Internal problem ID [15432]
Internal file name [OUTPUT/15433_Wednesday_May_08_2024_03_58_42_PM_47599306/index.tex]
Book: A book of problems in ordinary differential equations. M.L. KRASNOV, A.L. KISELYOV,
G.I. MARKARENKO. MIR, MOSCOW. 1983
Section: Chapter 2 (Higher order ODE’s). Section 15.5 Linear equations with variable coefficients.
The Lagrange method. Exercises page 148
Problem number: 668.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "kovacic", "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]]
Unable to solve or complete the solution.
\[ \boxed {4 x y^{\prime \prime }+2 y^{\prime }+y=\frac {6+x}{x^{2}}} \] With initial conditions \begin {align*} [y \left (\infty \right ) = 0] \end {align*}
This is a linear ODE. In canonical form it is written as \begin {align*} y^{\prime \prime } + p(x)y^{\prime } + q(x) y &= F \end {align*}
Where here \begin {align*} p(x) &=\frac {1}{2 x}\\ q(x) &=\frac {1}{4 x}\\ F &=\frac {6+x}{4 x^{3}} \end {align*}
Hence the ode is \begin {align*} y^{\prime \prime }+\frac {y^{\prime }}{2 x}+\frac {y}{4 x} = \frac {6+x}{4 x^{3}} \end {align*}
The domain of \(p(x)=\frac {1}{2 x}\) is \[
\{x <0\boldsymbol {\lor }0
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 \[
4 x y^{\prime \prime }+2 y^{\prime }+y = 0
\] In normal form the ode \begin {align*} 4 x y^{\prime \prime }+2 y^{\prime }+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 {1}{2 x}\\ q \left (x \right )&=\frac {1}{4 x} \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 {1}{2 x}d x \right )}d x\\ &= \int e^{-\frac {\ln \left (x \right )}{2}} \,dx\\ &= \int \frac {1}{\sqrt {x}}d x\\ &= 2 \sqrt {x}\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 {1}{4 x}}{\frac {1}{x}}\\ &= {\frac {1}{4}}\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 )+\frac {y \left (\tau \right )}{4}&=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 \(A=1, B=0, C={\frac {1}{4}}\). Let the solution be \(y \left (\tau \right )=e^{\lambda \tau }\). Substituting
this into the ODE gives \[ \lambda ^{2} {\mathrm e}^{\lambda \tau }+\frac {{\mathrm e}^{\lambda \tau }}{4} = 0 \tag {1} \] Since exponential function is never zero, then dividing Eq(2)
throughout by \(e^{\lambda \tau }\) gives \[ \lambda ^{2}+\frac {1}{4} = 0 \tag {2} \] 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 \(A=1, B=0, C={\frac {1}{4}}\) 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 ({\frac {1}{4}}\right )}\\ &= \pm \frac {i}{2} \end {align*}
Hence \begin {align*} \lambda _1 &= + \frac {i}{2}\\ \lambda _2 &= - \frac {i}{2} \end {align*}
Which simplifies to \begin{align*}
\lambda _1 &= \frac {i}{2} \\
\lambda _2 &= -\frac {i}{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 \(\beta ={\frac {1}{2}}\). 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 \[
y \left (\tau \right ) = e^{0}\left (c_{1} \cos \left (\frac {\tau }{2}\right )+c_{2} \sin \left (\frac {\tau }{2}\right )\right )
\] Or \[
y \left (\tau \right ) = c_{1} \cos \left (\frac {\tau }{2}\right )+c_{2} \sin \left (\frac {\tau }{2}\right )
\] The above solution is now transformed back to \(y\) using (6) which results
in \begin {align*} y &= c_{1} \cos \left (\sqrt {x}\right )+c_{2} \sin \left (\sqrt {x}\right ) \end {align*}
Therefore the homogeneous solution \(y_h\) is \[
y_h = c_{1} \cos \left (\sqrt {x}\right )+c_{2} \sin \left (\sqrt {x}\right )
\] 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 (\sqrt {x}\right ) \\
y_2 &= \sin \left (\sqrt {x}\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 \[ W = \begin {vmatrix} \cos \left (\sqrt {x}\right ) & \sin \left (\sqrt {x}\right ) \\ \frac {d}{dx}\left (\cos \left (\sqrt {x}\right )\right ) & \frac {d}{dx}\left (\sin \left (\sqrt {x}\right )\right ) \end {vmatrix} \] Which gives \[ W = \begin {vmatrix} \cos \left (\sqrt {x}\right ) & \sin \left (\sqrt {x}\right ) \\ -\frac {\sin \left (\sqrt {x}\right )}{2 \sqrt {x}} & \frac {\cos \left (\sqrt {x}\right )}{2 \sqrt {x}} \end {vmatrix} \] Therefore \[
W = \left (\cos \left (\sqrt {x}\right )\right )\left (\frac {\cos \left (\sqrt {x}\right )}{2 \sqrt {x}}\right ) - \left (\sin \left (\sqrt {x}\right )\right )\left (-\frac {\sin \left (\sqrt {x}\right )}{2 \sqrt {x}}\right )
\] Which
simplifies to \[
W = \frac {\cos \left (\sqrt {x}\right )^{2}+\sin \left (\sqrt {x}\right )^{2}}{2 \sqrt {x}}
\] Which simplifies to \[
W = \frac {1}{2 \sqrt {x}}
\] Therefore Eq. (2) becomes \[
u_1 = -\int \frac {\frac {\sin \left (\sqrt {x}\right ) \left (6+x \right )}{x^{2}}}{2 \sqrt {x}}\,dx
\] Which simplifies to \[
u_1 = - \int \frac {\sin \left (\sqrt {x}\right ) \left (6+x \right )}{2 x^{\frac {5}{2}}}d x
\]
Hence \[
