17.43 problem 43

Internal problem ID [14485]
Book : Ordinary Differential Equations. An introduction to the fundamentals. Kenneth B. Howell. second edition. CRC Press. FL, USA. 2020
Section : Chapter 25. Review exercises for part III. page 447
Problem number : 43
Date solved : Friday, October 18, 2024 at 11:35:51 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

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

17.43.1 Solved as second order Euler type ode

Time used: 0.223 (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 \(A=x^{2}, B=3 x, C=2, f(x)=6\). 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

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

Next, we find the particular solution to the ODE

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

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 {\cos \left (\ln \left (x \right )\right )}{x} \\ y_2 &= -\frac {\sin \left (\ln \left (x \right )\right )}{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

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

Which gives

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

Therefore

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

Which simplifies to

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

Which simplifies to

\[ W = -\frac {1}{x^{3}} \]

Therefore Eq. (2) becomes

\[ u_1 = -\int \frac {-\frac {6 \sin \left (\ln \left (x \right )\right )}{x}}{-\frac {1}{x}}\,dx \]

Which simplifies to

\[ u_1 = - \int 6 \sin \left (\ln \left (x \right )\right )d x \]

Hence

\[ u_1 = 3 \cos \left (\ln \left (x \right )\right ) x -3 \sin \left (\ln \left (x \right )\right ) x \]

And Eq. (3) becomes

\[ u_2 = \int \frac {\frac {6 \cos \left (\ln \left (x \right )\right )}{x}}{-\frac {1}{x}}\,dx \]

Which simplifies to

\[ u_2 = \int -6 \cos \left (\ln \left (x \right )\right )d x \]

Hence

\[ u_2 = -3 \cos \left (\ln \left (x \right )\right ) x -3 \sin \left (\ln \left (x \right )\right ) x \]

Therefore the particular solution, from equation (1) is

\[ y_p(x) = \frac {\left (3 \cos \left (\ln \left (x \right )\right ) x -3 \sin \left (\ln \left (x \right )\right ) x \right ) \cos \left (\ln \left (x \right )\right )}{x}-\frac {\sin \left (\ln \left (x \right )\right ) \left (-3 \cos \left (\ln \left (x \right )\right ) x -3 \sin \left (\ln \left (x \right )\right ) x \right )}{x} \]

Which simplifies to

\[ y_p(x) = 3 \]

Therefore the general solution is

\begin{align*} y &= y_h + y_p \\ &= \frac {c_1 \cos \left (\ln \left (x \right )\right )+c_2 \sin \left (\ln \left (x \right )\right )+3 x}{x} \end{align*}

Will add steps showing solving for IC soon.

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

Time used: 0.456 (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

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

In normal form the ode

\begin{align*} x^{2} y^{\prime \prime }+3 x y^{\prime }+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 {2}{x^{2}} \end{align*}

Applying change of variables \(\tau = g \left (x \right )\) to (2) gives

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

Where \(\tau \) is the new independent variable, and

\begin{align*} p_{1} \left (\tau \right ) &=\frac {\tau ^{\prime \prime }\left (x \right )+p \left (x \right ) \tau ^{\prime }\left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\tag {4} \\ q_{1} \left (\tau \right ) &=\frac {q \left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\tag {5} \end{align*}

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

\begin{align*} \tau ^{\prime \prime }\left (x \right )+p \left (x \right ) \tau ^{\prime }\left (x \right )&=0 \end{align*}

This ode is solved resulting in

\begin{align*} \tau &= \int {\mathrm e}^{-\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 {2}{x^{2}}}{\frac {1}{x^{6}}}\\ &= 2 x^{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 )+2 x^{4} y \left (\tau \right )&=0 \\ \end{align*}

But in terms of \(\tau \)

\begin{align*} 2 x^{4}&=\frac {1}{2 \tau ^{2}} \end{align*}

Hence the above ode becomes

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

The above ode is now solved for \(y \left (\tau \right )\). This is Euler second order ODE. Let the solution be \(y \left (\tau \right ) = \tau ^r\), then \(y'=r \tau ^{r-1}\) and \(y''=r(r-1) \tau ^{r-2}\). Substituting these back into the given ODE gives

\[ 2 \tau ^{2}(r(r-1))\tau ^{r-2}+0 r \tau ^{r-1}+\tau ^{r} = 0 \]

Simplifying gives

\[ 2 r \left (r -1\right )\tau ^{r}+0\,\tau ^{r}+\tau ^{r} = 0 \]

Since \(\tau ^{r}\neq 0\) then dividing throughout by \(\tau ^{r}\) gives

\[ 2 r \left (r -1\right )+0+1 = 0 \]

Or

\[ 2 r^{2}-2 r +1 = 0 \tag {1} \]

Equation (1) is the characteristic equation. Its roots determine the form of the general solution. Using the quadratic equation the roots are

\begin{align*} r_1 &= \frac {1}{2}-\frac {i}{2}\\ r_2 &= \frac {1}{2}+\frac {i}{2} \end{align*}

