 x^{2}}\right )-c_2 \sin \left (\frac {a}{2 x^{2}}\right ) \]

Therefore the homogeneous solution \(y_h\) is

\[ y_h = c_1 \cos \left (\frac {a}{2 x^{2}}\right )-c_2 \sin \left (\frac {a}{2 x^{2}}\right ) \]

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 &= \cos \left (\frac {a}{2 x^{2}}\right ) \\ y_2 &= \sin \left (\frac {a}{2 x^{2}}\right ) \\ \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

\[ W = \begin {vmatrix} \cos \left (\frac {a}{2 x^{2}}\right ) & \sin \left (\frac {a}{2 x^{2}}\right ) \\ \frac {d}{dx}\left (\cos \left (\frac {a}{2 x^{2}}\right )\right ) & \frac {d}{dx}\left (\sin \left (\frac {a}{2 x^{2}}\right )\right ) \end {vmatrix} \]

Which gives

\[ W = \begin {vmatrix} \cos \left (\frac {a}{2 x^{2}}\right ) & \sin \left (\frac {a}{2 x^{2}}\right ) \\ \frac {a \sin \left (\frac {a}{2 x^{2}}\right )}{x^{3}} & -\frac {a \cos \left (\frac {a}{2 x^{2}}\right )}{x^{3}} \end {vmatrix} \]

Therefore

\[ W = \left (\cos \left (\frac {a}{2 x^{2}}\right )\right )\left (-\frac {a \cos \left (\frac {a}{2 x^{2}}\right )}{x^{3}}\right ) - \left (\sin \left (\frac {a}{2 x^{2}}\right )\right )\left (\frac {a \sin \left (\frac {a}{2 x^{2}}\right )}{x^{3}}\right ) \]

Which simplifies to

\[ W = -\frac {a \left (\cos \left (\frac {a}{2 x^{2}}\right )^{2}+\sin \left (\frac {a}{2 x^{2}}\right )^{2}\right )}{x^{3}} \]

Which simplifies to

\[ W = -\frac {a}{x^{3}} \]

Therefore Eq. (2) becomes

\[ u_1 = -\int \frac {\frac {\sin \left (\frac {a}{2 x^{2}}\right )}{x^{2}}}{-a \,x^{3}}\,dx \]

Which simplifies to

\[ u_1 = - \int -\frac {\sin \left (\frac {a}{2 x^{2}}\right )}{x^{5} a}d x \]

Hence

\[ u_1 = \frac {\frac {\cos \left (\frac {a}{2 x^{2}}\right )}{a \,x^{2}}-\frac {2 \sin \left (\frac {a}{2 x^{2}}\right )}{a^{2}}}{a} \]

And Eq. (3) becomes

\[ u_2 = \int \frac {\frac {\cos \left (\frac {a}{2 x^{2}}\right )}{x^{2}}}{-a \,x^{3}}\,dx \]

Which simplifies to

\[ u_2 = \int -\frac {\cos \left (\frac {a}{2 x^{2}}\right )}{x^{5} a}d x \]

Hence

\[ u_2 = -\frac {-\frac {\sin \left (\frac {a}{2 x^{2}}\right )}{x^{2} a}-\frac {2 \cos \left (\frac {a}{2 x^{2}}\right )}{a^{2}}}{a} \]

Therefore the particular solution, from equation (1) is

\[ y_p(x) = \frac {\left (\frac {\cos \left (\frac {a}{2 x^{2}}\right )}{a \,x^{2}}-\frac {2 \sin \left (\frac {a}{2 x^{2}}\right )}{a^{2}}\right ) \cos \left (\frac {a}{2 x^{2}}\right )}{a}-\frac {\sin \left (\frac {a}{2 x^{2}}\right ) \left (-\frac {\sin \left (\frac {a}{2 x^{2}}\right )}{x^{2} a}-\frac {2 \cos \left (\frac {a}{2 x^{2}}\right )}{a^{2}}\right )}{a} \]

Which simplifies to

\[ y_p(x) = \frac {1}{a^{2} x^{2}} \]

Therefore the general solution is

\begin{align*} y &= y_h + y_p \\ &= \left (c_1 \cos \left (\frac {a}{2 x^{2}}\right )-c_2 \sin \left (\frac {a}{2 x^{2}}\right )\right ) + \left (\frac {1}{a^{2} x^{2}}\right ) \\ \end{align*}

Will add steps showing solving for IC soon.

Summary of solutions found

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

Solved as second order ode using change of variable on x method 1

Time used: 0.910 (sec)

This is second order non-homogeneous ODE. In standard form the ODE is

\[ A y''(x) + B y'(x) + C y(x) = f(x) \]

Where \(A=x^{6}, B=3 x^{5}, C=a^{2}, f(x)=\frac {1}{x^{2}}\). Let the solution be

\[ y = y_h + y_p \]

Where \(y_h\) is the solution to the homogeneous ODE \( A y''(x) + B y'(x) + C y(x) = 0\), and \(y_p\) is a particular solution to the non-homogeneous ODE \(A y''(x) + B y'(x) + C y(x) = f(x)\). Solving for \(y_h\) from

\[ x^{6} y^{\prime \prime }+3 x^{5} y^{\prime }+a^{2} y = 0 \]

