Internal problem ID [9503]
Internal file name [OUTPUT/8443_Monday_June_06_2022_03_05_22_AM_8323642/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1174.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "kovacic", "second_order_euler_ode", "second_order_change_of_variable_on_x_method_1", "second_order_change_of_variable_on_x_method_2", "second_order_change_of_variable_on_y_method_1", "second_order_change_of_variable_on_y_method_2", "linear_second_order_ode_solved_by_an_integrating_factor", "second_order_ode_non_constant_coeff_transformation_on_B"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {x^{2} y^{\prime \prime }-2 y^{\prime } x +2 y=x^{5} \ln \left (x \right )} \]
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 \(A=x^{2}, B=-2 x, C=2, f(x)=x^{5} \ln \left (x \right )\). 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 \[ x^{2} y^{\prime \prime }-2 y^{\prime } x +2 y = 0 \] This is Euler second order ODE. Let the solution be \(y = x^r\), then \(y'=r x^{r-1}\) and \(y''=r(r-1) x^{r-2}\). Substituting these back into the given ODE gives \[ x^{2}(r(r-1))x^{r-2}-2 x r x^{r-1}+2 x^{r} = 0 \] Simplifying gives \[ r \left (r -1\right )x^{r}-2 r\,x^{r}+2 x^{r} = 0 \] Since \(x^{r}\neq 0\) then dividing throughout by \(x^{r}\) gives \[ r \left (r -1\right )-2 r+2 = 0 \] Or \[ r^{2}-3 r +2 = 0 \tag {1} \] 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 &= 1\\ r_2 &= 2 \end {align*}
Since the roots are real and distinct, then the general solution is \[ y= c_{1} y_1 + c_{2} y_2 \] Where \(y_1 = x^{r_1}\) and \(y_2 = x^{r_2} \). Hence \[ y = c_{2} x^{2}+c_{1} x \] Next, we find the particular solution to the ODE \[ x^{2} y^{\prime \prime }-2 y^{\prime } x +2 y = x^{5} \ln \left (x \right ) \] 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 &= x \\ y_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 \[ W = \begin {vmatrix} x & x^{2} \\ \frac {d}{dx}\left (x\right ) & \frac {d}{dx}\left (x^{2}\right ) \end {vmatrix} \] Which gives \[ W = \begin {vmatrix} x & x^{2} \\ 1 & 2 x \end {vmatrix} \] Therefore \[ W = \left (x\right )\left (2 x\right ) - \left (x^{2}\right )\left (1\right ) \] Which simplifies to \[ W = x^{2} \] Which simplifies to \[ W = x^{2} \] Therefore Eq. (2) becomes \[ u_1 = -\int \frac {x^{7} \ln \left (x \right )}{x^{4}}\,dx \] Which simplifies to \[ u_1 = - \int x^{3} \ln \left (x \right )d x \] Hence \[ u_1 = -\frac {\ln \left (x \right ) x^{4}}{4}+\frac {x^{4}}{16} \] And Eq. (3) becomes \[ u_2 = \int \frac {x^{6} \ln \left (x \right )}{x^{4}}\,dx \] Which simplifies to \[ u_2 = \int \ln \left (x \right ) x^{2}d x \] Hence \[ u_2 = \frac {x^{3} \ln \left (x \right )}{3}-\frac {x^{3}}{9} \] Which simplifies to \begin{align*} u_1 &= -\frac {x^{4} \left (-1+4 \ln \left (x \right )\right )}{16} \\ u_2 &= \frac {x^{3} \left (-1+3 \ln \left (x \right )\right )}{9} \\ \end{align*} Therefore the particular solution, from equation (1) is \[ y_p(x) = -\frac {x^{5} \left (-1+4 \ln \left (x \right )\right )}{16}+\frac {x^{5} \left (-1+3 \ln \left (x \right )\right )}{9} \] Which simplifies to \[ y_p(x) = \frac {x^{5} \left (-7+12 \ln \left (x \right )\right )}{144} \] Therefore the general solution is \begin {align*} y &= y_h + y_p \\ &= \frac {x^{5} \left (-7+12 \ln \left (x \right )\right )}{144}+c_{2} x^{2}+c_{1} x \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {x^{5} \left (-7+12 \ln \left (x \right )\right )}{144}+c_{2} x^{2}+c_{1} x \\ \end{align*}
Verification of solutions
\[ y = \frac {x^{5} \left (-7+12 \ln \left (x \right )\right )}{144}+c_{2} x^{2}+c_{1} x \] Verified OK.