The roots are complex conjugate of each others. Let the roots be

\begin{align*} r_1 &= \alpha + i \beta \\ r_2 &= \alpha - i \beta \\ \end{align*}

Where in this case \(\alpha ={\frac {1}{2}}\) and \(\beta =-{\frac {1}{2}}\). Hence the solution becomes

\begin{align*} y \left (\tau \right ) &= c_1 \tau ^{r_1} + c_2 \tau ^{r_2} \\ &= c_1 \tau ^{\alpha + i \beta } + c_2 \tau ^{\alpha - i \beta } \\ &= \tau ^{\alpha } \left ( c_1 \tau ^{i \beta } + c_2 \tau ^{- i \beta }\right ) \\ &= \tau ^{\alpha } \left ( c_1 e^{\ln \left (\tau ^{i \beta }\right )} + c_2 e^{\ln \left (\tau ^{-i \beta }\right )}\right ) \\ &= \tau ^{\alpha } \left ( c_1 e^{i \left (\beta \ln {\tau }\right )} + c_2 e^{-i \left (\beta \ln {\tau }\right )}\right ) \end{align*}

Using the values for \(\alpha ={\frac {1}{2}}, \beta =-{\frac {1}{2}}\), the above becomes

\begin{align*} y \left (\tau \right )&= \tau ^{{\frac {1}{2}}} \left ( c_1 e^{-\frac {i \ln \left (\tau \right )}{2}} + c_2 e^{\frac {i \ln \left (\tau \right )}{2}} \right ) \end{align*}

Using Euler relation, the expression \(c_1 e^{i A}+ c_2 e^{-i A}\) is transformed to \( c_1 \cos A+ c_1 \sin A\) where the constants are free to change. Applying this to the above result gives

\begin{align*} y \left (\tau \right )&=\sqrt {\tau }\left (c_1 \cos \left (\frac {\ln \left (\tau \right )}{2}\right )+c_2 \sin \left (\frac {\ln \left (\tau \right )}{2}\right )\right ) \end{align*}

Will add steps showing solving for IC soon.

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

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

Therefore the homogeneous solution \(y_h\) is

\[ y_h = \sqrt {-\frac {1}{2 x^{2}}}\, \left (c_1 \cos \left (\frac {\ln \left (-\frac {1}{2 x^{2}}\right )}{2}\right )+c_2 \sin \left (\frac {\ln \left (-\frac {1}{2 x^{2}}\right )}{2}\right )\right ) \]

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

\[ 1 \]

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

\[ [\{1\}] \]

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

\[ \left \{\sqrt {-\frac {1}{2 x^{2}}}\, \cos \left (\frac {\ln \left (2\right )}{2}\right ) \cos \left (\frac {\ln \left (-\frac {1}{x^{2}}\right )}{2}\right )+\sqrt {-\frac {1}{2 x^{2}}}\, \sin \left (\frac {\ln \left (2\right )}{2}\right ) \sin \left (\frac {\ln \left (-\frac {1}{x^{2}}\right )}{2}\right ), \sqrt {-\frac {1}{2 x^{2}}}\, \cos \left (\frac {\ln \left (2\right )}{2}\right ) \sin \left (\frac {\ln \left (-\frac {1}{x^{2}}\right )}{2}\right )-\sqrt {-\frac {1}{2 x^{2}}}\, \cos \left (\frac {\ln \left (-\frac {1}{x^{2}}\right )}{2}\right ) \sin \left (\frac {\ln \left (2\right )}{2}\right )\right \} \]

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

\[ y_p = A_{1} \]

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

\[ 2 A_{1} = 6 \]

Solving for the unknowns by comparing coefficients results in

\[ [A_{1} = 3] \]

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

\[ y_p = 3 \]

Therefore the general solution is

\begin{align*} y &= y_h + y_p \\ &= \left (\sqrt {-\frac {1}{2 x^{2}}}\, \left (c_1 \cos \left (\frac {\ln \left (-\frac {1}{2 x^{2}}\right )}{2}\right )+c_2 \sin \left (\frac {\ln \left (-\frac {1}{2 x^{2}}\right )}{2}\right )\right )\right ) + \left (3\right ) \\ \end{align*}

Will add steps showing solving for IC soon.

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

Time used: 0.201 (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 \(A=x^{2}, B=3 x, C=2, f(x)=6\). 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

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

In normal form the ode

\begin{align*} x^{2} y^{\prime \prime }+3 x y^{\prime }+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 {2}{x^{2}} \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 {2}\, \sqrt {\frac {1}{x^{2}}}}{c}\tag {6} \\ \tau '' &= -\frac {\sqrt {2}}{c \sqrt {\frac {1}{x^{2}}}\, x^{3}} \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 {\sqrt {2}}{c \sqrt {\frac {1}{x^{2}}}\, x^{3}}+\frac {3}{x}\frac {\sqrt {2}\, \sqrt {\frac {1}{x^{2}}}}{c}}{\left (\frac {\sqrt {2}\, \sqrt {\frac {1}{x^{2}}}}{c}\right )^2} \\ &=c \sqrt {2} \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 \sqrt {2}\, \left (\frac {d}{d \tau }y \left (\tau \right )\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 ) &= {\mathrm e}^{-\frac {\sqrt {2}\, c \tau }{2}} \left (c_1 \cos \left (\frac {\sqrt {2}\, c \tau }{2}\right )+c_2 \sin \left (\frac {\sqrt {2}\, c \tau }{2}\right )\right ) \end{align*}