In normal form the ode

\begin{align*} x^{6} y^{\prime \prime }+3 x^{5} y^{\prime }+a^{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 {3}{x}\\ q \left (x \right )&=\frac {a^{2}}{x^{6}} \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 {\frac {a^{2}}{x^{6}}}}{c}\tag {6} \\ \tau '' &= -\frac {3 a^{2}}{c \sqrt {\frac {a^{2}}{x^{6}}}\, x^{7}} \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 {3 a^{2}}{c \sqrt {\frac {a^{2}}{x^{6}}}\, x^{7}}+\frac {3}{x}\frac {\sqrt {\frac {a^{2}}{x^{6}}}}{c}}{\left (\frac {\sqrt {\frac {a^{2}}{x^{6}}}}{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 {\frac {a^{2}}{x^{6}}}d x}{c}\\ &= -\frac {x \sqrt {\frac {a^{2}}{x^{6}}}}{2 c} \end{align*}

Substituting the above into the solution obtained gives

\[ y = c_1 \cos \left (\frac {x \sqrt {\frac {a^{2}}{x^{6}}}}{2}\right )-c_2 \sin \left (\frac {x \sqrt {\frac {a^{2}}{x^{6}}}}{2}\right ) \]

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 &= \cos \left (\frac {x \sqrt {\frac {a^{2}}{x^{6}}}}{2}\right ) \\ y_2 &= -\sin \left (\frac {x \sqrt {\frac {a^{2}}{x^{6}}}}{2}\right ) \\ \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

\[ W = \begin {vmatrix} \cos \left (\frac {x \sqrt {\frac {a^{2}}{x^{6}}}}{2}\right ) & -\sin \left (\frac {x \sqrt {\frac {a^{2}}{x^{6}}}}{2}\right ) \\ \frac {d}{dx}\left (\cos \left (\frac {x \sqrt {\frac {a^{2}}{x^{6}}}}{2}\right )\right ) & \frac {d}{dx}\left (-\sin \left (\frac {x \sqrt {\frac {a^{2}}{x^{6}}}}{2}\right )\right ) \end {vmatrix} \]

Which gives

\[ W = \begin {vmatrix} \cos \left (\frac {x \sqrt {\frac {a^{2}}{x^{6}}}}{2}\right ) & -\sin \left (\frac {x \sqrt {\frac {a^{2}}{x^{6}}}}{2}\right ) \\ -\left (\frac {\sqrt {\frac {a^{2}}{x^{6}}}}{2}-\frac {3 a^{2}}{2 x^{6} \sqrt {\frac {a^{2}}{x^{6}}}}\right ) \sin \left (\frac {x \sqrt {\frac {a^{2}}{x^{6}}}}{2}\right ) & -\left (\frac {\sqrt {\frac {a^{2}}{x^{6}}}}{2}-\frac {3 a^{2}}{2 x^{6} \sqrt {\frac {a^{2}}{x^{6}}}}\right ) \cos \left (\frac {x \sqrt {\frac {a^{2}}{x^{6}}}}{2}\right ) \end {vmatrix} \]

Therefore

\[ W = \left (\cos \left (\frac {x \sqrt {\frac {a^{2}}{x^{6}}}}{2}\right )\right )\left (-\left (\frac {\sqrt {\frac {a^{2}}{x^{6}}}}{2}-\frac {3 a^{2}}{2 x^{6} \sqrt {\frac {a^{2}}{x^{6}}}}\right ) \cos \left (\frac {x \sqrt {\frac {a^{2}}{x^{6}}}}{2}\right )\right ) - \left (-\sin \left (\frac {x \sqrt {\frac {a^{2}}{x^{6}}}}{2}\right )\right )\left (-\left (\frac {\sqrt {\frac {a^{2}}{x^{6}}}}{2}-\frac {3 a^{2}}{2 x^{6} \sqrt {\frac {a^{2}}{x^{6}}}}\right ) \sin \left (\frac {x \sqrt {\frac {a^{2}}{x^{6}}}}{2}\right )\right ) \]

Which simplifies to

\[ W = \frac {a^{2} \left ({\cos \left (\frac {x \sqrt {\frac {a^{2}}{x^{6}}}}{2}\right )}^{2}+{\sin \left (\frac {x \sqrt {\frac {a^{2}}{x^{6}}}}{2}\right )}^{2}\right )}{x^{6} \sqrt {\frac {a^{2}}{x^{6}}}} \]

Which simplifies to

\[ W = \frac {a^{2}}{x^{6} \sqrt {\frac {a^{2}}{x^{6}}}} \]

Therefore Eq. (2) becomes

\[ u_1 = -\int \frac {-\frac {\sin \left (\frac {x \sqrt {\frac {a^{2}}{x^{6}}}}{2}\right )}{x^{2}}}{\frac {a^{2}}{\sqrt {\frac {a^{2}}{x^{6}}}}}\,dx \]

Which simplifies to

\[ u_1 = - \int -\frac {\sin \left (\frac {x \sqrt {\frac {a^{2}}{x^{6}}}}{2}\right ) \sqrt {\frac {a^{2}}{x^{6}}}}{x^{2} a^{2}}d x \]

Hence

\[ u_1 = -\frac {4 x^{3} \sqrt {\pi }\, \sqrt {\frac {a^{2}}{x^{6}}}\, \left (-\frac {x^{7} \left (\frac {a^{2}}{x^{6}}\right )^{{3}/{2}} \cos \left (\frac {a}{2 x^{2}}\right )}{4 \sqrt {\pi }\, a^{2}}+\frac {x^{9} \left (\frac {a^{2}}{x^{6}}\right )^{{3}/{2}} \sin \left (\frac {a}{2 x^{2}}\right )}{2 \sqrt {\pi }\, a^{3}}\right )}{a^{4}} \]