u_1 = \frac {2 \sin \left (\sqrt {x}\right )}{x^{\frac {3}{2}}}+\frac {\cos \left (\sqrt {x}\right )}{x}
\] And Eq. (3) becomes \[
u_2 = \int \frac {\frac {\cos \left (\sqrt {x}\right ) \left (6+x \right )}{x^{2}}}{2 \sqrt {x}}\,dx
\] Which simplifies to \[
u_2 = \int \frac {\cos \left (\sqrt {x}\right ) \left (6+x \right )}{2 x^{\frac {5}{2}}}d x
\] Hence \[
u_2 = -\frac {2 \cos \left (\sqrt {x}\right )}{x^{\frac {3}{2}}}+\frac {\sin \left (\sqrt {x}\right )}{x}
\] Which simplifies to \begin{align*}
u_1 &= \frac {\cos \left (\sqrt {x}\right ) \sqrt {x}+2 \sin \left (\sqrt {x}\right )}{x^{\frac {3}{2}}} \\
u_2 &= \frac {\sin \left (\sqrt {x}\right ) \sqrt {x}-2 \cos \left (\sqrt {x}\right )}{x^{\frac {3}{2}}} \\
\end{align*}
Therefore the particular solution, from equation (1) is \[
y_p(x) = \frac {\left (\cos \left (\sqrt {x}\right ) \sqrt {x}+2 \sin \left (\sqrt {x}\right )\right ) \cos \left (\sqrt {x}\right )}{x^{\frac {3}{2}}}+\frac {\left (\sin \left (\sqrt {x}\right ) \sqrt {x}-2 \cos \left (\sqrt {x}\right )\right ) \sin \left (\sqrt {x}\right )}{x^{\frac {3}{2}}}
\] Which simplifies to \[
y_p(x) = \frac {1}{x}
\] Therefore
the general solution is \begin{align*}
y &= y_h + y_p \\
&= \left (c_{1} \cos \left (\sqrt {x}\right )+c_{2} \sin \left (\sqrt {x}\right )\right ) + \left (\frac {1}{x}\right ) \\
\end{align*} Initial conditions are used to solve for the constants of
integration.
Looking at the above solution \begin {align*} y = c_{1} \cos \left (\sqrt {x}\right )+c_{2} \sin \left (\sqrt {x}\right )+\frac {1}{x} \tag {1} \end {align*}
Initial conditions are now substituted in the above solution. This will generate the required
equations to solve for the integration constants. substituting \(y = 0\) and \(x = \infty \) in the above gives
\begin {align*} 0 = -{| c_{1} |}-{| c_{2} |}..{| c_{1} |}+{| c_{2} |}\tag {1A} \end {align*}
Equations {1A} are now solved for \(\{c_{1}, c_{2}\}\). There is no solution for the constants of integrations.
This solution is removed.
Verification of solutions N/A
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 \(A=4 x, B=2, C=1, f(x)=\frac {6+x}{x^{2}}\). 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 \[ 4 x y^{\prime \prime }+2 y^{\prime }+y = 0 \] In normal form the ode
\begin {align*} 4 x y^{\prime \prime }+2 y^{\prime }+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 {1}{2 x}\\ q \left (x \right )&=\frac {1}{4 x} \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 {1}{x}}}{2 c}\tag {6} \\ \tau '' &= -\frac {1}{4 c \sqrt {\frac {1}{x}}\, x^{2}} \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 {1}{4 c \sqrt {\frac {1}{x}}\, x^{2}}+\frac {1}{2 x}\frac {\sqrt {\frac {1}{x}}}{2 c}}{\left (\frac {\sqrt {\frac {1}{x}}}{2 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 \frac {\sqrt {\frac {1}{x}}}{2}d x}{c}\\ &= \frac {x \sqrt {\frac {1}{x}}}{c} \end {align*}
Substituting the above into the solution obtained gives \[
y = c_{1} \cos \left (\sqrt {x}\right )+c_{2} \sin \left (\sqrt {x}\right )
\] Now the particular solution to this
ODE is found \[ 4 x y^{\prime \prime }+2 y^{\prime }+y = \frac {6+x}{x^{2}} \] 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 &= \cos \left (x \sqrt {\frac {1}{x}}\right ) \\
y_2 &= \sin \left (x \sqrt {\frac {1}{x}}\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 \[ W = \begin {vmatrix} \cos \left (x \sqrt {\frac {1}{x}}\right ) & \sin \left (x \sqrt {\frac {1}{x}}\right ) \\ \frac {d}{dx}\left (\cos \left (x \sqrt {\frac {1}{x}}\right )\right ) & \frac {d}{dx}\left (\sin \left (x \sqrt {\frac {1}{x}}\right )\right ) \end {vmatrix} \] Which gives \[ W = \begin {vmatrix} \cos \left (x \sqrt {\frac {1}{x}}\right ) & \sin \left (x \sqrt {\frac {1}{x}}\right ) \\ -\left (\sqrt {\frac {1}{x}}-\frac {1}{2 x \sqrt {\frac {1}{x}}}\right ) \sin \left (x \sqrt {\frac {1}{x}}\right ) & \left (\sqrt {\frac {1}{x}}-\frac {1}{2 x \sqrt {\frac {1}{x}}}\right ) \cos \left (x \sqrt {\frac {1}{x}}\right ) \end {vmatrix} \] Therefore \[
W = \left (\cos \left (x \sqrt {\frac {1}{x}}\right )\right )\left (\left (\sqrt {\frac {1}{x}}-\frac {1}{2 x \sqrt {\frac {1}{x}}}\right ) \cos \left (x \sqrt {\frac {1}{x}}\right )\right ) - \left (\sin \left (x \sqrt {\frac {1}{x}}\right )\right )\left (-\left (\sqrt {\frac {1}{x}}-\frac {1}{2 x \sqrt {\frac {1}{x}}}\right ) \sin \left (x \sqrt {\frac {1}{x}}\right )\right )
\] Which simplifies to \[
W = \frac {\cos \left (x \sqrt {\frac {1}{x}}\right )^{2}+\sin \left (x \sqrt {\frac {1}{x}}\right )^{2}}{2 x \sqrt {\frac {1}{x}}}
\] Which
simplifies to \[
W = \frac {1}{2 x \sqrt {\frac {1}{x}}}
\] Therefore Eq. (2) becomes \[