Now from (6)

\begin{align*} \tau &= \int \frac {1}{c} \sqrt q \,dx \\ &= \frac {\int \sqrt {2}\, \sqrt {\frac {1}{x^{2}}}d x}{c}\\ &= \frac {\sqrt {2}\, \ln \left (x \right )}{c} \end{align*}

Substituting the above into the solution obtained gives

\[ y = \frac {c_1 \cos \left (\ln \left (x \right )\right )+c_2 \sin \left (\ln \left (x \right )\right )}{x} \]

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

\[ 1 \]

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

\[ [\{1\}] \]

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

\[ \left \{\frac {\cos \left (\ln \left (x \right )\right )}{x}, \frac {\sin \left (\ln \left (x \right )\right )}{x}\right \} \]

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

\[ y_p = A_{1} \]

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

\[ 2 A_{1} = 6 \]

Solving for the unknowns by comparing coefficients results in

\[ [A_{1} = 3] \]

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

\[ y_p = 3 \]

Therefore the general solution is

\begin{align*} y &= y_h + y_p \\ &= \left (\frac {c_1 \cos \left (\ln \left (x \right )\right )+c_2 \sin \left (\ln \left (x \right )\right )}{x}\right ) + \left (3\right ) \\ \end{align*}

Will add steps showing solving for IC soon.

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

Time used: 0.211 (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 \(A=x^{2}, B=3 x, C=2, f(x)=6\). 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

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

In normal form the ode

\begin{align*} x^{2} y^{\prime \prime }+3 x y^{\prime }+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 {2}{x^{2}} \end{align*}

Applying change of variables on the depndent variable \(y = v \left (x \right ) x^{n}\) to (2) gives the following ode where the dependent variables is \(v \left (x \right )\) and not \(y\).

\begin{align*} v^{\prime \prime }\left (x \right )+\left (\frac {2 n}{x}+p \right ) v^{\prime }\left (x \right )+\left (\frac {n \left (n -1\right )}{x^{2}}+\frac {n p}{x}+q \right ) v \left (x \right )&=0 \tag {3} \end{align*}

Let the coefficient of \(v \left (x \right )\) above be zero. Hence

\begin{align*} \frac {n \left (n -1\right )}{x^{2}}+\frac {n p}{x}+q&=0 \tag {4} \end{align*}

Substituting the earlier values found for \(p \left (x \right )\) and \(q \left (x \right )\) into (4) gives

\begin{align*} \frac {n \left (n -1\right )}{x^{2}}+\frac {3 n}{x^{2}}+\frac {2}{x^{2}}&=0 \tag {5} \end{align*}

Solving (5) for \(n\) gives

\begin{align*} n&=-1+i \tag {6} \end{align*}

Substituting this value in (3) gives

\begin{align*} v^{\prime \prime }\left (x \right )+\left (\frac {-2+2 i}{x}+\frac {3}{x}\right ) v^{\prime }\left (x \right )&=0 \\ v^{\prime \prime }\left (x \right )+\frac {\left (1+2 i\right ) v^{\prime }\left (x \right )}{x}&=0 \tag {7} \\ \end{align*}

Using the substitution

\begin{align*} u \left (x \right ) = v^{\prime }\left (x \right ) \end{align*}

Then (7) becomes

\begin{align*} u^{\prime }\left (x \right )+\frac {\left (1+2 i\right ) u \left (x \right )}{x} = 0 \tag {8} \\ \end{align*}

The above is now solved for \(u \left (x \right )\). In canonical form a linear first order is

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

Comparing the above to the given ode shows that

\begin{align*} q(x) &=\frac {1+2 i}{x}\\ p(x) &=0 \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,dx}}\\ &= {\mathrm e}^{\int \frac {1+2 i}{x}d x}\\ &= x^{1+2 i} \end{align*}

The ode becomes

\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \mu u &= 0 \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (u \,x^{1+2 i}\right ) &= 0 \end{align*}

Integrating gives

\begin{align*} u \,x^{1+2 i}&= \int {0 \,dx} + c_1 \\ &=c_1 \end{align*}

Dividing throughout by the integrating factor \(x^{1+2 i}\) gives the final solution

\[ u \left (x \right ) = x^{-1-2 i} c_1 \]

Now that \(u \left (x \right )\) is known, then

\begin{align*} v^{\prime }\left (x \right )&= u \left (x \right )\\ v \left (x \right )&= \int u \left (x \right )d x +c_2\\ &= \frac {i x^{-2 i} c_1}{2}+c_2 \end{align*}