And Eq. (3) becomes

\[ u_2 = \int \frac {\frac {\cos \left (\frac {x \sqrt {\frac {a^{2}}{x^{6}}}}{2}\right )}{x^{2}}}{\frac {a^{2}}{\sqrt {\frac {a^{2}}{x^{6}}}}}\,dx \]

Which simplifies to

\[ u_2 = \int \frac {\cos \left (\frac {x \sqrt {\frac {a^{2}}{x^{6}}}}{2}\right ) \sqrt {\frac {a^{2}}{x^{6}}}}{x^{2} a^{2}}d x \]

Hence

\[ u_2 = -\frac {4 \sqrt {\pi }\, \sqrt {\frac {a^{2}}{x^{6}}}\, x^{3} \left (-\frac {1}{2 \sqrt {\pi }}+\frac {\cos \left (\frac {a}{2 x^{2}}\right )}{2 \sqrt {\pi }}+\frac {a \sin \left (\frac {a}{2 x^{2}}\right )}{4 \sqrt {\pi }\, x^{2}}\right )}{a^{4}} \]

Therefore the particular solution, from equation (1) is

\[ y_p(x) = -\frac {4 x^{3} \sqrt {\pi }\, \sqrt {\frac {a^{2}}{x^{6}}}\, \left (-\frac {x^{7} \left (\frac {a^{2}}{x^{6}}\right )^{{3}/{2}} \cos \left (\frac {a}{2 x^{2}}\right )}{4 \sqrt {\pi }\, a^{2}}+\frac {x^{9} \left (\frac {a^{2}}{x^{6}}\right )^{{3}/{2}} \sin \left (\frac {a}{2 x^{2}}\right )}{2 \sqrt {\pi }\, a^{3}}\right ) \cos \left (\frac {x \sqrt {\frac {a^{2}}{x^{6}}}}{2}\right )}{a^{4}}+\frac {4 \sin \left (\frac {x \sqrt {\frac {a^{2}}{x^{6}}}}{2}\right ) \sqrt {\pi }\, \sqrt {\frac {a^{2}}{x^{6}}}\, x^{3} \left (-\frac {1}{2 \sqrt {\pi }}+\frac {\cos \left (\frac {a}{2 x^{2}}\right )}{2 \sqrt {\pi }}+\frac {a \sin \left (\frac {a}{2 x^{2}}\right )}{4 \sqrt {\pi }\, x^{2}}\right )}{a^{4}} \]

Which simplifies to

\[ y_p(x) = \frac {x^{3} \sqrt {\frac {a^{2}}{x^{6}}}\, \left (2 x^{2} \cos \left (\frac {a}{2 x^{2}}\right )+a \sin \left (\frac {a}{2 x^{2}}\right )-2 x^{2}\right ) \sin \left (\frac {x \sqrt {\frac {a^{2}}{x^{6}}}}{2}\right )+a \cos \left (\frac {x \sqrt {\frac {a^{2}}{x^{6}}}}{2}\right ) \left (-2 x^{2} \sin \left (\frac {a}{2 x^{2}}\right )+a \cos \left (\frac {a}{2 x^{2}}\right )\right )}{x^{2} a^{4}} \]

Therefore the general solution is

\begin{align*} y &= y_h + y_p \\ &= \left (c_1 \cos \left (\frac {x \sqrt {\frac {a^{2}}{x^{6}}}}{2}\right )-c_2 \sin \left (\frac {x \sqrt {\frac {a^{2}}{x^{6}}}}{2}\right )\right ) + \left (\frac {x^{3} \sqrt {\frac {a^{2}}{x^{6}}}\, \left (2 x^{2} \cos \left (\frac {a}{2 x^{2}}\right )+a \sin \left (\frac {a}{2 x^{2}}\right )-2 x^{2}\right ) \sin \left (\frac {x \sqrt {\frac {a^{2}}{x^{6}}}}{2}\right )+a \cos \left (\frac {x \sqrt {\frac {a^{2}}{x^{6}}}}{2}\right ) \left (-2 x^{2} \sin \left (\frac {a}{2 x^{2}}\right )+a \cos \left (\frac {a}{2 x^{2}}\right )\right )}{x^{2} a^{4}}\right ) \\ \end{align*}

Will add steps showing solving for IC soon.

Summary of solutions found

\begin{align*} y &= c_1 \cos \left (\frac {x \sqrt {\frac {a^{2}}{x^{6}}}}{2}\right )-c_2 \sin \left (\frac {x \sqrt {\frac {a^{2}}{x^{6}}}}{2}\right )+\frac {x^{3} \sqrt {\frac {a^{2}}{x^{6}}}\, \left (2 x^{2} \cos \left (\frac {a}{2 x^{2}}\right )+a \sin \left (\frac {a}{2 x^{2}}\right )-2 x^{2}\right ) \sin \left (\frac {x \sqrt {\frac {a^{2}}{x^{6}}}}{2}\right )+a \cos \left (\frac {x \sqrt {\frac {a^{2}}{x^{6}}}}{2}\right ) \left (-2 x^{2} \sin \left (\frac {a}{2 x^{2}}\right )+a \cos \left (\frac {a}{2 x^{2}}\right )\right )}{x^{2} a^{4}} \\ \end{align*}

