Internal problem ID [7459]
Internal file name [OUTPUT/6426_Sunday_June_19_2022_05_02_04_PM_23103942/index.tex]
Book: Second order enumerated odes
Section: section 2
Problem number: 19.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_bessel_ode", "second_order_change_of_variable_on_x_method_1", "second_order_change_of_variable_on_x_method_2"
Maple gives the following as the ode type
[[_2nd_order, _linear, _nonhomogeneous]]
\[ \boxed {x^{6} y^{\prime \prime }+3 x^{5} y^{\prime }+a^{2} y=\frac {1}{x^{2}}} \]
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}^{-\left (\int p \left (x \right )d x \right )}d x\\ &= \int {\mathrm e}^{-\left (\int \frac {3}{x}d x \right )}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 \[ \lambda ^{2} {\mathrm e}^{\lambda \tau }+a^{2} {\mathrm e}^{\lambda \tau } = 0 \tag {1} \] 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 &= \sqrt {-a^{2}} \\ \lambda _2 &= -\sqrt {-a^{2}} \\ \end{align*} Since roots are real and 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 {-a^{2}}\right )\tau } +c_{2} e^{\left (-\sqrt {-a^{2}}\right )\tau } \\ \end{align*} Or \[ y \left (\tau \right ) =c_{1} {\mathrm e}^{\sqrt {-a^{2}}\, \tau }+c_{2} {\mathrm e}^{-\sqrt {-a^{2}}\, \tau } \] The above solution is now transformed back to \(y\) using (6) which results in \begin {align*} y &= c_{1} {\mathrm e}^{-\frac {\sqrt {-a^{2}}}{2 x^{2}}}+c_{2} {\mathrm e}^{\frac {\sqrt {-a^{2}}}{2 x^{2}}} \end {align*}
Therefore the homogeneous solution \(y_h\) is \[ y_h = c_{1} {\mathrm e}^{-\frac {\sqrt {-a^{2}}}{2 x^{2}}}+c_{2} {\mathrm e}^{\frac {\sqrt {-a^{2}}}{2 x^{2}}} \] 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 &= {\mathrm e}^{-\frac {\sqrt {-a^{2}}}{2 x^{2}}} \\ y_2 &= {\mathrm e}^{\frac {\sqrt {-a^{2}}}{2 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 \[ W = \begin {vmatrix} {\mathrm e}^{-\frac {\sqrt {-a^{2}}}{2 x^{2}}} & {\mathrm e}^{\frac {\sqrt {-a^{2}}}{2 x^{2}}} \\ \frac {d}{dx}\left ({\mathrm e}^{-\frac {\sqrt {-a^{2}}}{2 x^{2}}}\right ) & \frac {d}{dx}\left ({\mathrm e}^{\frac {\sqrt {-a^{2}}}{2 x^{2}}}\right ) \end {vmatrix} \] Which gives \[ W = \begin {vmatrix} {\mathrm e}^{-\frac {\sqrt {-a^{2}}}{2 x^{2}}} & {\mathrm e}^{\frac {\sqrt {-a^{2}}}{2 x^{2}}} \\ \frac {\sqrt {-a^{2}}\, {\mathrm e}^{-\frac {\sqrt {-a^{2}}}{2 x^{2}}}}{x^{3}} & -\frac {\sqrt {-a^{2}}\, {\mathrm e}^{\frac {\sqrt {-a^{2}}}{2 x^{2}}}}{x^{3}} \end {vmatrix} \] Therefore \[ W = \left ({\mathrm e}^{-\frac {\sqrt {-a^{2}}}{2 x^{2}}}\right )\left (-\frac {\sqrt {-a^{2}}\, {\mathrm e}^{\frac {\sqrt {-a^{2}}}{2 x^{2}}}}{x^{3}}\right ) - \left ({\mathrm e}^{\frac {\sqrt {-a^{2}}}{2 x^{2}}}\right )\left (\frac {\sqrt {-a^{2}}\, {\mathrm e}^{-\frac {\sqrt {-a^{2}}}{2 x^{2}}}}{x^{3}}\right ) \] Which simplifies to \[ W = -\frac {2 \,{\mathrm e}^{-\frac {\sqrt {-a^{2}}}{2 x^{2}}} \sqrt {-a^{2}}\, {\mathrm e}^{\frac {\sqrt {-a^{2}}}{2 x^{2}}}}{x^{3}} \] Which simplifies to \[ W = -\frac {2 \sqrt {-a^{2}}}{x^{3}} \] Therefore Eq. (2) becomes \[ u_1 = -\int \frac {\frac {{\mathrm e}^{\frac {\sqrt {-a^{2}}}{2 x^{2}}}}{x^{2}}}{-2 x^{3} \sqrt {-a^{2}}}\,dx \] Which simplifies to \[ u_1 = - \int -\frac {{\mathrm e}^{\frac {\sqrt {-a^{2}}}{2 x^{2}}}}{2 x^{5} \sqrt {-a^{2}}}d x \] Hence \[ u_1 = \frac {-\frac {{\mathrm e}^{\frac {\sqrt {-a^{2}}}{2 x^{2}}}}{x^{2} \sqrt {-a^{2}}}-\frac {2 \,{\mathrm e}^{\frac {\sqrt {-a^{2}}}{2 x^{2}}}}{a^{2}}}{2 \sqrt {-a^{2}}} \] And Eq. (3) becomes \[ u_2 = \int \frac {\frac {{\mathrm e}^{-\frac {\sqrt {-a^{2}}}{2 x^{2}}}}{x^{2}}}{-2 x^{3} \sqrt {-a^{2}}}\,dx \] Which simplifies to \[ u_2 = \int -\frac {{\mathrm e}^{-\frac {\sqrt {-a^{2}}}{2 x^{2}}}}{2 x^{5} \sqrt {-a^{2}}}d x \] Hence \[ u_2 = -\frac {\frac {{\mathrm e}^{-\frac {\sqrt {-a^{2}}}{2 x^{2}}}}{x^{2} \sqrt {-a^{2}}}-\frac {2 \,{\mathrm e}^{-\frac {\sqrt {-a^{2}}}{2 x^{2}}}}{a^{2}}}{2 \sqrt {-a^{2}}} \] Which simplifies to \begin{align*} u_1 &= \frac {{\mathrm e}^{\frac {\sqrt {-a^{2}}}{2 x^{2}}} \left (2 x^{2} \sqrt {-a^{2}}+a^{2}\right )}{2 a^{4} x^{2}} \\ u_2 &= \frac {{\mathrm e}^{-\frac {\sqrt {-a^{2}}}{2 x^{2}}} \left (-2 x^{2} \sqrt {-a^{2}}+a^{2}\right )}{2 a^{4} x^{2}} \\ \end{align*} Therefore the particular solution, from equation (1) is \[ y_p(x) = \frac {{\mathrm e}^{\frac {\sqrt {-a^{2}}}{2 x^{2}}} \left (2 x^{2} \sqrt {-a^{2}}+a^{2}\right ) {\mathrm e}^{-\frac {\sqrt {-a^{2}}}{2 x^{2}}}}{2 a^{4} x^{2}}+\frac {{\mathrm e}^{-\frac {\sqrt {-a^{2}}}{2 x^{2}}} \left (-2 x^{2} \sqrt {-a^{2}}+a^{2}\right ) {\mathrm e}^{\frac {\sqrt {-a^{2}}}{2 x^{2}}}}{2 a^{4} x^{2}} \] 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 (c_{1} {\mathrm e}^{-\frac {\sqrt {-a^{2}}}{2 x^{2}}}+c_{2} {\mathrm e}^{\frac {\sqrt {-a^{2}}}{2 x^{2}}}\right ) + \left (\frac {1}{a^{2} x^{2}}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} {\mathrm e}^{-\frac {\sqrt {-a^{2}}}{2 x^{2}}}+c_{2} {\mathrm e}^{\frac {\sqrt {-a^{2}}}{2 x^{2}}}+\frac {1}{a^{2} x^{2}} \\ \end{align*}
Verification of solutions
\[ y = c_{1} {\mathrm e}^{-\frac {\sqrt {-a^{2}}}{2 x^{2}}}+c_{2} {\mathrm e}^{\frac {\sqrt {-a^{2}}}{2 x^{2}}}+\frac {1}{a^{2} x^{2}} \] Verified OK.