Hence

\begin{align*} y&= v \left (x \right ) x^{n}\\ &= \left (\frac {i x^{-2 i} c_1}{2}+c_2 \right ) x^{-1+i}\\ &= \frac {\left (i x^{-2 i} c_1 +2 c_2 \right ) x^{-1+i}}{2}\\ \end{align*}

Now the particular solution to this ODE is found

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

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 {x^{i}}{x} \\ y_2 &= \frac {x^{i} x^{-2 i}}{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

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

Which gives

\[ W = \begin {vmatrix} \frac {x^{i}}{x} & \frac {x^{i} x^{-2 i}}{x} \\ -\frac {x^{i}}{x^{2}}+\frac {i x^{i}}{x^{2}} & -\frac {x^{i} x^{-2 i}}{x^{2}}-\frac {i x^{i} x^{-2 i}}{x^{2}} \end {vmatrix} \]

Therefore

\[ W = \left (\frac {x^{i}}{x}\right )\left (-\frac {x^{i} x^{-2 i}}{x^{2}}-\frac {i x^{i} x^{-2 i}}{x^{2}}\right ) - \left (\frac {x^{i} x^{-2 i}}{x}\right )\left (-\frac {x^{i}}{x^{2}}+\frac {i x^{i}}{x^{2}}\right ) \]

Which simplifies to

\[ W = -\frac {2 i x^{2 i} x^{-2 i}}{x^{3}} \]

Which simplifies to

\[ W = -\frac {2 i}{x^{3}} \]

Therefore Eq. (2) becomes

\[ u_1 = -\int \frac {\frac {6 x^{i} x^{-2 i}}{x}}{-\frac {2 i}{x}}\,dx \]

Which simplifies to

\[ u_1 = - \int 3 i x^{-i}d x \]

Hence

\[ u_1 = \left (\frac {3}{2}-\frac {3 i}{2}\right ) x^{1-i} \]

And Eq. (3) becomes

\[ u_2 = \int \frac {\frac {6 x^{i}}{x}}{-\frac {2 i}{x}}\,dx \]

Which simplifies to

\[ u_2 = \int 3 i x^{i}d x \]

Hence

\[ u_2 = \left (\frac {3}{2}+\frac {3 i}{2}\right ) x^{1+i} \]

Therefore the particular solution, from equation (1) is

\[ y_p(x) = \frac {\left (\frac {3}{2}-\frac {3 i}{2}\right ) x^{1-i} x^{i}}{x}+\frac {\left (\frac {3}{2}+\frac {3 i}{2}\right ) x^{i} x^{-2 i} x^{1+i}}{x} \]

Which simplifies to

\[ y_p(x) = 3 \]

Therefore the general solution is

\begin{align*} y &= y_h + y_p \\ &= \left (\left (\frac {i x^{-2 i} c_1}{2}+c_2 \right ) x^{-1+i}\right ) + \left (3\right ) \\ \end{align*}

Will add steps showing solving for IC soon.

17.43.5 Solved as second order ode using Kovacic algorithm

Time used: 0.276 (sec)

Writing the ode as

\begin{align*} x^{2} y^{\prime \prime }+3 x y^{\prime }+2 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 &= x^{2} \\ B &= 3 x\tag {3} \\ C &= 2 \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 {-5}{4 x^{2}}\tag {6} \end{align*}

Comparing the above to (5) shows that

\begin{align*} s &= -5\\ t &= 4 x^{2} \end{align*}

Therefore eq. (4) becomes

\begin{align*} z''(x) &= \left ( -\frac {5}{4 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	\(\left \{ 1,2\right \} \)	\(\left \{ 2,3,4,5,6,7,\cdots \right \} \)

Table 243: Necessary conditions for each Kovacic case

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 - 0 \\ &= 2 \end{align*}

The poles of \(r\) in eq. (7) and the order of each pole are determined by solving for the roots of \(t=4 x^{2}\). There is a pole at \(x=0\) of order \(2\). Since there is no odd order pole larger than \(2\) and the order at \(\infty \) is \(2\) then the necessary conditions for case one are met. Since there is a pole of order \(2\) then necessary conditions for case two are met. Since pole order is not larger than \(2\) and the order at \(\infty \) is \(2\) then the necessary conditions for case three are met. Therefore

\begin{align*} L &= [1, 2, 4, 6, 12] \end{align*}

Attempting to find a solution using case \(n=1\).