Solved as second order Bessel ode

Time used: 0.364 (sec)

Writing the ode as

\begin{align*} x^{2} y^{\prime \prime }+3 y^{\prime } x +\frac {a^{2} y}{x^{4}} = \frac {1}{x^{6}}\tag {1} \end{align*}

Let the solution be

\begin{align*} y &= y_h + y_p \end{align*}

Where \(y_h\) is the solution to the homogeneous ODE and \(y_p\) is a particular solution to the non-homogeneous ODE. Bessel ode has the form

\begin{align*} x^{2} y^{\prime \prime }+y^{\prime } x +\left (-n^{2}+x^{2}\right ) y = 0\tag {2} \end{align*}

The generalized form of Bessel ode is given by Bowman (1958) as the following

\begin{align*} x^{2} y^{\prime \prime }+\left (1-2 \alpha \right ) x y^{\prime }+\left (\beta ^{2} \gamma ^{2} x^{2 \gamma }-n^{2} \gamma ^{2}+\alpha ^{2}\right ) y = 0\tag {3} \end{align*}

With the standard solution

\begin{align*} y&=x^{\alpha } \left (c_1 \operatorname {BesselJ}\left (n , \beta \,x^{\gamma }\right )+c_2 \operatorname {BesselY}\left (n , \beta \,x^{\gamma }\right )\right )\tag {4} \end{align*}

Comparing (3) to (1) and solving for \(\alpha ,\beta ,n,\gamma \) gives

\begin{align*} \alpha &= -1\\ \beta &= \frac {a}{2}\\ n &= {\frac {1}{2}}\\ \gamma &= -2 \end{align*}

Substituting all the above into (4) gives the solution as

\begin{align*} y = \frac {2 c_1 \sin \left (\frac {a}{2 x^{2}}\right )}{x \sqrt {\pi }\, \sqrt {\frac {a}{x^{2}}}}-\frac {2 c_2 \cos \left (\frac {a}{2 x^{2}}\right )}{x \sqrt {\pi }\, \sqrt {\frac {a}{x^{2}}}} \end{align*}

Therefore the homogeneous solution \(y_h\) is

\[ y_h = \frac {2 c_1 \sin \left (\frac {a}{2 x^{2}}\right )}{x \sqrt {\pi }\, \sqrt {\frac {a}{x^{2}}}}-\frac {2 c_2 \cos \left (\frac {a}{2 x^{2}}\right )}{x \sqrt {\pi }\, \sqrt {\frac {a}{x^{2}}}} \]

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 &= \frac {2 \sin \left (\frac {a}{2 x^{2}}\right )}{x \sqrt {\pi }\, \sqrt {\frac {a}{x^{2}}}} \\ y_2 &= -\frac {2 \cos \left (\frac {a}{2 x^{2}}\right )}{x \sqrt {\pi }\, \sqrt {\frac {a}{x^{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

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

Which gives

\[ W = \begin {vmatrix} \frac {2 \sin \left (\frac {a}{2 x^{2}}\right )}{x \sqrt {\pi }\, \sqrt {\frac {a}{x^{2}}}} & -\frac {2 \cos \left (\frac {a}{2 x^{2}}\right )}{x \sqrt {\pi }\, \sqrt {\frac {a}{x^{2}}}} \\ -\frac {2 \sin \left (\frac {a}{2 x^{2}}\right )}{x^{2} \sqrt {\pi }\, \sqrt {\frac {a}{x^{2}}}}+\frac {2 \sin \left (\frac {a}{2 x^{2}}\right ) a}{x^{4} \sqrt {\pi }\, \left (\frac {a}{x^{2}}\right )^{{3}/{2}}}-\frac {2 a \cos \left (\frac {a}{2 x^{2}}\right )}{x^{4} \sqrt {\pi }\, \sqrt {\frac {a}{x^{2}}}} & \frac {2 \cos \left (\frac {a}{2 x^{2}}\right )}{x^{2} \sqrt {\pi }\, \sqrt {\frac {a}{x^{2}}}}-\frac {2 \cos \left (\frac {a}{2 x^{2}}\right ) a}{x^{4} \sqrt {\pi }\, \left (\frac {a}{x^{2}}\right )^{{3}/{2}}}-\frac {2 a \sin \left (\frac {a}{2 x^{2}}\right )}{x^{4} \sqrt {\pi }\, \sqrt {\frac {a}{x^{2}}}} \end {vmatrix} \]

Therefore

\[ W = \left (\frac {2 \sin \left (\frac {a}{2 x^{2}}\right )}{x \sqrt {\pi }\, \sqrt {\frac {a}{x^{2}}}}\right )\left (\frac {2 \cos \left (\frac {a}{2 x^{2}}\right )}{x^{2} \sqrt {\pi }\, \sqrt {\frac {a}{x^{2}}}}-\frac {2 \cos \left (\frac {a}{2 x^{2}}\right ) a}{x^{4} \sqrt {\pi }\, \left (\frac {a}{x^{2}}\right )^{{3}/{2}}}-\frac {2 a \sin \left (\frac {a}{2 x^{2}}\right )}{x^{4} \sqrt {\pi }\, \sqrt {\frac {a}{x^{2}}}}\right ) - \left (-\frac {2 \cos \left (\frac {a}{2 x^{2}}\right )}{x \sqrt {\pi }\, \sqrt {\frac {a}{x^{2}}}}\right )\left (-\frac {2 \sin \left (\frac {a}{2 x^{2}}\right )}{x^{2} \sqrt {\pi }\, \sqrt {\frac {a}{x^{2}}}}+\frac {2 \sin \left (\frac {a}{2 x^{2}}\right ) a}{x^{4} \sqrt {\pi }\, \left (\frac {a}{x^{2}}\right )^{{3}/{2}}}-\frac {2 a \cos \left (\frac {a}{2 x^{2}}\right )}{x^{4} \sqrt {\pi }\, \sqrt {\frac {a}{x^{2}}}}\right ) \]