u_1 = -\int \frac {\frac {\sin \left (x \sqrt {\frac {1}{x}}\right ) \left (6+x \right )}{x^{2}}}{\frac {2}{\sqrt {\frac {1}{x}}}}\,dx
\] Which simplifies to \[
u_1 = - \int \frac {\sin \left (x \sqrt {\frac {1}{x}}\right ) \left (6+x \right ) \sqrt {\frac {1}{x}}}{2 x^{2}}d x
\] Hence \[
u_1 = -\frac {3 \sqrt {\pi }\, \left (-\frac {8}{\sqrt {\pi }\, x}-\frac {4 \left (2 \mcoloneq \gamma -\frac {11}{3}+\ln \left (x \right )+2 \ln \left (\sqrt {\frac {1}{x}}\, \sqrt {x}\right )\right )}{3 \sqrt {\pi }}+\frac {-\frac {44 x}{9}+8}{\sqrt {\pi }\, x}+\frac {8 \mcoloneq \gamma }{3 \sqrt {\pi }}+\frac {8 \ln \left (2\right )}{3 \sqrt {\pi }}+\frac {8 \ln \left (\frac {\sqrt {x}}{2}\right )}{3 \sqrt {\pi }}-\frac {8 \cos \left (\sqrt {x}\right )}{3 \sqrt {\pi }\, x}-\frac {16 \left (-\frac {5 x}{2}+5\right ) \sin \left (\sqrt {x}\right )}{15 \sqrt {\pi }\, x^{\frac {3}{2}}}-\frac {8 \,\operatorname {Ci}\left (\sqrt {x}\right )}{3 \sqrt {\pi }}\right )}{8}-\frac {\sqrt {\pi }\, \left (\frac {4 \mcoloneq \gamma -4+2 \ln \left (x \right )+4 \ln \left (\sqrt {\frac {1}{x}}\, \sqrt {x}\right )}{\sqrt {\pi }}+\frac {4}{\sqrt {\pi }}-\frac {4 \mcoloneq \gamma }{\sqrt {\pi }}-\frac {4 \ln \left (2\right )}{\sqrt {\pi }}-\frac {4 \ln \left (\frac {\sqrt {x}}{2}\right )}{\sqrt {\pi }}-\frac {4 \sin \left (\sqrt {x}\right )}{\sqrt {\pi }\, \sqrt {x}}+\frac {4 \,\operatorname {Ci}\left (\sqrt {x}\right )}{\sqrt {\pi }}\right )}{4}
\] And Eq. (3)
becomes \[
u_2 = \int \frac {\frac {\cos \left (x \sqrt {\frac {1}{x}}\right ) \left (6+x \right )}{x^{2}}}{\frac {2}{\sqrt {\frac {1}{x}}}}\,dx
\] Which simplifies to \[
u_2 = \int \frac {\cos \left (x \sqrt {\frac {1}{x}}\right ) \left (6+x \right ) \sqrt {\frac {1}{x}}}{2 x^{2}}d x
\] Hence \[
u_2 = \frac {3 \sqrt {\pi }\, \sqrt {\frac {1}{x}}\, \sqrt {x}\, \left (-\frac {8 \left (-x +2\right ) \cos \left (\sqrt {x}\right )}{3 x^{\frac {3}{2}} \sqrt {\pi }}+\frac {8 \sin \left (\sqrt {x}\right )}{3 x \sqrt {\pi }}+\frac {8 \,\operatorname {Si}\left (\sqrt {x}\right )}{3 \sqrt {\pi }}\right )}{8}+\frac {\sqrt {\pi }\, \sqrt {\frac {1}{x}}\, \sqrt {x}\, \left (-\frac {4 \cos \left (\sqrt {x}\right )}{\sqrt {x}\, \sqrt {\pi }}-\frac {4 \,\operatorname {Si}\left (\sqrt {x}\right )}{\sqrt {\pi }}\right )}{4}
\] Which simplifies to \begin{align*}
u_1 &= \frac {\cos \left (\sqrt {x}\right ) \sqrt {x}+2 \sin \left (\sqrt {x}\right )}{x^{\frac {3}{2}}} \\
u_2 &= \frac {\sqrt {\frac {1}{x}}\, \left (\sin \left (\sqrt {x}\right ) \sqrt {x}-2 \cos \left (\sqrt {x}\right )\right )}{x} \\
\end{align*} Therefore the particular
solution, from equation (1) is \[
y_p(x) = \frac {\left (\cos \left (\sqrt {x}\right ) \sqrt {x}+2 \sin \left (\sqrt {x}\right )\right ) \cos \left (x \sqrt {\frac {1}{x}}\right )}{x^{\frac {3}{2}}}+\frac {\sqrt {\frac {1}{x}}\, \left (\sin \left (\sqrt {x}\right ) \sqrt {x}-2 \cos \left (\sqrt {x}\right )\right ) \sin \left (x \sqrt {\frac {1}{x}}\right )}{x}
\] Which simplifies to \[
y_p(x) = \frac {\left (\cos \left (\sqrt {x}\right ) \sqrt {x}+2 \sin \left (\sqrt {x}\right )\right ) \cos \left (x \sqrt {\frac {1}{x}}\right )+\sqrt {\frac {1}{x}}\, \sin \left (x \sqrt {\frac {1}{x}}\right ) \left (\sin \left (\sqrt {x}\right ) x -2 \cos \left (\sqrt {x}\right ) \sqrt {x}\right )}{x^{\frac {3}{2}}}
\] Therefore the general solution
is \begin{align*}
y &= y_h + y_p \\
&= \left (c_{1} \cos \left (\sqrt {x}\right )+c_{2} \sin \left (\sqrt {x}\right )\right ) + \left (\frac {\left (\cos \left (\sqrt {x}\right ) \sqrt {x}+2 \sin \left (\sqrt {x}\right )\right ) \cos \left (x \sqrt {\frac {1}{x}}\right )+\sqrt {\frac {1}{x}}\, \sin \left (x \sqrt {\frac {1}{x}}\right ) \left (\sin \left (\sqrt {x}\right ) x -2 \cos \left (\sqrt {x}\right ) \sqrt {x}\right )}{x^{\frac {3}{2}}}\right ) \\
&= \frac {\left (\cos \left (\sqrt {x}\right ) \sqrt {x}+2 \sin \left (\sqrt {x}\right )\right ) \cos \left (x \sqrt {\frac {1}{x}}\right )+\sqrt {\frac {1}{x}}\, \sin \left (x \sqrt {\frac {1}{x}}\right ) \left (\sin \left (\sqrt {x}\right ) x -2 \cos \left (\sqrt {x}\right ) \sqrt {x}\right )}{x^{\frac {3}{2}}}+c_{1} \cos \left (\sqrt {x}\right )+c_{2} \sin \left (\sqrt {x}\right ) \\
\end{align*} Which simplifies to \[
y = \frac {\left (\cos \left (\sqrt {x}\right ) \sqrt {x}+2 \sin \left (\sqrt {x}\right )\right ) \cos \left (x \sqrt {\frac {1}{x}}\right )+\sqrt {\frac {1}{x}}\, \sin \left (x \sqrt {\frac {1}{x}}\right ) \left (\sin \left (\sqrt {x}\right ) x -2 \cos \left (\sqrt {x}\right ) \sqrt {x}\right )}{x^{\frac {3}{2}}}+c_{1} \cos \left (\sqrt {x}\right )+c_{2} \sin \left (\sqrt {x}\right )
\] Initial conditions are used to solve for the constants of
integration.