Looking at poles of order 2. The partial fractions decomposition of \(r\) is

\[ r = -\frac {5}{4 x^{2}} \]

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 \(b=-{\frac {5}{4}}\). Hence

\begin{alignat*}{2} [\sqrt r]_c &= 0 \\ \alpha _c^{+} &= \frac {1}{2} + \sqrt {1+4 b} &&= \frac {1}{2}+i\\ \alpha _c^{-} &= \frac {1}{2} - \sqrt {1+4 b} &&= \frac {1}{2}-i \end{alignat*}

Since the order of \(r\) at \(\infty \) is 2 then \([\sqrt r]_\infty =0\). Let \(b\) be the coefficient of \(\frac {1}{x^{2}}\) in the Laurent series expansion of \(r\) at \(\infty \). which can be found by dividing the leading coefficient of \(s\) by the leading coefficient of \(t\) from

\begin{alignat*}{2} r &= \frac {s}{t} &&= -\frac {5}{4 x^{2}} \end{alignat*}

Since the \(\text {gcd}(s,t)=1\). This gives \(b=-{\frac {5}{4}}\). Hence

\begin{alignat*}{2} [\sqrt r]_\infty &= 0 \\ \alpha _{\infty }^{+} &= \frac {1}{2} + \sqrt {1+4 b} &&= \frac {1}{2}+i\\ \alpha _{\infty }^{-} &= \frac {1}{2} - \sqrt {1+4 b} &&= \frac {1}{2}-i \end{alignat*}

The following table summarizes the findings so far for poles and for the order of \(r\) at \(\infty \) where \(r\) is

\[ r=-\frac {5}{4 x^{2}} \]

pole \(c\) location	pole order	\([\sqrt r]_c\)	\(\alpha _c^{+}\)	\(\alpha _c^{-}\)
\(0\)	\(2\)	\(0\)	\(\frac {1}{2}+i\)	\(\frac {1}{2}-i\)

Order of \(r\) at \(\infty \)	\([\sqrt r]_\infty \)	\(\alpha _\infty ^{+}\)	\(\alpha _\infty ^{-}\)
\(2\)	\(0\)	\(\frac {1}{2}+i\)	\(\frac {1}{2}-i\)

Now that the all \([\sqrt r]_c\) and its associated \(\alpha _c^{\pm }\) have been determined for all the poles in the set \(\Gamma \) and \([\sqrt r]_\infty \) and its associated \(\alpha _\infty ^{\pm }\) have also been found, the next step is to determine possible non negative integer \(d\) from these using

\begin{align*} d &= \alpha _\infty ^{s(\infty )} - \sum _{c \in \Gamma } \alpha _c^{s(c)} \end{align*}

Where \(s(c)\) is either \(+\) or \(-\) and \(s(\infty )\) is the sign of \(\alpha _\infty ^{\pm }\). This is done by trial over all set of families \(s=(s(c))_{c \in \Gamma \cup {\infty }}\) until such \(d\) is found to work in finding candidate \(\omega \). Trying \(\alpha _\infty ^{-} = \frac {1}{2}-i\) then

\begin{align*} d &= \alpha _\infty ^{-} - \left ( \alpha _{c_1}^{-} \right ) \\ &= \frac {1}{2}-i - \left ( \frac {1}{2}-i \right ) \\ &= 0 \end{align*}

Since \(d\) an integer and \(d \geq 0\) then it can be used to find \(\omega \) using

\begin{align*} \omega &= \sum _{c \in \Gamma } \left ( s(c) [\sqrt r]_c + \frac {\alpha _c^{s(c)}}{x-c} \right ) + s(\infty ) [\sqrt r]_\infty \end{align*}

The above gives

\begin{align*} \omega &= \left ( (-)[\sqrt r]_{c_1} + \frac { \alpha _{c_1}^{-} }{x- c_1}\right ) + (-) [\sqrt r]_\infty \\ &= \frac {\frac {1}{2}-i}{x} + (-) \left ( 0 \right ) \\ &= \frac {\frac {1}{2}-i}{x}\\ &= \frac {\frac {1}{2}-i}{x} \end{align*}

Now that \(\omega \) is determined, the next step is find a corresponding minimal polynomial \(p(x)\) of degree \(d=0\) to solve the ode. The polynomial \(p(x)\) needs to satisfy the equation

\begin{align*} p'' + 2 \omega p' + \left ( \omega ' +\omega ^2 -r\right ) p = 0 \tag {1A} \end{align*}

Let

\begin{align*} p(x) &= 1\tag {2A} \end{align*}

Substituting the above in eq. (1A) gives

\begin{align*} \left (0\right ) + 2 \left (\frac {\frac {1}{2}-i}{x}\right ) \left (0\right ) + \left ( \left (\frac {-\frac {1}{2}+i}{x^{2}}\right ) + \left (\frac {\frac {1}{2}-i}{x}\right )^2 - \left (-\frac {5}{4 x^{2}}\right ) \right ) &= 0\\ 0 = 0 \end{align*}

The equation is satisfied since both sides are zero. Therefore the first solution to the ode \(z'' = r z\) is

\begin{align*} z_1(x) &= p e^{ \int \omega \,dx} \\ &= {\mathrm e}^{\int \frac {\frac {1}{2}-i}{x}d x}\\ &= x^{\frac {1}{2}-i} \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 {3 x}{x^{2}} \,dx} \\ &= z_1 e^{-\frac {3 \ln \left (x \right )}{2}} \\ &= z_1 \left (\frac {1}{x^{{3}/{2}}}\right ) \\ \end{align*}

Which simplifies to

\[ y_1 = x^{-1-i} \]