The ode satisfies this form \[ y^{\prime \prime }+p \left (x \right ) y^{\prime }+\frac {\left (p \left (x \right )^{2}+p^{\prime }\left (x \right )\right ) y}{2} = f \left (x \right ) \] Where \( p(x) = -\frac {2}{x}\). Therefore, there is an integrating factor given by \begin {align*} M(x) &= e^{\frac {1}{2} \int p \, dx} \\ &= e^{ \int -\frac {2}{x} \, dx} \\ &= \frac {1}{x} \end {align*}
Multiplying both sides of the ODE by the integrating factor \(M(x)\) makes the left side of the ODE a complete differential \begin{align*} \left ( M(x) y \right )'' &= \ln \left (x \right ) x^{2} \\ \left ( \frac {y}{x} \right )'' &= \ln \left (x \right ) x^{2} \\ \end{align*} Integrating once gives \[ \left ( \frac {y}{x} \right )' = \frac {x^{3} \left (-1+3 \ln \left (x \right )\right )}{9}+c_{1} \] Integrating again gives \[ \left ( \frac {y}{x} \right ) = -\frac {7 x^{4}}{144}+\frac {\ln \left (x \right ) x^{4}}{12}+c_{1} x +c_{2} \] Hence the solution is \begin{align*} y &= \frac {-\frac {7 x^{4}}{144}+\frac {\ln \left (x \right ) x^{4}}{12}+c_{1} x +c_{2}}{\frac {1}{x}} \\ \end{align*} Or \[ y = \frac {x^{5} \ln \left (x \right )}{12}-\frac {7 x^{5}}{144}+c_{1} x^{2}+c_{2} x \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {x^{5} \ln \left (x \right )}{12}-\frac {7 x^{5}}{144}+c_{1} x^{2}+c_{2} x \\ \end{align*}
Verification of solutions
\[ y = \frac {x^{5} \ln \left (x \right )}{12}-\frac {7 x^{5}}{144}+c_{1} x^{2}+c_{2} x \] Verified OK.
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 \[ x^{2} y^{\prime \prime }-2 y^{\prime } x +2 y = 0 \] In normal form the ode \begin {align*} x^{2} y^{\prime \prime }-2 y^{\prime } x +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 {2}{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 {2}{x}d x \right )}d x\\ &= \int e^{2 \ln \left (x \right )} \,dx\\ &= \int x^{2}d x\\ &= \frac {x^{3}}{3}\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}}}{x^{4}}\\ &= \frac {2}{x^{6}}\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 {2 y \left (\tau \right )}{x^{6}}&=0 \\ \end {align*}
But in terms of \(\tau \) \begin {align*} \frac {2}{x^{6}}&=\frac {2}{9 \tau ^{2}} \end {align*}
Hence the above ode becomes \begin {align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+\frac {2 y \left (\tau \right )}{9 \tau ^{2}}&=0 \end {align*}
The above ode is now solved for \(y \left (\tau \right )\). The ode can be written as \[ 9 \left (\frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )\right ) \tau ^{2}+2 y \left (\tau \right ) = 0 \] Which shows it is a Euler ODE. 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 \[ 9 \tau ^{2}(r(r-1))\tau ^{r-2}+0 r \tau ^{r-1}+2 \tau ^{r} = 0 \] Simplifying gives \[ 9 r \left (r -1\right )\tau ^{r}+0\,\tau ^{r}+2 \tau ^{r} = 0 \] Since \(\tau ^{r}\neq 0\) then dividing throughout by \(\tau ^{r}\) gives \[ 9 r \left (r -1\right )+0+2 = 0 \] Or \[ 9 r^{2}-9 r +2 = 0 \tag {1} \] 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}{3}}\\ r_2 &= {\frac {2}{3}} \end {align*}