Looking at the above solution \begin {align*} y = \frac {\left (\cos \left (\sqrt {x}\right ) \sqrt {x}+2 \sin \left (\sqrt {x}\right )\right ) \cos \left (x \sqrt {\frac {1}{x}}\right )+\sqrt {\frac {1}{x}}\, \sin \left (x \sqrt {\frac {1}{x}}\right ) \left (\sin \left (\sqrt {x}\right ) x -2 \cos \left (\sqrt {x}\right ) \sqrt {x}\right )}{x^{\frac {3}{2}}}+c_{1} \cos \left (\sqrt {x}\right )+c_{2} \sin \left (\sqrt {x}\right ) \tag {1} \end {align*}
Initial conditions are now substituted in the above solution. This will generate the required
equations to solve for the integration constants. substituting \(y = 0\) and \(x = \infty \) in the above gives
\begin {align*} 0 = -{| c_{1} |}-{| c_{2} |}..{| c_{1} |}+{| c_{2} |}\tag {1A} \end {align*}
Equations {1A} are now solved for \(\{c_{1}, c_{2}\}\). There is no solution for the constants of integrations.
This solution is removed.
Verification of solutions N/A
Writing the ode as \begin {align*} x^{2} y^{\prime \prime }+\frac {y^{\prime } x}{2}+\frac {y x}{4} = \frac {6+x}{4 x}\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 &= {\frac {1}{4}}\\ \beta &= 1\\ n &= {\frac {1}{2}}\\ \gamma &= {\frac {1}{2}} \end {align*}
Substituting all the above into (4) gives the solution as \begin {align*} y = \frac {c_{1} \sqrt {2}\, \sin \left (\sqrt {x}\right )}{\sqrt {\pi }}-\frac {c_{2} \sqrt {2}\, \cos \left (\sqrt {x}\right )}{\sqrt {\pi }} \end {align*}
Therefore the homogeneous solution \(y_h\) is \[
y_h = \frac {c_{1} \sqrt {2}\, \sin \left (\sqrt {x}\right )}{\sqrt {\pi }}-\frac {c_{2} \sqrt {2}\, \cos \left (\sqrt {x}\right )}{\sqrt {\pi }}
\] 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 {\sqrt {2}\, \sin \left (\sqrt {x}\right )}{\sqrt {\pi }} \\
y_2 &= -\frac {\sqrt {2}\, \cos \left (\sqrt {x}\right )}{\sqrt {\pi }} \\
\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 \[ W = \begin {vmatrix} \frac {\sqrt {2}\, \sin \left (\sqrt {x}\right )}{\sqrt {\pi }} & -\frac {\sqrt {2}\, \cos \left (\sqrt {x}\right )}{\sqrt {\pi }} \\ \frac {d}{dx}\left (\frac {\sqrt {2}\, \sin \left (\sqrt {x}\right )}{\sqrt {\pi }}\right ) & \frac {d}{dx}\left (-\frac {\sqrt {2}\, \cos \left (\sqrt {x}\right )}{\sqrt {\pi }}\right ) \end {vmatrix} \] Which gives \[ W = \begin {vmatrix} \frac {\sqrt {2}\, \sin \left (\sqrt {x}\right )}{\sqrt {\pi }} & -\frac {\sqrt {2}\, \cos \left (\sqrt {x}\right )}{\sqrt {\pi }} \\ \frac {\sqrt {2}\, \cos \left (\sqrt {x}\right )}{2 \sqrt {\pi }\, \sqrt {x}} & \frac {\sqrt {2}\, \sin \left (\sqrt {x}\right )}{2 \sqrt {\pi }\, \sqrt {x}} \end {vmatrix} \]
Therefore \[
W = \left (\frac {\sqrt {2}\, \sin \left (\sqrt {x}\right )}{\sqrt {\pi }}\right )\left (\frac {\sqrt {2}\, \sin \left (\sqrt {x}\right )}{2 \sqrt {\pi }\, \sqrt {x}}\right ) - \left (-\frac {\sqrt {2}\, \cos \left (\sqrt {x}\right )}{\sqrt {\pi }}\right )\left (\frac {\sqrt {2}\, \cos \left (\sqrt {x}\right )}{2 \sqrt {\pi }\, \sqrt {x}}\right )
\] Which simplifies to \[
W = \frac {\cos \left (\sqrt {x}\right )^{2}+\sin \left (\sqrt {x}\right )^{2}}{\pi \sqrt {x}}
\] Which simplifies to \[
W = \frac {1}{\pi \sqrt {x}}
\] Therefore Eq. (2) becomes \[
u_1 = -\int \frac {-\frac {\sqrt {2}\, \cos \left (\sqrt {x}\right ) \left (6+x \right )}{4 \sqrt {\pi }\, x}}{\frac {x^{\frac {3}{2}}}{\pi }}\,dx
\]
Which simplifies to \[
u_1 = - \int -\frac {\sqrt {2}\, \sqrt {\pi }\, \cos \left (\sqrt {x}\right ) \left (6+x \right )}{4 x^{\frac {5}{2}}}d x
\] Hence \[
u_1 = \frac {\sqrt {2}\, \sqrt {\pi }\, \left (-\frac {4 \cos \left (\sqrt {x}\right )}{x^{\frac {3}{2}}}+\frac {2 \sin \left (\sqrt {x}\right )}{x}\right )}{4}
\] And Eq. (3) becomes \[
u_2 = \int \frac {\frac {\sqrt {2}\, \sin \left (\sqrt {x}\right ) \left (6+x \right )}{4 \sqrt {\pi }\, x}}{\frac {x^{\frac {3}{2}}}{\pi }}\,dx
\] Which simplifies to \[
u_2 = \int \frac {\sqrt {2}\, \sqrt {\pi }\, \sin \left (\sqrt {x}\right ) \left (6+x \right )}{4 x^{\frac {5}{2}}}d x
\] Hence \[