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 {3 x}{x^{2}} \,dx}}{\left (y_1\right )^2} \,dx \\ &= y_1 \int \frac { e^{-3 \ln \left (x \right )}}{\left (y_1\right )^2} \,dx \\ &= y_1 \left (-\frac {i x^{2+2 i}}{2 x^{2}}\right ) \\ \end{align*}

Therefore the solution is

\begin{align*} y &= c_1 y_1 + c_2 y_2 \\ &= c_1 \left (x^{-1-i}\right ) + c_2 \left (x^{-1-i}\left (-\frac {i x^{2+2 i}}{2 x^{2}}\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

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

The homogeneous solution is found using the Kovacic algorithm which results in

\[ y_h = c_1 \,x^{-1-i}-\frac {i c_2 \,x^{-1+i}}{2} \]

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 &= x^{-1-i} \\ y_2 &= -\frac {i x^{-1+i}}{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

\[ W = \begin {vmatrix} x^{-1-i} & -\frac {i x^{-1+i}}{2} \\ \frac {d}{dx}\left (x^{-1-i}\right ) & \frac {d}{dx}\left (-\frac {i x^{-1+i}}{2}\right ) \end {vmatrix} \]

Which gives

\[ W = \begin {vmatrix} x^{-1-i} & -\frac {i x^{-1+i}}{2} \\ \frac {\left (-1-i\right ) x^{-1-i}}{x} & \frac {\left (\frac {1}{2}+\frac {i}{2}\right ) x^{-1+i}}{x} \end {vmatrix} \]

Therefore

\[ W = \left (x^{-1-i}\right )\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) x^{-1+i}}{x}\right ) - \left (-\frac {i x^{-1+i}}{2}\right )\left (\frac {\left (-1-i\right ) x^{-1-i}}{x}\right ) \]

Which simplifies to

\[ W = \frac {x^{-1-i} x^{-1+i}}{x} \]

Which simplifies to

\[ W = \frac {1}{x^{3}} \]

Therefore Eq. (2) becomes

\[ u_1 = -\int \frac {-3 i x^{-1+i}}{\frac {1}{x}}\,dx \]

Which simplifies to

\[ u_1 = - \int -3 i x^{i}d x \]

Hence

\[ u_1 = \left (\frac {3}{2}+\frac {3 i}{2}\right ) x^{1+i} \]

And Eq. (3) becomes

\[ u_2 = \int \frac {6 x^{-1-i}}{\frac {1}{x}}\,dx \]

Which simplifies to

\[ u_2 = \int 6 x^{-i}d x \]

Hence

\[ u_2 = \left (3+3 i\right ) x^{1-i} \]

Therefore the particular solution, from equation (1) is

\[ y_p(x) = \left (\frac {3}{2}+\frac {3 i}{2}\right ) x^{1+i} x^{-1-i}+\left (\frac {3}{2}-\frac {3 i}{2}\right ) x^{-1+i} x^{1-i} \]

Which simplifies to

\[ y_p(x) = 3 \]

Therefore the general solution is

\begin{align*} y &= y_h + y_p \\ &= \left (c_1 \,x^{-1-i}-\frac {i c_2 \,x^{-1+i}}{2}\right ) + \left (3\right ) \\ \end{align*}

Will add steps showing solving for IC soon.

17.43.6 Solved as second order ode adjoint method

Time used: 1.539 (sec)

In normal form the ode

\begin{align*} x^{2} y^{\prime \prime }+3 x y^{\prime }+2 y = 6 \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 {2}{x^{2}}\\ r \left (x \right )&=\frac {6}{x^{2}} \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 {2 \xi \left (x \right )}{x^{2}}\right ) &= 0\\ \xi ^{\prime \prime }\left (x \right )+\frac {5 \xi \left (x \right )}{x^{2}}-\frac {3 \xi ^{\prime }\left (x \right )}{x}&= 0 \end{align*}

Which is solved for \(\xi (x)\). This is Euler second order ODE. Let the solution be \(\xi = x^r\), then \(\xi '=r x^{r-1}\) and \(\xi ''=r(r-1) x^{r-2}\). Substituting these back into the given ODE gives

\[ x^{2}(r(r-1))x^{r-2}-3 x r x^{r-1}+5 x^{r} = 0 \]

Simplifying gives

\[ r \left (r -1\right )x^{r}-3 r\,x^{r}+5 x^{r} = 0 \]

Since \(x^{r}\neq 0\) then dividing throughout by \(x^{r}\) gives

\[ r \left (r -1\right )-3 r+5 = 0 \]

Or

\[ r^{2}-4 r +5 = 0 \tag {1} \]

Equation (1) is the characteristic equation. Its roots determine the form of the general solution. Using the quadratic equation the roots are

\begin{align*} r_1 &= 2-i\\ r_2 &= 2+i \end{align*}

The roots are complex conjugate of each others. Let the roots be

\begin{align*} r_1 &= \alpha + i \beta \\ r_2 &= \alpha - i \beta \\ \end{align*}