Which simplifies to

\[ W = -\frac {4 \left (\cos \left (\frac {a}{2 x^{2}}\right )^{2}+\sin \left (\frac {a}{2 x^{2}}\right )^{2}\right )}{x^{3} \pi } \]

Which simplifies to

\[ W = -\frac {4}{x^{3} \pi } \]

Therefore Eq. (2) becomes

\[ u_1 = -\int \frac {-\frac {2 \cos \left (\frac {a}{2 x^{2}}\right )}{x^{7} \sqrt {\pi }\, \sqrt {\frac {a}{x^{2}}}}}{-\frac {4}{x \pi }}\,dx \]

Which simplifies to

\[ u_1 = - \int \frac {\sqrt {\pi }\, \cos \left (\frac {a}{2 x^{2}}\right )}{2 x^{6} \sqrt {\frac {a}{x^{2}}}}d x \]

Hence

\[ u_1 = \frac {\sqrt {\pi }\, \cos \left (\frac {a}{2 x^{2}}\right )}{x \,a^{2} \sqrt {\frac {a}{x^{2}}}}+\frac {\sqrt {\pi }\, \sin \left (\frac {a}{2 x^{2}}\right )}{2 a \,x^{3} \sqrt {\frac {a}{x^{2}}}} \]

And Eq. (3) becomes

\[ u_2 = \int \frac {\frac {2 \sin \left (\frac {a}{2 x^{2}}\right )}{x^{7} \sqrt {\pi }\, \sqrt {\frac {a}{x^{2}}}}}{-\frac {4}{x \pi }}\,dx \]

Which simplifies to

\[ u_2 = \int -\frac {\sqrt {\pi }\, \sin \left (\frac {a}{2 x^{2}}\right )}{2 x^{6} \sqrt {\frac {a}{x^{2}}}}d x \]

Hence

\[ u_2 = -\frac {\sqrt {\pi }\, \cos \left (\frac {a}{2 x^{2}}\right )}{2 a \,x^{3} \sqrt {\frac {a}{x^{2}}}}+\frac {\sqrt {\pi }\, \sin \left (\frac {a}{2 x^{2}}\right )}{x \,a^{2} \sqrt {\frac {a}{x^{2}}}} \]

Therefore the particular solution, from equation (1) is

\[ y_p(x) = \frac {2 \left (\frac {\sqrt {\pi }\, \cos \left (\frac {a}{2 x^{2}}\right )}{x \,a^{2} \sqrt {\frac {a}{x^{2}}}}+\frac {\sqrt {\pi }\, \sin \left (\frac {a}{2 x^{2}}\right )}{2 a \,x^{3} \sqrt {\frac {a}{x^{2}}}}\right ) \sin \left (\frac {a}{2 x^{2}}\right )}{x \sqrt {\pi }\, \sqrt {\frac {a}{x^{2}}}}-\frac {2 \cos \left (\frac {a}{2 x^{2}}\right ) \left (-\frac {\sqrt {\pi }\, \cos \left (\frac {a}{2 x^{2}}\right )}{2 a \,x^{3} \sqrt {\frac {a}{x^{2}}}}+\frac {\sqrt {\pi }\, \sin \left (\frac {a}{2 x^{2}}\right )}{x \,a^{2} \sqrt {\frac {a}{x^{2}}}}\right )}{x \sqrt {\pi }\, \sqrt {\frac {a}{x^{2}}}} \]

Which simplifies to

\[ y_p(x) = \frac {1}{a^{2} x^{2}} \]

Therefore the general solution is

\begin{align*} y &= y_h + y_p \\ &= \left (\frac {2 c_1 \sin \left (\frac {a}{2 x^{2}}\right )}{x \sqrt {\pi }\, \sqrt {\frac {a}{x^{2}}}}-\frac {2 c_2 \cos \left (\frac {a}{2 x^{2}}\right )}{x \sqrt {\pi }\, \sqrt {\frac {a}{x^{2}}}}\right ) + \left (\frac {1}{a^{2} x^{2}}\right ) \\ \end{align*}

Will add steps showing solving for IC soon.