u_2 = \frac {\sqrt {2}\, \sqrt {\pi }\, \left (-\frac {4 \sin \left (\sqrt {x}\right )}{x^{\frac {3}{2}}}-\frac {2 \cos \left (\sqrt {x}\right )}{x}\right )}{4}
\]
Therefore the particular solution, from equation (1) is \[
y_p(x) = \frac {\left (-\frac {4 \cos \left (\sqrt {x}\right )}{x^{\frac {3}{2}}}+\frac {2 \sin \left (\sqrt {x}\right )}{x}\right ) \sin \left (\sqrt {x}\right )}{2}-\frac {\left (-\frac {4 \sin \left (\sqrt {x}\right )}{x^{\frac {3}{2}}}-\frac {2 \cos \left (\sqrt {x}\right )}{x}\right ) \cos \left (\sqrt {x}\right )}{2}
\] Which simplifies to \[
y_p(x) = \frac {1}{x}
\] Therefore
the general solution is \begin{align*}
y &= y_h + y_p \\
&= \left (\frac {c_{1} \sqrt {2}\, \sin \left (\sqrt {x}\right )}{\sqrt {\pi }}-\frac {c_{2} \sqrt {2}\, \cos \left (\sqrt {x}\right )}{\sqrt {\pi }}\right ) + \left (\frac {1}{x}\right ) \\
\end{align*} Initial conditions are used to solve for the constants of
integration.
Looking at the above solution \begin {align*} y = \frac {c_{1} \sqrt {2}\, \sin \left (\sqrt {x}\right )}{\sqrt {\pi }}-\frac {c_{2} \sqrt {2}\, \cos \left (\sqrt {x}\right )}{\sqrt {\pi }}+\frac {1}{x} \tag {1} \end {align*}
Initial conditions are now substituted in the above solution. This will generate the required
equations to solve for the integration constants. substituting \(y = 0\) and \(x = \infty \) in the above gives
\begin {align*} 0 = \frac {\left (-{| c_{1} |}-{| c_{2} |}\right ) \sqrt {2}}{\sqrt {\pi }}..\frac {\sqrt {2}\, \left ({| c_{1} |}+{| c_{2} |}\right )}{\sqrt {\pi }}\tag {1A} \end {align*}
Equations {1A} are now solved for \(\{c_{1}, c_{2}\}\). There is no solution for the constants of integrations.
This solution is removed.
Verification of solutions N/A
Writing the ode as \begin {align*} 4 x y^{\prime \prime }+2 y^{\prime }+y &= 0 \tag {1} \\ A y^{\prime \prime } + B y^{\prime } + C y &= 0 \tag {2} \end {align*}
Comparing (1) and (2) shows that \begin {align*} A &= 4 x \\ B &= 2\tag {3} \\ C &= 1 \end {align*}
Applying the Liouville transformation on the dependent variable gives \begin {align*} z(x) &= y e^{\int \frac {B}{2 A} \,dx} \end {align*}
Then (2) becomes \begin {align*} z''(x) = r z(x)\tag {4} \end {align*}
Where \(r\) is given by \begin {align*} r &= \frac {s}{t}\tag {5} \\ &= \frac {2 A B' - 2 B A' + B^2 - 4 A C}{4 A^2} \end {align*}
Substituting the values of \(A,B,C\) from (3) in the above and simplifying gives \begin {align*} r &= \frac {-4 x -3}{16 x^{2}}\tag {6} \end {align*}
Comparing the above to (5) shows that \begin {align*} s &= -4 x -3\\ t &= 16 x^{2} \end {align*}
Therefore eq. (4) becomes \begin {align*} z''(x) &= \left ( \frac {-4 x -3}{16 x^{2}}\right ) z(x)\tag {7} \end {align*}
Equation (7) is now solved. After finding \(z(x)\) then \(y\) is found using the inverse transformation
\begin {align*} y &= z \left (x \right ) e^{-\int \frac {B}{2 A} \,dx} \end {align*}
The first step is to determine the case of Kovacic algorithm this ode belongs to. There are 3
cases depending on the order of poles of \(r\) and the order of \(r\) at \(\infty \). The following table
summarizes these cases. Case Allowed pole order for \(r\) Allowed value for \(\mathcal {O}(\infty )\) 1 \(\left \{ 0,1,2,4,6,8,\cdots \right \} \) \(\left \{ \cdots ,-6,-4,-2,0,2,3,4,5,6,\cdots \right \} \) 2
Need to have at least one pole
that is either order \(2\) or odd order
greater than \(2\). Any other pole order
is allowed as long as the above
condition is satisfied. Hence the
following set of pole orders are all
allowed. \(\{1,2\}\),\(\{1,3\}\),\(\{2\}\),\(\{3\}\),\(\{3,4\}\),\(\{1,2,5\}\). no condition 3 \(\left \{ 1,2\right \} \) \(\left \{ 2,3,4,5,6,7,\cdots \right \} \) The order of \(r\) at \(\infty \) is the degree of \(t\) minus the degree of \(s\). Therefore \begin {align*} O\left (\infty \right ) &= \text {deg}(t) - \text {deg}(s) \\ &= 2 - 1 \\ &= 1 \end {align*}
The poles of \(r\) in eq. (7) and the order of each pole are determined by solving for the roots of \(t=16 x^{2}\).