Where in this case \(\alpha =2\) and \(\beta =-1\). Hence the solution becomes

\begin{align*} \xi &= c_1 x^{r_1} + c_2 x^{r_2} \\ &= c_1 x^{\alpha + i \beta } + c_2 x^{\alpha - i \beta } \\ &= x^{\alpha } \left ( c_1 x^{i \beta } + c_2 x^{- i \beta }\right ) \\ &= x^{\alpha } \left ( c_1 e^{\ln \left (x^{i \beta }\right )} + c_2 e^{\ln \left (x^{-i \beta }\right )}\right ) \\ &= x^{\alpha } \left ( c_1 e^{i \left (\beta \ln {x}\right )} + c_2 e^{-i \left (\beta \ln {x}\right )}\right ) \end{align*}

Using the values for \(\alpha =2, \beta =-1\), the above becomes

\begin{align*} \xi &= x^{2} \left ( c_1 e^{-i \ln \left (x \right )} + c_2 e^{i \ln \left (x \right )} \right ) \end{align*}

Using Euler relation, the expression \(c_1 e^{i A}+ c_2 e^{-i A}\) is transformed to \( c_1 \cos A+ c_1 \sin A\) where the constants are free to change. Applying this to the above result gives

\begin{align*} \xi &=x^{2}\left (c_1 \cos \left (\ln \left (x \right )\right )+c_2 \sin \left (\ln \left (x \right )\right )\right ) \end{align*}

Will add steps showing solving for IC soon.

The original ode (2) 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 )}\\ y^{\prime }+y \left (\frac {3}{x}-\frac {2 x \left (c_1 \cos \left (\ln \left (x \right )\right )+c_2 \sin \left (\ln \left (x \right )\right )\right )+x^{2} \left (-\frac {c_1 \sin \left (\ln \left (x \right )\right )}{x}+\frac {c_2 \cos \left (\ln \left (x \right )\right )}{x}\right )}{x^{2} \left (c_1 \cos \left (\ln \left (x \right )\right )+c_2 \sin \left (\ln \left (x \right )\right )\right )}\right )&=\frac {3 c_1 \cos \left (\ln \left (x \right )\right ) x +3 c_1 \sin \left (\ln \left (x \right )\right ) x -3 c_2 \cos \left (\ln \left (x \right )\right ) x +3 c_2 \sin \left (\ln \left (x \right )\right ) x}{x^{2} \left (c_1 \cos \left (\ln \left (x \right )\right )+c_2 \sin \left (\ln \left (x \right )\right )\right )} \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 {c_1 \cos \left (\ln \left (x \right )\right )-c_2 \cos \left (\ln \left (x \right )\right )+c_1 \sin \left (\ln \left (x \right )\right )+c_2 \sin \left (\ln \left (x \right )\right )}{x \left (c_1 \cos \left (\ln \left (x \right )\right )+c_2 \sin \left (\ln \left (x \right )\right )\right )}\\ p(x) &=\frac {3 c_1 \cos \left (\ln \left (x \right )\right )-3 c_2 \cos \left (\ln \left (x \right )\right )+3 c_1 \sin \left (\ln \left (x \right )\right )+3 c_2 \sin \left (\ln \left (x \right )\right )}{x \left (c_1 \cos \left (\ln \left (x \right )\right )+c_2 \sin \left (\ln \left (x \right )\right )\right )} \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,dx}}\\ &= {\mathrm e}^{\int \frac {c_1 \cos \left (\ln \left (x \right )\right )-c_2 \cos \left (\ln \left (x \right )\right )+c_1 \sin \left (\ln \left (x \right )\right )+c_2 \sin \left (\ln \left (x \right )\right )}{x \left (c_1 \cos \left (\ln \left (x \right )\right )+c_2 \sin \left (\ln \left (x \right )\right )\right )}d x}\\ &= {\mathrm e}^{-\ln \left (c_2 \tan \left (\ln \left (x \right )\right )+c_1 \right )+\frac {\ln \left (\tan \left (\ln \left (x \right )\right )^{2}+1\right )}{2}+\ln \left (x \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 {3 c_1 \cos \left (\ln \left (x \right )\right )-3 c_2 \cos \left (\ln \left (x \right )\right )+3 c_1 \sin \left (\ln \left (x \right )\right )+3 c_2 \sin \left (\ln \left (x \right )\right )}{x \left (c_1 \cos \left (\ln \left (x \right )\right )+c_2 \sin \left (\ln \left (x \right )\right )\right )}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (y \,{\mathrm e}^{-\ln \left (c_2 \tan \left (\ln \left (x \right )\right )+c_1 \right )+\frac {\ln \left (\tan \left (\ln \left (x \right )\right )^{2}+1\right )}{2}+\ln \left (x \right )}\right ) &= \left ({\mathrm e}^{-\ln \left (c_2 \tan \left (\ln \left (x \right )\right )+c_1 \right )+\frac {\ln \left (\tan \left (\ln \left (x \right )\right )^{2}+1\right )}{2}+\ln \left (x \right )}\right ) \left (\frac {3 c_1 \cos \left (\ln \left (x \right )\right )-3 c_2 \cos \left (\ln \left (x \right )\right )+3 c_1 \sin \left (\ln \left (x \right )\right )+3 c_2 \sin \left (\ln \left (x \right )\right )}{x \left (c_1 \cos \left (\ln \left (x \right )\right )+c_2 \sin \left (\ln \left (x \right )\right )\right )}\right ) \\ \mathrm {d} \left (y \,{\mathrm e}^{-\ln \left (c_2 \tan \left (\ln \left (x \right )\right )+c_1 \right )+\frac {\ln \left (\tan \left (\ln \left (x \right )\right )^{2}+1\right )}{2}+\ln \left (x \right )}\right ) &= \left (\frac {3 \left (c_1 \cos \left (\ln \left (x \right )\right )-c_2 \cos \left (\ln \left (x \right )\right )+c_1 \sin \left (\ln \left (x \right )\right )+c_2 \sin \left (\ln \left (x \right )\right )\right ) {\mathrm e}^{-\ln \left (c_2 \tan \left (\ln \left (x \right )\right )+c_1 \right )+\frac {\ln \left (\tan \left (\ln \left (x \right )\right )^{2}+1\right )}{2}+\ln \left (x \right )}}{x \left (c_1 \cos \left (\ln \left (x \right )\right )+c_2 \sin \left (\ln \left (x \right )\right )\right )}\right )\, \mathrm {d} x \\ \end{align*}