Summary of solutions found

\begin{align*} y &= \frac {2 c_1 \sin \left (\frac {a}{2 x^{2}}\right )}{x \sqrt {\pi }\, \sqrt {\frac {a}{x^{2}}}}-\frac {2 c_2 \cos \left (\frac {a}{2 x^{2}}\right )}{x \sqrt {\pi }\, \sqrt {\frac {a}{x^{2}}}}+\frac {1}{a^{2} x^{2}} \\ \end{align*}

Solved as second order ode adjoint method

Time used: 2.253 (sec)

In normal form the ode

\begin{align*} x^{6} y^{\prime \prime }+3 x^{5} y^{\prime }+a^{2} y = \frac {1}{x^{2}} \tag {1} \end{align*}

Becomes

\begin{align*} y^{\prime \prime }+p \left (x \right ) y^{\prime }+q \left (x \right ) y&=r \left (x \right ) \tag {2} \end{align*}

Where

\begin{align*} p \left (x \right )&=\frac {3}{x}\\ q \left (x \right )&=\frac {a^{2}}{x^{6}}\\ r \left (x \right )&=\frac {1}{x^{8}} \end{align*}

The Lagrange adjoint ode is given by

\begin{align*} \xi ^{''}-(\xi \, p)'+\xi q &= 0\\ \xi ^{''}-\left (\frac {3 \xi \left (x \right )}{x}\right )' + \left (\frac {a^{2} \xi \left (x \right )}{x^{6}}\right ) &= 0\\ \xi ^{\prime \prime }\left (x \right )-\frac {3 \xi ^{\prime }\left (x \right )}{x}+\frac {\left (3 x^{4}+a^{2}\right ) \xi \left (x \right )}{x^{6}}&= 0 \end{align*}

Which is solved for \(\xi (x)\). Writing the ode as

\begin{align*} x^{2} \xi ^{\prime \prime }-3 \xi ^{\prime } x +\left (3+\frac {a^{2}}{x^{4}}\right ) \xi = 0\tag {1} \end{align*}

Bessel ode has the form

\begin{align*} x^{2} \xi ^{\prime \prime }+\xi ^{\prime } x +\left (-n^{2}+x^{2}\right ) \xi = 0\tag {2} \end{align*}

The generalized form of Bessel ode is given by Bowman (1958) as the following

\begin{align*} x^{2} \xi ^{\prime \prime }+\left (1-2 \alpha \right ) x \xi ^{\prime }+\left (\beta ^{2} \gamma ^{2} x^{2 \gamma }-n^{2} \gamma ^{2}+\alpha ^{2}\right ) \xi = 0\tag {3} \end{align*}

With the standard solution

\begin{align*} \xi &=x^{\alpha } \left (c_3 \operatorname {BesselJ}\left (n , \beta \,x^{\gamma }\right )+c_4 \operatorname {BesselY}\left (n , \beta \,x^{\gamma }\right )\right )\tag {4} \end{align*}

Comparing (3) to (1) and solving for \(\alpha ,\beta ,n,\gamma \) gives

\begin{align*} \alpha &= 2\\ \beta &= \frac {a}{2}\\ n &= -{\frac {1}{2}}\\ \gamma &= -2 \end{align*}

Substituting all the above into (4) gives the solution as

\begin{align*} \xi = \frac {2 c_3 \,x^{2} \cos \left (\frac {a}{2 x^{2}}\right )}{\sqrt {\pi }\, \sqrt {\frac {a}{x^{2}}}}+\frac {2 c_4 \,x^{2} \sin \left (\frac {a}{2 x^{2}}\right )}{\sqrt {\pi }\, \sqrt {\frac {a}{x^{2}}}} \end{align*}

Will add steps showing solving for IC soon.

The original ode now reduces to first order ode

\begin{align*} \xi \left (x \right ) y^{\prime }-y \xi ^{\prime }\left (x \right )+\xi \left (x \right ) p \left (x \right ) y&=\int \xi \left (x \right ) r \left (x \right )d x\\ y^{\prime }+y \left (p \left (x \right )-\frac {\xi ^{\prime }\left (x \right )}{\xi \left (x \right )}\right )&=\frac {\int \xi \left (x \right ) r \left (x \right )d x}{\xi \left (x \right )} \end{align*}

Or

\begin{align*} y^{\prime }+y \left (\frac {3}{x}-\frac {\frac {4 c_3 x \cos \left (\frac {a}{2 x^{2}}\right )}{\sqrt {\pi }\, \sqrt {\frac {a}{x^{2}}}}+\frac {2 c_3 \cos \left (\frac {a}{2 x^{2}}\right ) a}{x \sqrt {\pi }\, \left (\frac {a}{x^{2}}\right )^{{3}/{2}}}+\frac {2 c_3 a \sin \left (\frac {a}{2 x^{2}}\right )}{x \sqrt {\pi }\, \sqrt {\frac {a}{x^{2}}}}+\frac {4 c_4 x \sin \left (\frac {a}{2 x^{2}}\right )}{\sqrt {\pi }\, \sqrt {\frac {a}{x^{2}}}}+\frac {2 c_4 \sin \left (\frac {a}{2 x^{2}}\right ) a}{x \sqrt {\pi }\, \left (\frac {a}{x^{2}}\right )^{{3}/{2}}}-\frac {2 c_4 a \cos \left (\frac {a}{2 x^{2}}\right )}{x \sqrt {\pi }\, \sqrt {\frac {a}{x^{2}}}}}{\frac {2 c_3 \,x^{2} \cos \left (\frac {a}{2 x^{2}}\right )}{\sqrt {\pi }\, \sqrt {\frac {a}{x^{2}}}}+\frac {2 c_4 \,x^{2} \sin \left (\frac {a}{2 x^{2}}\right )}{\sqrt {\pi }\, \sqrt {\frac {a}{x^{2}}}}}\right )&=\frac {\frac {2 \left (-2 c_3 \,x^{2}+c_4 a \right ) \cos \left (\frac {a}{2 x^{2}}\right )}{\sqrt {\pi }\, a^{2} x^{3} \sqrt {\frac {a}{x^{2}}}}-\frac {2 \left (2 c_4 \,x^{2}+c_3 a \right ) \sin \left (\frac {a}{2 x^{2}}\right )}{\sqrt {\pi }\, a^{2} x^{3} \sqrt {\frac {a}{x^{2}}}}}{\frac {2 c_3 \,x^{2} \cos \left (\frac {a}{2 x^{2}}\right )}{\sqrt {\pi }\, \sqrt {\frac {a}{x^{2}}}}+\frac {2 c_4 \,x^{2} \sin \left (\frac {a}{2 x^{2}}\right )}{\sqrt {\pi }\, \sqrt {\frac {a}{x^{2}}}}} \end{align*}