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 \[ y = c_{1} \cos \left (\frac {a}{2 x^{2}}\right )-c_{2} \sin \left (\frac {a}{2 x^{2}}\right ) \] Now the particular solution to this ODE is found \[ x^{6} y^{\prime \prime }+3 x^{5} y^{\prime }+a^{2} y = \frac {1}{x^{2}} \] 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 &= {\mathrm e}^{-\frac {\sqrt {-a^{2}}}{2 x^{2}}} \\ y_2 &= {\mathrm e}^{\frac {\sqrt {-a^{2}}}{2 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 \[ W = \begin {vmatrix} {\mathrm e}^{-\frac {\sqrt {-a^{2}}}{2 x^{2}}} & {\mathrm e}^{\frac {\sqrt {-a^{2}}}{2 x^{2}}} \\ \frac {d}{dx}\left ({\mathrm e}^{-\frac {\sqrt {-a^{2}}}{2 x^{2}}}\right ) & \frac {d}{dx}\left ({\mathrm e}^{\frac {\sqrt {-a^{2}}}{2 x^{2}}}\right ) \end {vmatrix} \] Which gives \[ W = \begin {vmatrix} {\mathrm e}^{-\frac {\sqrt {-a^{2}}}{2 x^{2}}} & {\mathrm e}^{\frac {\sqrt {-a^{2}}}{2 x^{2}}} \\ \frac {\sqrt {-a^{2}}\, {\mathrm e}^{-\frac {\sqrt {-a^{2}}}{2 x^{2}}}}{x^{3}} & -\frac {\sqrt {-a^{2}}\, {\mathrm e}^{\frac {\sqrt {-a^{2}}}{2 x^{2}}}}{x^{3}} \end {vmatrix} \] Therefore \[ W = \left ({\mathrm e}^{-\frac {\sqrt {-a^{2}}}{2 x^{2}}}\right )\left (-\frac {\sqrt {-a^{2}}\, {\mathrm e}^{\frac {\sqrt {-a^{2}}}{2 x^{2}}}}{x^{3}}\right ) - \left ({\mathrm e}^{\frac {\sqrt {-a^{2}}}{2 x^{2}}}\right )\left (\frac {\sqrt {-a^{2}}\, {\mathrm e}^{-\frac {\sqrt {-a^{2}}}{2 x^{2}}}}{x^{3}}\right ) \] Which simplifies to \[ W = -\frac {2 \,{\mathrm e}^{-\frac {\sqrt {-a^{2}}}{2 x^{2}}} \sqrt {-a^{2}}\, {\mathrm e}^{\frac {\sqrt {-a^{2}}}{2 x^{2}}}}{x^{3}} \] Which simplifies to \[ W = -\frac {2 \sqrt {-a^{2}}}{x^{3}} \] Therefore Eq. (2) becomes \[ u_1 = -\int \frac {\frac {{\mathrm e}^{\frac {\sqrt {-a^{2}}}{2 x^{2}}}}{x^{2}}}{-2 x^{3} \sqrt {-a^{2}}}\,dx \] Which simplifies to \[ u_1 = - \int -\frac {{\mathrm e}^{\frac {\sqrt {-a^{2}}}{2 x^{2}}}}{2 x^{5} \sqrt {-a^{2}}}d x \] Hence \[ u_1 = \frac {-\frac {{\mathrm e}^{\frac {\sqrt {-a^{2}}}{2 x^{2}}}}{x^{2} \sqrt {-a^{2}}}-\frac {2 \,{\mathrm e}^{\frac {\sqrt {-a^{2}}}{2 x^{2}}}}{a^{2}}}{2 \sqrt {-a^{2}}} \] And Eq. (3) becomes \[ u_2 = \int \frac {\frac {{\mathrm e}^{-\frac {\sqrt {-a^{2}}}{2 x^{2}}}}{x^{2}}}{-2 x^{3} \sqrt {-a^{2}}}\,dx \] Which simplifies to \[ u_2 = \int -\frac {{\mathrm e}^{-\frac {\sqrt {-a^{2}}}{2 x^{2}}}}{2 x^{5} \sqrt {-a^{2}}}d x \] Hence \[ u_2 = -\frac {\frac {{\mathrm e}^{-\frac {\sqrt {-a^{2}}}{2 x^{2}}}}{x^{2} \sqrt {-a^{2}}}-\frac {2 \,{\mathrm e}^{-\frac {\sqrt {-a^{2}}}{2 x^{2}}}}{a^{2}}}{2 \sqrt {-a^{2}}} \] Which simplifies to \begin{align*} u_1 &= \frac {{\mathrm e}^{\frac {\sqrt {-a^{2}}}{2 x^{2}}} \left (2 x^{2} \sqrt {-a^{2}}+a^{2}\right )}{2 a^{4} x^{2}} \\ u_2 &= \frac {{\mathrm e}^{-\frac {\sqrt {-a^{2}}}{2 x^{2}}} \left (-2 x^{2} \sqrt {-a^{2}}+a^{2}\right )}{2 a^{4} x^{2}} \\ \end{align*} Therefore the particular solution, from equation (1) is \[ y_p(x) = \frac {{\mathrm e}^{\frac {\sqrt {-a^{2}}}{2 x^{2}}} \left (2 x^{2} \sqrt {-a^{2}}+a^{2}\right ) {\mathrm e}^{-\frac {\sqrt {-a^{2}}}{2 x^{2}}}}{2 a^{4} x^{2}}+\frac {{\mathrm e}^{-\frac {\sqrt {-a^{2}}}{2 x^{2}}} \left (-2 x^{2} \sqrt {-a^{2}}+a^{2}\right ) {\mathrm e}^{\frac {\sqrt {-a^{2}}}{2 x^{2}}}}{2 a^{4} x^{2}} \] 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 (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 ) \\ &= \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*} Which simplifies to \[ y = \frac {1}{a^{2} x^{2}}+c_{1} \cos \left (\frac {a}{2 x^{2}}\right )-c_{2} \sin \left (\frac {a}{2 x^{2}}\right ) \]
The solution(s) found are the following \begin{align*} \tag{1} 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*}
Verification of solutions
\[ y = \frac {1}{a^{2} x^{2}}+c_{1} \cos \left (\frac {a}{2 x^{2}}\right )-c_{2} \sin \left (\frac {a}{2 x^{2}}\right ) \] Verified OK.
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 &= {\mathrm e}^{-\frac {\sqrt {-a^{2}}}{2 x^{2}}} \\ y_2 &= {\mathrm e}^{\frac {\sqrt {-a^{2}}}{2 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 \[ W = \begin {vmatrix} {\mathrm e}^{-\frac {\sqrt {-a^{2}}}{2 x^{2}}} & {\mathrm e}^{\frac {\sqrt {-a^{2}}}{2 x^{2}}} \\ \frac {d}{dx}\left ({\mathrm e}^{-\frac {\sqrt {-a^{2}}}{2 x^{2}}}\right ) & \frac {d}{dx}\left ({\mathrm e}^{\frac {\sqrt {-a^{2}}}{2 x^{2}}}\right ) \end {vmatrix} \] Which gives \[ W = \begin {vmatrix} {\mathrm e}^{-\frac {\sqrt {-a^{2}}}{2 x^{2}}} & {\mathrm e}^{\frac {\sqrt {-a^{2}}}{2 x^{2}}} \\ \frac {\sqrt {-a^{2}}\, {\mathrm e}^{-\frac {\sqrt {-a^{2}}}{2 x^{2}}}}{x^{3}} & -\frac {\sqrt {-a^{2}}\, {\mathrm e}^{\frac {\sqrt {-a^{2}}}{2 x^{2}}}}{x^{3}} \end {vmatrix} \] Therefore \[ W = \left ({\mathrm e}^{-\frac {\sqrt {-a^{2}}}{2 x^{2}}}\right )\left (-\frac {\sqrt {-a^{2}}\, {\mathrm e}^{\frac {\sqrt {-a^{2}}}{2 x^{2}}}}{x^{3}}\right ) - \left ({\mathrm e}^{\frac {\sqrt {-a^{2}}}{2 x^{2}}}\right )\left (\frac {\sqrt {-a^{2}}\, {\mathrm e}^{-\frac {\sqrt {-a^{2}}}{2 x^{2}}}}{x^{3}}\right ) \] Which simplifies to \[ W = -\frac {2 \,{\mathrm e}^{-\frac {\sqrt {-a^{2}}}{2 x^{2}}} \sqrt {-a^{2}}\, {\mathrm e}^{\frac {\sqrt {-a^{2}}}{2 x^{2}}}}{x^{3}} \] Which simplifies to \[ W = -\frac {2 \sqrt {-a^{2}}}{x^{3}} \] Therefore Eq. (2) becomes \[ u_1 = -\int \frac {\frac {{\mathrm e}^{\frac {\sqrt {-a^{2}}}{2 x^{2}}}}{x^{6}}}{-\frac {2 \sqrt {-a^{2}}}{x}}\,dx \] Which simplifies to \[ u_1 = - \int -\frac {{\mathrm e}^{\frac {\sqrt {-a^{2}}}{2 x^{2}}}}{2 x^{5} \sqrt {-a^{2}}}d x \] Hence \[ u_1 = \frac {-\frac {{\mathrm e}^{\frac {\sqrt {-a^{2}}}{2 x^{2}}}}{x^{2} \sqrt {-a^{2}}}-\frac {2 \,{\mathrm e}^{\frac {\sqrt {-a^{2}}}{2 x^{2}}}}{a^{2}}}{2 \sqrt {-a^{2}}} \] And Eq. (3) becomes \[ u_2 = \int \frac {\frac {{\mathrm e}^{-\frac {\sqrt {-a^{2}}}{2 x^{2}}}}{x^{6}}}{-\frac {2 \sqrt {-a^{2}}}{x}}\,dx \] Which simplifies to \[ u_2 = \int -\frac {{\mathrm e}^{-\frac {\sqrt {-a^{2}}}{2 x^{2}}}}{2 x^{5} \sqrt {-a^{2}}}d x \] Hence \[ u_2 = -\frac {\frac {{\mathrm e}^{-\frac {\sqrt {-a^{2}}}{2 x^{2}}}}{x^{2} \sqrt {-a^{2}}}-\frac {2 \,{\mathrm e}^{-\frac {\sqrt {-a^{2}}}{2 x^{2}}}}{a^{2}}}{2 \sqrt {-a^{2}}} \] Which simplifies to \begin{align*} u_1 &= \frac {{\mathrm e}^{\frac {\sqrt {-a^{2}}}{2 x^{2}}} \left (2 x^{2} \sqrt {-a^{2}}+a^{2}\right )}{2 a^{4} x^{2}} \\ u_2 &= \frac {{\mathrm e}^{-\frac {\sqrt {-a^{2}}}{2 x^{2}}} \left (-2 x^{2} \sqrt {-a^{2}}+a^{2}\right )}{2 a^{4} x^{2}} \\ \end{align*} Therefore the particular solution, from equation (1) is \[ y_p(x) = \frac {{\mathrm e}^{\frac {\sqrt {-a^{2}}}{2 x^{2}}} \left (2 x^{2} \sqrt {-a^{2}}+a^{2}\right ) {\mathrm e}^{-\frac {\sqrt {-a^{2}}}{2 x^{2}}}}{2 a^{4} x^{2}}+\frac {{\mathrm e}^{-\frac {\sqrt {-a^{2}}}{2 x^{2}}} \left (-2 x^{2} \sqrt {-a^{2}}+a^{2}\right ) {\mathrm e}^{\frac {\sqrt {-a^{2}}}{2 x^{2}}}}{2 a^{4} x^{2}} \] 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*}
The solution(s) found are the following \begin{align*} \tag{1} 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*}
Verification of solutions
\[ y = \frac {2 c_{1} \sin \left (\frac {a}{2 x^{2}}\right )}{x \sqrt {\pi }\, \sqrt {\frac {a}{x^{2}}}}-\frac {2 c_{2} \cos \left (\frac {a}{2 x^{2}}\right )}{x \sqrt {\pi }\, \sqrt {\frac {a}{x^{2}}}}+\frac {1}{a^{2} x^{2}} \] Verified OK.
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`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 30
dsolve(x^6*diff(y(x),x$2)+3*x^5*diff(y(x),x)+a^2*y(x)=1/x^2,y(x), singsol=all)
\[ y \left (x \right ) = \sin \left (\frac {a}{2 x^{2}}\right ) c_{2} +\cos \left (\frac {a}{2 x^{2}}\right ) c_{1} +\frac {1}{a^{2} x^{2}} \]
✓ Solution by Mathematica
Time used: 0.073 (sec). Leaf size: 38
DSolve[x^6*y''[x]+3*x^5*y'[x]+a^2*y[x]==1/x^2,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {1}{a^2 x^2}+c_1 \cos \left (\frac {a}{2 x^2}\right )-c_2 \sin \left (\frac {a}{2 x^2}\right ) \]