Integrating gives

\begin{align*} y \,{\mathrm e}^{-\ln \left (c_2 \tan \left (\ln \left (x \right )\right )+c_1 \right )+\frac {\ln \left (\tan \left (\ln \left (x \right )\right )^{2}+1\right )}{2}+\ln \left (x \right )}&= \int {\frac {3 \left (c_1 \cos \left (\ln \left (x \right )\right )-c_2 \cos \left (\ln \left (x \right )\right )+c_1 \sin \left (\ln \left (x \right )\right )+c_2 \sin \left (\ln \left (x \right )\right )\right ) {\mathrm e}^{-\ln \left (c_2 \tan \left (\ln \left (x \right )\right )+c_1 \right )+\frac {\ln \left (\tan \left (\ln \left (x \right )\right )^{2}+1\right )}{2}+\ln \left (x \right )}}{x \left (c_1 \cos \left (\ln \left (x \right )\right )+c_2 \sin \left (\ln \left (x \right )\right )\right )} \,dx} \\ &=3 \,{\mathrm e}^{-\ln \left (c_2 \tan \left (\ln \left (x \right )\right )+c_1 \right )+\frac {\ln \left (\tan \left (\ln \left (x \right )\right )^{2}+1\right )}{2}+\ln \left (x \right )} + c_3 \end{align*}

Dividing throughout by the integrating factor \({\mathrm e}^{-\ln \left (c_2 \tan \left (\ln \left (x \right )\right )+c_1 \right )+\frac {\ln \left (\tan \left (\ln \left (x \right )\right )^{2}+1\right )}{2}+\ln \left (x \right )}\) gives the final solution

\[ y = \frac {\left (c_2 \tan \left (\ln \left (x \right )\right )+c_1 \right ) \left (3 \,{\mathrm e}^{-\ln \left (c_2 \tan \left (\ln \left (x \right )\right )+c_1 \right )+\frac {\ln \left (\tan \left (\ln \left (x \right )\right )^{2}+1\right )}{2}+\ln \left (x \right )}+c_3 \right )}{x \sqrt {\tan \left (\ln \left (x \right )\right )^{2}+1}} \]

Hence, the solution found using Lagrange adjoint equation method is

\begin{align*} y &= \frac {\left (c_2 \tan \left (\ln \left (x \right )\right )+c_1 \right ) \left (3 \,{\mathrm e}^{-\ln \left (c_2 \tan \left (\ln \left (x \right )\right )+c_1 \right )+\frac {\ln \left (\tan \left (\ln \left (x \right )\right )^{2}+1\right )}{2}+\ln \left (x \right )}+c_3 \right )}{x \sqrt {\tan \left (\ln \left (x \right )\right )^{2}+1}} \\ \end{align*}

Will add steps showing solving for IC soon.

17.43.7 Maple step by step solution

17.43.8 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 
   <- LODE of Euler type successful 
<- solving first the homogeneous part of the ODE successful`
 

17.43.9 Maple dsolve solution

Solving time : 0.007 (sec)
Leaf size : 22

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

\[ y \left (x \right ) = \frac {c_{2} \sin \left (\ln \left (x \right )\right )+c_{1} \cos \left (\ln \left (x \right )\right )+3 x}{x} \]

17.43.10 Mathematica DSolve solution

Solving time : 0.024 (sec)
Leaf size : 25

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

\[ y(x)\to \frac {3 x+c_2 \cos (\log (x))+c_1 \sin (\log (x))}{x} \]