There is a pole at \(x=0\) of order \(2\). Since there is a pole of order \(2\) then necessary conditions for case
two are met. Therefore \begin {align*} L &= [2] \end {align*}
Attempting to find a solution using case \(n=2\).
Looking at poles of order 2. The partial fractions decomposition of \(r\) is \[
r = -\frac {3}{16 x^{2}}-\frac {1}{4 x}
\] For the pole at \(x=0\) let \(b\)
be the coefficient of \(\frac {1}{ x^{2}}\) in the partial fractions decomposition of \(r\) given above. Therefore \(b=-{\frac {3}{16}}\). Hence
\begin {align*} E_c &= \{2, 2+2\sqrt {1+4 b}, 2-2\sqrt {1+4 b} \} \\ &= \{1, 2, 3\} \end {align*}
Since the order of \(r\) at \(\infty \) is \(1 <2\) then \begin {align*} E_\infty = \{1\} \end {align*}
The following table summarizes the findings so far for poles and for the order of \(r\) at \(\infty \) for case
2 of Kovacic algorithm.
Using the family \(\{e_1,e_2,\dots ,e_\infty \}\) given by \[ e_1=1,\hspace {3pt} e_\infty =1 \] Gives a non negative integer \(d\) (the degree of the polynomial \(p(x)\)),
which is generated using \begin {align*} d &= \frac {1}{2} \left ( e_\infty - \sum _{c \in \Gamma } e_c \right )\\ &= \frac {1}{2} \left ( 1 - \left (1\right )\right )\\ &= 0 \end {align*}
We now form the following rational function \begin {align*} \theta &= \frac {1}{2} \sum _{c \in \Gamma } \frac {e_c}{x-c} \\ &= \frac {1}{2} \left (\frac {1}{\left (x-\left (0\right )\right )}\right ) \\ &= \frac {1}{2 x} \end {align*}
Now we search for a monic polynomial \(p(x)\) of degree \(d=0\) such that \[ p'''+3 \theta p'' + \left (3 \theta ^2 + 3 \theta ' - 4 r\right )p' + \left (\theta '' + 3 \theta \theta ' + \theta ^3 - 4 r \theta - 2 r' \right ) p = 0 \tag {1A} \] Since \(d=0\), then letting \[ p = 1\tag {2A} \]
Substituting \(p\) and \(\theta \) into Eq. (1A) gives \[
0 = 0
\] And solving for \(p\) gives \[ p = 1 \] Now that \(p(x)\) is found let
\begin {align*} \phi &= \theta + \frac {p'}{p}\\ &= \frac {1}{2 x} \end {align*}
Let \(\omega \) be the solution of \begin {align*} \omega ^2 - \phi \omega + \left ( \frac {1}{2} \phi ' + \frac {1}{2} \phi ^2 - r \right ) &= 0 \end {align*}
Substituting the values for \(\phi \) and \(r\) into the above equation gives \[
w^{2}-\frac {w}{2 x}+\frac {1+4 x}{16 x^{2}} = 0
\] Solving for \(\omega \) gives
\begin {align*} \omega &= \frac {1+2 \sqrt {-x}}{4 x} \end {align*}
Therefore the first solution to the ode \(z'' = r z\) is \begin {align*} z_1(x) &= e^{ \int \omega \,dx} \\ &= {\mathrm e}^{\int \frac {1+2 \sqrt {-x}}{4 x}d x}\\ &= x^{\frac {1}{4}} {\mathrm e}^{\sqrt {-x}} \end {align*}
The first solution to the original ode in \(y\) is found from \begin{align*}
y_1 &= z_1 e^{ \int -\frac {1}{2} \frac {B}{A} \,dx} \\
&= z_1 e^{ -\int \frac {1}{2} \frac {2}{4 x} \,dx} \\
&= z_1 e^{-\frac {\ln \left (x \right )}{4}} \\
&= z_1 \left (\frac {1}{x^{\frac {1}{4}}}\right ) \\
\end{align*} Which simplifies to \[
y_1 = {\mathrm e}^{\sqrt {-x}}
\] The second
solution \(y_2\) to the original ode is found using reduction of order \[ y_2 = y_1 \int \frac { e^{\int -\frac {B}{A} \,dx}}{y_1^2} \,dx \] Substituting gives \begin{align*}
y_2 &= y_1 \int \frac { e^{\int -\frac {2}{4 x} \,dx}}{\left (y_1\right )^2} \,dx \\
&= y_1 \int \frac { e^{-\frac {\ln \left (x \right )}{2}}}{\left (y_1\right )^2} \,dx \\
&= y_1 \left (\frac {\sqrt {-x}\, \left (-1+{\mathrm e}^{-2 \sqrt {-x}}\right )}{\sqrt {x}}\right ) \\
\end{align*} Therefore
the solution is
\begin{align*}
y &= c_{1} y_1 + c_{2} y_2 \\
&= c_{1} \left ({\mathrm e}^{\sqrt {-x}}\right ) + c_{2} \left ({\mathrm e}^{\sqrt {-x}}\left (\frac {\sqrt {-x}\, \left (-1+{\mathrm e}^{-2 \sqrt {-x}}\right )}{\sqrt {x}}\right )\right ) \\
\end{align*} This is second order nonhomogeneous 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 nonhomogeneous ODE \(A y''(x) + B y'(x) + C y(x) = f(x)\). \(y_h\) is the