Since the roots are real and distinct, then the general solution is \[ y \left (\tau \right )= c_{1} y_1 + c_{2} y_2 \] Where \(y_1 = \tau ^{r_1}\) and \(y_2 = \tau ^{r_2} \). Hence \[ y \left (\tau \right ) = c_{1} \tau ^{\frac {1}{3}}+c_{2} \tau ^{\frac {2}{3}} \] The above solution is now transformed back to \(y\) using (6) which results in \begin {align*} y &= \frac {c_{1} 3^{\frac {2}{3}} \left (x^{3}\right )^{\frac {1}{3}}}{3}+\frac {c_{2} 3^{\frac {1}{3}} \left (x^{3}\right )^{\frac {2}{3}}}{3} \end {align*}
Therefore the homogeneous solution \(y_h\) is \[ y_h = \frac {c_{1} 3^{\frac {2}{3}} \left (x^{3}\right )^{\frac {1}{3}}}{3}+\frac {c_{2} 3^{\frac {1}{3}} \left (x^{3}\right )^{\frac {2}{3}}}{3} \] 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 &= \left (x^{3}\right )^{\frac {1}{3}} \\ y_2 &= \left (x^{3}\right )^{\frac {2}{3}} \\ \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 \[ W = \begin {vmatrix} \left (x^{3}\right )^{\frac {1}{3}} & \left (x^{3}\right )^{\frac {2}{3}} \\ \frac {d}{dx}\left (\left (x^{3}\right )^{\frac {1}{3}}\right ) & \frac {d}{dx}\left (\left (x^{3}\right )^{\frac {2}{3}}\right ) \end {vmatrix} \] Which gives \[ W = \begin {vmatrix} \left (x^{3}\right )^{\frac {1}{3}} & \left (x^{3}\right )^{\frac {2}{3}} \\ \frac {x^{2}}{\left (x^{3}\right )^{\frac {2}{3}}} & \frac {2 x^{2}}{\left (x^{3}\right )^{\frac {1}{3}}} \end {vmatrix} \] Therefore \[ W = \left (\left (x^{3}\right )^{\frac {1}{3}}\right )\left (\frac {2 x^{2}}{\left (x^{3}\right )^{\frac {1}{3}}}\right ) - \left (\left (x^{3}\right )^{\frac {2}{3}}\right )\left (\frac {x^{2}}{\left (x^{3}\right )^{\frac {2}{3}}}\right ) \] Which simplifies to \[ W = x^{2} \] Which simplifies to \[ W = x^{2} \] Therefore Eq. (2) becomes \[ u_1 = -\int \frac {\left (x^{3}\right )^{\frac {2}{3}} x^{5} \ln \left (x \right )}{x^{4}}\,dx \] Which simplifies to \[ u_1 = - \int \left (x^{3}\right )^{\frac {2}{3}} x \ln \left (x \right )d x \] Hence \[ u_1 = -\frac {\left (x^{3}\right )^{\frac {2}{3}} x^{2} \ln \left (x \right )}{4}+\frac {\left (x^{3}\right )^{\frac {2}{3}} x^{2}}{16} \] And Eq. (3) becomes \[ u_2 = \int \frac {\left (x^{3}\right )^{\frac {1}{3}} x^{5} \ln \left (x \right )}{x^{4}}\,dx \] Which simplifies to \[ u_2 = \int \left (x^{3}\right )^{\frac {1}{3}} x \ln \left (x \right )d x \] Hence \[ u_2 = \frac {\left (x^{3}\right )^{\frac {1}{3}} x^{2} \ln \left (x \right )}{3}-\frac {\left (x^{3}\right )^{\frac {1}{3}} x^{2}}{9} \] Which simplifies to \begin{align*} u_1 &= -\frac {\left (x^{3}\right )^{\frac {2}{3}} x^{2} \left (-1+4 \ln \left (x \right )\right )}{16} \\ u_2 &= \frac {\left (x^{3}\right )^{\frac {1}{3}} x^{2} \left (-1+3 \ln \left (x \right )\right )}{9} \\ \end{align*} Therefore the particular solution, from equation (1) is \[ y_p(x) = -\frac {x^{5} \left (-1+4 \ln \left (x \right )\right )}{16}+\frac {x^{5} \left (-1+3 \ln \left (x \right )\right )}{9} \] Which simplifies to \[ y_p(x) = \frac {x^{5} \left (-7+12 \ln \left (x \right )\right )}{144} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (\frac {c_{1} 3^{\frac {2}{3}} \left (x^{3}\right )^{\frac {1}{3}}}{3}+\frac {c_{2} 3^{\frac {1}{3}} \left (x^{3}\right )^{\frac {2}{3}}}{3}\right ) + \left (\frac {x^{5} \left (-7+12 \ln \left (x \right )\right )}{144}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {c_{1} 3^{\frac {2}{3}} \left (x^{3}\right )^{\frac {1}{3}}}{3}+\frac {c_{2} 3^{\frac {1}{3}} \left (x^{3}\right )^{\frac {2}{3}}}{3}+\frac {x^{5} \left (-7+12 \ln \left (x \right )\right )}{144} \\ \end{align*}
Verification of solutions
\[ y = \frac {c_{1} 3^{\frac {2}{3}} \left (x^{3}\right )^{\frac {1}{3}}}{3}+\frac {c_{2} 3^{\frac {1}{3}} \left (x^{3}\right )^{\frac {2}{3}}}{3}+\frac {x^{5} \left (-7+12 \ln \left (x \right )\right )}{144} \] 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 \(A=x^{2}, B=-2 x, C=2, f(x)=x^{5} \ln \left (x \right )\). 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 \[ x^{2} y^{\prime \prime }-2 y^{\prime } x +2 y = 0 \] In normal form the ode \begin {align*} x^{2} y^{\prime \prime }-2 y^{\prime } x +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 {2}{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 {2}{x}\frac {\sqrt {2}\, \sqrt {\frac {1}{x^{2}}}}{c}}{\left (\frac {\sqrt {2}\, \sqrt {\frac {1}{x^{2}}}}{c}\right )^2} \\ &=-\frac {3 c \sqrt {2}}{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 )-\frac {3 c \sqrt {2}\, \left (\frac {d}{d \tau }y \left (\tau \right )\right )}{2}+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 {3 \sqrt {2}\, c \tau }{4}} \left (c_{1} \cosh \left (\frac {\sqrt {2}\, c \tau }{4}\right )+i c_{2} \sinh \left (\frac {\sqrt {2}\, c \tau }{4}\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}\, \sqrt {\frac {1}{x^{2}}}\, x \ln \left (x \right )}{c} \end {align*}
Substituting the above into the solution obtained gives \[ y = x^{\frac {3}{2}} \left (c_{1} \cosh \left (\frac {\ln \left (x \right )}{2}\right )+i c_{2} \sinh \left (\frac {\ln \left (x \right )}{2}\right )\right ) \] Now the particular solution to this ODE is found \[ x^{2} y^{\prime \prime }-2 y^{\prime } x +2 y = x^{5} \ln \left (x \right ) \] 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 &= \left (x^{3}\right )^{\frac {1}{3}} \\ y_2 &= \left (x^{3}\right )^{\frac {2}{3}} \\ \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 \[ W = \begin {vmatrix} \left (x^{3}\right )^{\frac {1}{3}} & \left (x^{3}\right )^{\frac {2}{3}} \\ \frac {d}{dx}\left (\left (x^{3}\right )^{\frac {1}{3}}\right ) & \frac {d}{dx}\left (\left (x^{3}\right )^{\frac {2}{3}}\right ) \end {vmatrix} \] Which gives \[ W = \begin {vmatrix} \left (x^{3}\right )^{\frac {1}{3}} & \left (x^{3}\right )^{\frac {2}{3}} \\ \frac {x^{2}}{\left (x^{3}\right )^{\frac {2}{3}}} & \frac {2 x^{2}}{\left (x^{3}\right )^{\frac {1}{3}}} \end {vmatrix} \] Therefore \[ W = \left (\left (x^{3}\right )^{\frac {1}{3}}\right )\left (\frac {2 x^{2}}{\left (x^{3}\right )^{\frac {1}{3}}}\right ) - \left (\left (x^{3}\right )^{\frac {2}{3}}\right )\left (\frac {x^{2}}{\left (x^{3}\right )^{\frac {2}{3}}}\right ) \] Which simplifies to \[ W = x^{2} \] Which simplifies to \[ W = x^{2} \] Therefore Eq. (2) becomes \[ u_1 = -\int \frac {\left (x^{3}\right )^{\frac {2}{3}} x^{5} \ln \left (x \right )}{x^{4}}\,dx \] Which simplifies to \[ u_1 = - \int \left (x^{3}\right )^{\frac {2}{3}} x \ln \left (x \right )d x \] Hence \[ u_1 = -\frac {\left (x^{3}\right )^{\frac {2}{3}} x^{2} \ln \left (x \right )}{4}+\frac {\left (x^{3}\right )^{\frac {2}{3}} x^{2}}{16} \] And Eq. (3) becomes \[ u_2 = \int \frac {\left (x^{3}\right )^{\frac {1}{3}} x^{5} \ln \left (x \right )}{x^{4}}\,dx \] Which simplifies to \[ u_2 = \int \left (x^{3}\right )^{\frac {1}{3}} x \ln \left (x \right )d x \] Hence \[ u_2 = \frac {\left (x^{3}\right )^{\frac {1}{3}} x^{2} \ln \left (x \right )}{3}-\frac {\left (x^{3}\right )^{\frac {1}{3}} x^{2}}{9} \] Which simplifies to \begin{align*} u_1 &= -\frac {\left (x^{3}\right )^{\frac {2}{3}} x^{2} \left (-1+4 \ln \left (x \right )\right )}{16} \\ u_2 &= \frac {\left (x^{3}\right )^{\frac {1}{3}} x^{2} \left (-1+3 \ln \left (x \right )\right )}{9} \\ \end{align*} Therefore the particular solution, from equation (1) is \[ y_p(x) = -\frac {x^{5} \left (-1+4 \ln \left (x \right )\right )}{16}+\frac {x^{5} \left (-1+3 \ln \left (x \right )\right )}{9} \] Which simplifies to \[ y_p(x) = \frac {x^{5} \left (-7+12 \ln \left (x \right )\right )}{144} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (x^{\frac {3}{2}} \left (c_{1} \cosh \left (\frac {\ln \left (x \right )}{2}\right )+i c_{2} \sinh \left (\frac {\ln \left (x \right )}{2}\right )\right )\right ) + \left (\frac {x^{5} \left (-7+12 \ln \left (x \right )\right )}{144}\right ) \\ &= \frac {x^{5} \left (-7+12 \ln \left (x \right )\right )}{144}+x^{\frac {3}{2}} \left (c_{1} \cosh \left (\frac {\ln \left (x \right )}{2}\right )+i c_{2} \sinh \left (\frac {\ln \left (x \right )}{2}\right )\right ) \\ \end{align*} Which simplifies to \[ y = \frac {x^{5} \left (-7+12 \ln \left (x \right )\right )}{144}+x^{\frac {3}{2}} \left (c_{1} \cosh \left (\frac {\ln \left (x \right )}{2}\right )+i c_{2} \sinh \left (\frac {\ln \left (x \right )}{2}\right )\right ) \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {x^{5} \left (-7+12 \ln \left (x \right )\right )}{144}+x^{\frac {3}{2}} \left (c_{1} \cosh \left (\frac {\ln \left (x \right )}{2}\right )+i c_{2} \sinh \left (\frac {\ln \left (x \right )}{2}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = \frac {x^{5} \left (-7+12 \ln \left (x \right )\right )}{144}+x^{\frac {3}{2}} \left (c_{1} \cosh \left (\frac {\ln \left (x \right )}{2}\right )+i c_{2} \sinh \left (\frac {\ln \left (x \right )}{2}\right )\right ) \] Verified OK.