Which is now a first order ode. This is now solved for \(y\). In canonical form a linear first order is

\begin{align*} y^{\prime } + q(x)y &= p(x) \end{align*}

Comparing the above to the given ode shows that

\begin{align*} q(x) &=-\frac {a \left (\sin \left (\frac {a}{2 x^{2}}\right ) c_3 -\cos \left (\frac {a}{2 x^{2}}\right ) c_4 \right )}{x^{3} \left (c_4 \sin \left (\frac {a}{2 x^{2}}\right )+c_3 \cos \left (\frac {a}{2 x^{2}}\right )\right )}\\ p(x) &=\frac {\left (-2 c_4 \,x^{2}-c_3 a \right ) \sin \left (\frac {a}{2 x^{2}}\right )+\left (-2 c_3 \,x^{2}+c_4 a \right ) \cos \left (\frac {a}{2 x^{2}}\right )}{\left (c_4 \sin \left (\frac {a}{2 x^{2}}\right )+c_3 \cos \left (\frac {a}{2 x^{2}}\right )\right ) x^{5} a^{2}} \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,dx}}\\ &= {\mathrm e}^{\int -\frac {a \left (\sin \left (\frac {a}{2 x^{2}}\right ) c_3 -\cos \left (\frac {a}{2 x^{2}}\right ) c_4 \right )}{x^{3} \left (c_4 \sin \left (\frac {a}{2 x^{2}}\right )+c_3 \cos \left (\frac {a}{2 x^{2}}\right )\right )}d x}\\ &= \frac {1}{c_4 \sin \left (\frac {a}{2 x^{2}}\right )+c_3 \cos \left (\frac {a}{2 x^{2}}\right )} \end{align*}

The ode becomes

\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \mu p \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \left (\mu \right ) \left (\frac {\left (-2 c_4 \,x^{2}-c_3 a \right ) \sin \left (\frac {a}{2 x^{2}}\right )+\left (-2 c_3 \,x^{2}+c_4 a \right ) \cos \left (\frac {a}{2 x^{2}}\right )}{\left (c_4 \sin \left (\frac {a}{2 x^{2}}\right )+c_3 \cos \left (\frac {a}{2 x^{2}}\right )\right ) x^{5} a^{2}}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (\frac {y}{c_4 \sin \left (\frac {a}{2 x^{2}}\right )+c_3 \cos \left (\frac {a}{2 x^{2}}\right )}\right ) &= \left (\frac {1}{c_4 \sin \left (\frac {a}{2 x^{2}}\right )+c_3 \cos \left (\frac {a}{2 x^{2}}\right )}\right ) \left (\frac {\left (-2 c_4 \,x^{2}-c_3 a \right ) \sin \left (\frac {a}{2 x^{2}}\right )+\left (-2 c_3 \,x^{2}+c_4 a \right ) \cos \left (\frac {a}{2 x^{2}}\right )}{\left (c_4 \sin \left (\frac {a}{2 x^{2}}\right )+c_3 \cos \left (\frac {a}{2 x^{2}}\right )\right ) x^{5} a^{2}}\right ) \\ \mathrm {d} \left (\frac {y}{c_4 \sin \left (\frac {a}{2 x^{2}}\right )+c_3 \cos \left (\frac {a}{2 x^{2}}\right )}\right ) &= \left (\frac {\left (-2 c_4 \,x^{2}-c_3 a \right ) \sin \left (\frac {a}{2 x^{2}}\right )+\left (-2 c_3 \,x^{2}+c_4 a \right ) \cos \left (\frac {a}{2 x^{2}}\right )}{{\left (c_4 \sin \left (\frac {a}{2 x^{2}}\right )+c_3 \cos \left (\frac {a}{2 x^{2}}\right )\right )}^{2} x^{5} a^{2}}\right )\, \mathrm {d} x \\ \end{align*}

Integrating gives

\begin{align*} \frac {y}{c_4 \sin \left (\frac {a}{2 x^{2}}\right )+c_3 \cos \left (\frac {a}{2 x^{2}}\right )}&= \int {\frac {\left (-2 c_4 \,x^{2}-c_3 a \right ) \sin \left (\frac {a}{2 x^{2}}\right )+\left (-2 c_3 \,x^{2}+c_4 a \right ) \cos \left (\frac {a}{2 x^{2}}\right )}{{\left (c_4 \sin \left (\frac {a}{2 x^{2}}\right )+c_3 \cos \left (\frac {a}{2 x^{2}}\right )\right )}^{2} x^{5} a^{2}} \,dx} \\ &=-\frac {2 \,{\mathrm e}^{\frac {i a}{2 x^{2}}}}{x^{2} \left (\left (i c_4 -c_3 \right ) {\mathrm e}^{\frac {i a}{x^{2}}}-i c_4 -c_3 \right ) a^{2}} + c_5 \end{align*}