solution to \[
4 x y^{\prime \prime }+2 y^{\prime }+y = 0
\] The homogeneous solution is found using the Kovacic algorithm which results in \[
y_h = c_{1} {\mathrm e}^{\sqrt {-x}}+\frac {c_{2} \sqrt {-x}\, \left (-{\mathrm e}^{\sqrt {-x}}+{\mathrm e}^{-\sqrt {-x}}\right )}{\sqrt {x}}
\]
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}^{\sqrt {-x}} \\
y_2 &= \frac {\sqrt {-x}\, \left (-{\mathrm e}^{\sqrt {-x}}+{\mathrm e}^{-\sqrt {-x}}\right )}{\sqrt {x}} \\
\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 \[ W = \begin {vmatrix} {\mathrm e}^{\sqrt {-x}} & \frac {\sqrt {-x}\, \left (-{\mathrm e}^{\sqrt {-x}}+{\mathrm e}^{-\sqrt {-x}}\right )}{\sqrt {x}} \\ \frac {d}{dx}\left ({\mathrm e}^{\sqrt {-x}}\right ) & \frac {d}{dx}\left (\frac {\sqrt {-x}\, \left (-{\mathrm e}^{\sqrt {-x}}+{\mathrm e}^{-\sqrt {-x}}\right )}{\sqrt {x}}\right ) \end {vmatrix} \] Which gives \[ W = \begin {vmatrix} {\mathrm e}^{\sqrt {-x}} & \frac {\sqrt {-x}\, \left (-{\mathrm e}^{\sqrt {-x}}+{\mathrm e}^{-\sqrt {-x}}\right )}{\sqrt {x}} \\ -\frac {{\mathrm e}^{\sqrt {-x}}}{2 \sqrt {-x}} & -\frac {-{\mathrm e}^{\sqrt {-x}}+{\mathrm e}^{-\sqrt {-x}}}{2 \sqrt {-x}\, \sqrt {x}}+\frac {\sqrt {-x}\, \left (\frac {{\mathrm e}^{\sqrt {-x}}}{2 \sqrt {-x}}+\frac {{\mathrm e}^{-\sqrt {-x}}}{2 \sqrt {-x}}\right )}{\sqrt {x}}-\frac {\sqrt {-x}\, \left (-{\mathrm e}^{\sqrt {-x}}+{\mathrm e}^{-\sqrt {-x}}\right )}{2 x^{\frac {3}{2}}} \end {vmatrix} \] Therefore \[
W = \left ({\mathrm e}^{\sqrt {-x}}\right )\left (-\frac {-{\mathrm e}^{\sqrt {-x}}+{\mathrm e}^{-\sqrt {-x}}}{2 \sqrt {-x}\, \sqrt {x}}+\frac {\sqrt {-x}\, \left (\frac {{\mathrm e}^{\sqrt {-x}}}{2 \sqrt {-x}}+\frac {{\mathrm e}^{-\sqrt {-x}}}{2 \sqrt {-x}}\right )}{\sqrt {x}}-\frac {\sqrt {-x}\, \left (-{\mathrm e}^{\sqrt {-x}}+{\mathrm e}^{-\sqrt {-x}}\right )}{2 x^{\frac {3}{2}}}\right ) - \left (\frac {\sqrt {-x}\, \left (-{\mathrm e}^{\sqrt {-x}}+{\mathrm e}^{-\sqrt {-x}}\right )}{\sqrt {x}}\right )\left (-\frac {{\mathrm e}^{\sqrt {-x}}}{2 \sqrt {-x}}\right )
\] Which
simplifies to \[
W = \frac {{\mathrm e}^{\sqrt {-x}} {\mathrm e}^{-\sqrt {-x}}}{\sqrt {x}}
\] Which simplifies to \[
W = \frac {1}{\sqrt {x}}
\] Therefore Eq. (2) becomes \[
u_1 = -\int \frac {\frac {\sqrt {-x}\, \left (-{\mathrm e}^{\sqrt {-x}}+{\mathrm e}^{-\sqrt {-x}}\right ) \left (6+x \right )}{x^{\frac {5}{2}}}}{4 \sqrt {x}}\,dx
\] Which simplifies to \[
u_1 = - \int \frac {\left (6+x \right ) \left ({\mathrm e}^{\sqrt {-x}}-{\mathrm e}^{-\sqrt {-x}}\right )}{4 \left (-x \right )^{\frac {5}{2}}}d x
\]
Hence \[
u_1 = \frac {{\mathrm e}^{-\sqrt {-x}}}{\left (-x \right )^{\frac {3}{2}}}+\frac {{\mathrm e}^{-\sqrt {-x}}}{2 x}-\frac {{\mathrm e}^{\sqrt {-x}}}{\left (-x \right )^{\frac {3}{2}}}+\frac {{\mathrm e}^{\sqrt {-x}}}{2 x}
\] And Eq. (3) becomes \[
u_2 = \int \frac {\frac {{\mathrm e}^{\sqrt {-x}} \left (6+x \right )}{x^{2}}}{4 \sqrt {x}}\,dx
\] Which simplifies to \[
u_2 = \int \frac {{\mathrm e}^{\sqrt {-x}} \left (6+x \right )}{4 x^{\frac {5}{2}}}d x
\] Hence \[
u_2 = \frac {3 \left (-x \right )^{\frac {5}{2}} \left (\frac {1}{3 \left (-x \right )^{\frac {3}{2}}}-\frac {1}{2 x}+\frac {1}{2 \sqrt {-x}}+\frac {11}{36}-\frac {\ln \left (x \right )}{12}-\frac {\ln \left (-\frac {\sqrt {-x}}{\sqrt {x}}\right )}{6}-\frac {\sqrt {-x}\, \left (22 \left (-x \right )^{\frac {3}{2}}-36 x +36 \sqrt {-x}+24\right )}{72 x^{2}}+\frac {\sqrt {-x}\, \left (-4 x +4 \sqrt {-x}+8\right ) {\mathrm e}^{\sqrt {-x}}}{24 x^{2}}+\frac {\ln \left (-\sqrt {-x}\right )}{6}+\frac {\operatorname {expIntegral}_{1}\left (-\sqrt {-x}\right )}{6}\right )}{x^{\frac {5}{2}}}+\frac {\left (-x \right )^{\frac {3}{2}} \left (\frac {1}{\sqrt {-x}}+1-\frac {\ln \left (x \right )}{2}-\ln \left (-\frac {\sqrt {-x}}{\sqrt {x}}\right )+\frac {\sqrt {-x}\, \left (2 \sqrt {-x}+2\right )}{2 x}-\frac {\sqrt {-x}\, {\mathrm e}^{\sqrt {-x}}}{x}+\ln \left (-\sqrt {-x}\right )+\operatorname {expIntegral}_{1}\left (-\sqrt {-x}\right )\right )}{2 x^{\frac {3}{2}}}
\] Which simplifies to \begin{align*}
u_1 &= -\frac {\left (\sqrt {-x}-2\right ) {\mathrm e}^{-\sqrt {-x}}+{\mathrm e}^{\sqrt {-x}} \left (\sqrt {-x}+2\right )}{2 \left (-x \right )^{\frac {3}{2}}} \\
u_2 &= -\frac {{\mathrm e}^{\sqrt {-x}} \left (\sqrt {-x}+2\right )}{2 x^{\frac {3}{2}}} \\
\end{align*}
Therefore the particular solution, from equation (1) is \[
y_p(x) = -\frac {\left (\left (\sqrt {-x}-2\right ) {\mathrm e}^{-\sqrt {-x}}+{\mathrm e}^{\sqrt {-x}} \left (\sqrt {-x}+2\right )\right ) {\mathrm e}^{\sqrt {-x}}}{2 \left (-x \right )^{\frac {3}{2}}}-\frac {{\mathrm e}^{\sqrt {-x}} \left (\sqrt {-x}+2\right ) \sqrt {-x}\, \left (-{\mathrm e}^{\sqrt {-x}}+{\mathrm e}^{-\sqrt {-x}}\right )}{2 x^{2}}
\] Which simplifies to \[
y_p(x) = \frac {1}{x}
\] Therefore
the general solution is \begin{align*}
y &= y_h + y_p \\
&= \left (c_{1} {\mathrm e}^{\sqrt {-x}}+\frac {c_{2} \sqrt {-x}\, \left (-{\mathrm e}^{\sqrt {-x}}+{\mathrm e}^{-\sqrt {-x}}\right )}{\sqrt {x}}\right ) + \left (\frac {1}{x}\right ) \\
\end{align*} Initial conditions are used to solve for the constants of
integration.
Looking at the above solution \begin {align*} y = c_{1} {\mathrm e}^{\sqrt {-x}}+\frac {c_{2} \sqrt {-x}\, \left (-{\mathrm e}^{\sqrt {-x}}+{\mathrm e}^{-\sqrt {-x}}\right )}{\sqrt {x}}+\frac {1}{x} \tag {1} \end {align*}
Initial conditions are now substituted in the above solution. This will generate the required
equations to solve for the integration constants. substituting \(y = 0\) and \(x = \infty \) in the above gives
\begin {align*} 0 = -{| \Re \left (\left (1+i\right ) c_{1} \right )|}-i {| \Im \left (\left (1+i\right ) c_{1} \right )|}+2 \min \left (-\Re \left (\left (1-i\right ) c_{2} \right ), -\Re \left (\left (1+i\right ) c_{2} \right ), \Re \left (\left (1-i\right ) c_{2} \right ), \Re \left (\left (1+i\right ) c_{2} \right )\right )+2 i \min \left (-\Im \left (\left (1-i\right ) c_{2} \right ), -\Im \left (\left (1+i\right ) c_{2} \right ), \Im \left (\left (1-i\right ) c_{2} \right ), \Im \left (\left (1+i\right ) c_{2} \right )\right )..{| \Re \left (\left (1+i\right ) c_{1} \right )|}+i {| \Im \left (\left (1+i\right ) c_{1} \right )|}+2 \max \left (-\Re \left (\left (1-i\right ) c_{2} \right ), -\Re \left (\left (1+i\right ) c_{2} \right ), \Re \left (\left (1-i\right ) c_{2} \right ), \Re \left (\left (1+i\right ) c_{2} \right )\right )+2 i \max \left (-\Im \left (\left (1-i\right ) c_{2} \right ), -\Im \left (\left (1+i\right ) c_{2} \right ), \Im \left (\left (1-i\right ) c_{2} \right ), \Im \left (\left (1+i\right ) c_{2} \right )\right )\tag {1A} \end {align*}
Equations {1A} are now solved for \(\{c_{1}, c_{2}\}\). There is no solution for the constants of integrations.
This solution is removed.
Verification of solutions N/A Maple trace
✓ Solution by Maple
Time used: 0.75 (sec). Leaf size: 41
\[
y \left (x \right ) = \textit {undefined}
\]
✓ Solution by Mathematica
Time used: 0.192 (sec). Leaf size: 27
\[
y(x)\to \frac {1}{x}+c_1 \cos \left (\sqrt {x}\right )+c_2 \sin \left (\sqrt {x}\right )
\]
20.29.2 Solving as second order change of variable on x method 2 ode
20.29.3 Solving as second order change of variable on x method 1 ode
20.29.4 Solving as second order bessel ode ode
20.29.5 Solving using Kovacic algorithm
pole \(c\) location
pole order
\(E_c\)
\(0\) \(2\) \(\{1, 2, 3\}\)
Order of \(r\) at \(\infty \)
\(E_\infty \) \(1\)
\(\{1\}\)
`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`
dsolve([4*x*diff(y(x),x$2)+2*diff(y(x),x)+y(x)=(6+x)/x^2,y(infinity) = 0],y(x), singsol=all)
DSolve[{4*x*y''[x]+2*y'[x]+y[x]==(6+x)/x^2,{y[Infinity]==0}},y[x],x,IncludeSingularSolutions -> True]