Dividing throughout by the integrating factor \(\frac {1}{c_4 \sin \left (\frac {a}{2 x^{2}}\right )+c_3 \cos \left (\frac {a}{2 x^{2}}\right )}\) gives the final solution

\[ y = \frac {\left (c_4 \sin \left (\frac {a}{2 x^{2}}\right )+c_3 \cos \left (\frac {a}{2 x^{2}}\right )\right ) \left (-2 \,{\mathrm e}^{\frac {i a}{2 x^{2}}}+c_5 \left (\left (i c_4 -c_3 \right ) {\mathrm e}^{\frac {i a}{x^{2}}}-i c_4 -c_3 \right ) x^{2} a^{2}\right )}{\left (\left (i c_4 -c_3 \right ) {\mathrm e}^{\frac {i a}{x^{2}}}-i c_4 -c_3 \right ) x^{2} a^{2}} \]

Hence, the solution found using Lagrange adjoint equation method is

\begin{align*} y &= \frac {\left (c_4 \sin \left (\frac {a}{2 x^{2}}\right )+c_3 \cos \left (\frac {a}{2 x^{2}}\right )\right ) \left (-2 \,{\mathrm e}^{\frac {i a}{2 x^{2}}}+c_5 \left (\left (i c_4 -c_3 \right ) {\mathrm e}^{\frac {i a}{x^{2}}}-i c_4 -c_3 \right ) x^{2} a^{2}\right )}{\left (\left (i c_4 -c_3 \right ) {\mathrm e}^{\frac {i a}{x^{2}}}-i c_4 -c_3 \right ) x^{2} a^{2}} \\ \end{align*}

The constants can be merged to give

\[ y = \frac {\left (c_4 \sin \left (\frac {a}{2 x^{2}}\right )+c_3 \cos \left (\frac {a}{2 x^{2}}\right )\right ) \left (-2 \,{\mathrm e}^{\frac {i a}{2 x^{2}}}+\left (\left (i c_4 -c_3 \right ) {\mathrm e}^{\frac {i a}{x^{2}}}-i c_4 -c_3 \right ) x^{2} a^{2}\right )}{\left (\left (i c_4 -c_3 \right ) {\mathrm e}^{\frac {i a}{x^{2}}}-i c_4 -c_3 \right ) x^{2} a^{2}} \]

Will add steps showing solving for IC soon.

Summary of solutions found

\begin{align*} y &= \frac {\left (c_4 \sin \left (\frac {a}{2 x^{2}}\right )+c_3 \cos \left (\frac {a}{2 x^{2}}\right )\right ) \left (-2 \,{\mathrm e}^{\frac {i a}{2 x^{2}}}+\left (\left (i c_4 -c_3 \right ) {\mathrm e}^{\frac {i a}{x^{2}}}-i c_4 -c_3 \right ) x^{2} a^{2}\right )}{\left (\left (i c_4 -c_3 \right ) {\mathrm e}^{\frac {i a}{x^{2}}}-i c_4 -c_3 \right ) x^{2} a^{2}} \\ \end{align*}

Maple step by step solution

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)] 
-> Try solving first the homogeneous part of the ODE 
   checking if the LODE has constant coefficients 
   checking if the LODE is of Euler type 
   trying a symmetry of the form [xi=0, eta=F(x)] 
   <- linear_1 successful 
<- solving first the homogeneous part of the ODE successful`
 

Maple dsolve solution

Solving time : 0.006 (sec)
Leaf size : 30

dsolve(diff(diff(y(x),x),x)*x^6+3*x^5*diff(y(x),x)+a^2*y(x) = 1/x^2,y(x),singsol=all)
 

\[ y = \sin \left (\frac {a}{2 x^{2}}\right ) c_{2} +\cos \left (\frac {a}{2 x^{2}}\right ) c_{1} +\frac {1}{a^{2} x^{2}} \]

✓Mathematica DSolve solution

Solving time : 0.071 (sec)
Leaf size : 104

DSolve[{x^6*D[y[x],{x,2}]+3*x^5*D[y[x],x]+a^2*y[x]==1/x^2,{}},y[x],x,IncludeSingularSolutions->True]
 

\[ y(x)\to -\sin \left (\frac {a}{2 x^2}\right ) \int _1^x\frac {\cos \left (\frac {a}{2 K[1]^2}\right )}{a K[1]^5}dK[1]+\frac {-2 a^3 c_2 x^2 \sin \left (\frac {a}{2 x^2}\right )-2 x^2 \sin \left (\frac {a}{x^2}\right )+a \cos \left (\frac {a}{x^2}\right )+a}{2 a^3 x^2}+c_1 \cos \left (\frac {a}{2 x^2}\right ) \]

✗Sympy solution

Solving time : 0.000 (sec)
Leaf size : 0

Python version: 3.13.1 (main, Dec  4 2024, 18:05:56) [GCC 14.2.1 20240910] 
Sympy version 1.13.3
 

from sympy import * 
x = symbols("x") 
a = symbols("a") 
y = Function("y") 
ode = Eq(a**2*y(x) + x**6*Derivative(y(x), (x, 2)) + 3*x**5*Derivative(y(x), x) - 1/x**2,0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

NotImplementedError : The given ODE Derivative(y(x), x) - (-a**2*x**2*y(x) - x**8*Derivative(y(x), (x, 2)) + 1)/(3*x**7) cannot be solved by the factorable group method