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 \[ x^{2} y^{\prime \prime }-2 y^{\prime } x +2 y = 0 \] In normal form the given ode is written as \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 {2}{x}\\ q \left (x \right )&=\frac {2}{x^{2}} \end {align*}
Calculating the Liouville ode invariant \(Q\) given by \begin {align*} Q &= q - \frac {p'}{2}- \frac {p^2}{4} \\ &= \frac {2}{x^{2}} - \frac {\left (-\frac {2}{x}\right )'}{2}- \frac {\left (-\frac {2}{x}\right )^2}{4} \\ &= \frac {2}{x^{2}} - \frac {\left (\frac {2}{x^{2}}\right )}{2}- \frac {\left (\frac {4}{x^{2}}\right )}{4} \\ &= \frac {2}{x^{2}} - \left (\frac {1}{x^{2}}\right )-\frac {1}{x^{2}}\\ &= 0 \end {align*}
Since the Liouville ode invariant does not depend on the independent variable \(x\) then the transformation \begin {align*} y = v \left (x \right ) z \left (x \right )\tag {3} \end {align*}
is used to change the original ode to a constant coefficients ode in \(v\). In (3) the term \(z \left (x \right )\) is given by \begin {align*} z \left (x \right )&={\mathrm e}^{-\left (\int \frac {p \left (x \right )}{2}d x \right )}\\ &= e^{-\int \frac {-\frac {2}{x}}{2} }\\ &= x\tag {5} \end {align*}
Hence (3) becomes \begin {align*} y = v \left (x \right ) x\tag {4} \end {align*}
Applying this change of variable to the original ode results in \begin {align*} v^{\prime \prime }\left (x \right ) = \ln \left (x \right ) x^{2} \end {align*}
Which is now solved for \(v \left (x \right )\) Integrating once gives \[ v^{\prime }\left (x \right )= \frac {x^{3} \ln \left (x \right )}{3}-\frac {x^{3}}{9} + c_{1} \] Integrating again gives \[ v \left (x \right )= -\frac {7 x^{4}}{144}+\frac {\ln \left (x \right ) x^{4}}{12} + c_{1} x + c_{2} \] Now that \(v \left (x \right )\) is known, then \begin {align*} y&= v \left (x \right ) z \left (x \right )\\ &= \left (-\frac {7 x^{4}}{144}+\frac {\ln \left (x \right ) x^{4}}{12}+c_{1} x +c_{2}\right ) \left (z \left (x \right )\right )\tag {7} \end {align*}
But from (5) \begin {align*} z \left (x \right )&= x \end {align*}
Hence (7) becomes \begin {align*} y = \left (-\frac {7 x^{4}}{144}+\frac {\ln \left (x \right ) x^{4}}{12}+c_{1} x +c_{2} \right ) x \end {align*}
Therefore the homogeneous solution \(y_h\) is \[ y_h = \left (-\frac {7 x^{4}}{144}+\frac {\ln \left (x \right ) x^{4}}{12}+c_{1} x +c_{2} \right ) x \] 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 &= \left (x^{3}\right )^{\frac {1}{3}} \\ y_2 &= \left (x^{3}\right )^{\frac {2}{3}} \\ \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 \[ W = \begin {vmatrix} \left (x^{3}\right )^{\frac {1}{3}} & \left (x^{3}\right )^{\frac {2}{3}} \\ \frac {d}{dx}\left (\left (x^{3}\right )^{\frac {1}{3}}\right ) & \frac {d}{dx}\left (\left (x^{3}\right )^{\frac {2}{3}}\right ) \end {vmatrix} \] Which gives \[ W = \begin {vmatrix} \left (x^{3}\right )^{\frac {1}{3}} & \left (x^{3}\right )^{\frac {2}{3}} \\ \frac {x^{2}}{\left (x^{3}\right )^{\frac {2}{3}}} & \frac {2 x^{2}}{\left (x^{3}\right )^{\frac {1}{3}}} \end {vmatrix} \] Therefore \[ W = \left (\left (x^{3}\right )^{\frac {1}{3}}\right )\left (\frac {2 x^{2}}{\left (x^{3}\right )^{\frac {1}{3}}}\right ) - \left (\left (x^{3}\right )^{\frac {2}{3}}\right )\left (\frac {x^{2}}{\left (x^{3}\right )^{\frac {2}{3}}}\right ) \] Which simplifies to \[ W = x^{2} \] Which simplifies to \[ W = x^{2} \] Therefore Eq. (2) becomes \[ u_1 = -\int \frac {\left (x^{3}\right )^{\frac {2}{3}} x^{5} \ln \left (x \right )}{x^{4}}\,dx \] Which simplifies to \[ u_1 = - \int \left (x^{3}\right )^{\frac {2}{3}} x \ln \left (x \right )d x \] Hence \[ u_1 = -\frac {\left (x^{3}\right )^{\frac {2}{3}} x^{2} \ln \left (x \right )}{4}+\frac {\left (x^{3}\right )^{\frac {2}{3}} x^{2}}{16} \] And Eq. (3) becomes \[ u_2 = \int \frac {\left (x^{3}\right )^{\frac {1}{3}} x^{5} \ln \left (x \right )}{x^{4}}\,dx \] Which simplifies to \[ u_2 = \int \left (x^{3}\right )^{\frac {1}{3}} x \ln \left (x \right )d x \] Hence \[ u_2 = \frac {\left (x^{3}\right )^{\frac {1}{3}} x^{2} \ln \left (x \right )}{3}-\frac {\left (x^{3}\right )^{\frac {1}{3}} x^{2}}{9} \] Which simplifies to \begin{align*} u_1 &= -\frac {\left (x^{3}\right )^{\frac {2}{3}} x^{2} \left (-1+4 \ln \left (x \right )\right )}{16} \\ u_2 &= \frac {\left (x^{3}\right )^{\frac {1}{3}} x^{2} \left (-1+3 \ln \left (x \right )\right )}{9} \\ \end{align*} Therefore the particular solution, from equation (1) is \[ y_p(x) = -\frac {x^{5} \left (-1+4 \ln \left (x \right )\right )}{16}+\frac {x^{5} \left (-1+3 \ln \left (x \right )\right )}{9} \] Which simplifies to \[ y_p(x) = \frac {x^{5} \left (-7+12 \ln \left (x \right )\right )}{144} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (\left (-\frac {7 x^{4}}{144}+\frac {\ln \left (x \right ) x^{4}}{12}+c_{1} x +c_{2} \right ) x\right ) + \left (\frac {x^{5} \left (-7+12 \ln \left (x \right )\right )}{144}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= \left (-\frac {7 x^{4}}{144}+\frac {\ln \left (x \right ) x^{4}}{12}+c_{1} x +c_{2} \right ) x +\frac {x^{5} \left (-7+12 \ln \left (x \right )\right )}{144} \\ \end{align*}
Verification of solutions
\[ y = \left (-\frac {7 x^{4}}{144}+\frac {\ln \left (x \right ) x^{4}}{12}+c_{1} x +c_{2} \right ) x +\frac {x^{5} \left (-7+12 \ln \left (x \right )\right )}{144} \] 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 \(A=x^{2}, B=-2 x, C=2, f(x)=x^{5} \ln \left (x \right )\). 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 \[ x^{2} y^{\prime \prime }-2 y^{\prime } x +2 y = 0 \] In normal form the ode \begin {align*} x^{2} y^{\prime \prime }-2 y^{\prime } x +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 {2}{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 {2 n}{x^{2}}+\frac {2}{x^{2}}&=0 \tag {5} \end {align*}
Solving (5) for \(n\) gives \begin {align*} n&=2 \tag {6} \end {align*}
Substituting this value in (3) gives \begin {align*} v^{\prime \prime }\left (x \right )+\frac {2 v^{\prime }\left (x \right )}{x}&=0 \\ v^{\prime \prime }\left (x \right )+\frac {2 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 {2 u \left (x \right )}{x} = 0 \tag {8} \\ \end {align*}
The above is now solved for \(u \left (x \right )\). In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= -\frac {2 u}{x} \end {align*}
Where \(f(x)=-\frac {2}{x}\) and \(g(u)=u\). Integrating both sides gives \begin {align*} \frac {1}{u} \,du &= -\frac {2}{x} \,d x\\ \int { \frac {1}{u} \,du} &= \int {-\frac {2}{x} \,d x}\\ \ln \left (u \right )&=-2 \ln \left (x \right )+c_{1}\\ u&={\mathrm e}^{-2 \ln \left (x \right )+c_{1}}\\ &=\frac {c_{1}}{x^{2}} \end {align*}
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 {c_{1}}{x}+c_{2} \end {align*}
Hence \begin {align*} y&= v \left (x \right ) x^{n}\\ &= \left (-\frac {c_{1}}{x}+c_{2} \right ) x^{2}\\ &= \left (c_{2} x -c_{1} \right ) x\\ \end {align*}
Now the particular solution to this ODE is found \[ x^{2} y^{\prime \prime }-2 y^{\prime } x +2 y = x^{5} \ln \left (x \right ) \] 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 &= x \\ y_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 \[ W = \begin {vmatrix} x & x^{2} \\ \frac {d}{dx}\left (x\right ) & \frac {d}{dx}\left (x^{2}\right ) \end {vmatrix} \] Which gives \[ W = \begin {vmatrix} x & x^{2} \\ 1 & 2 x \end {vmatrix} \] Therefore \[ W = \left (x\right )\left (2 x\right ) - \left (x^{2}\right )\left (1\right ) \] Which simplifies to \[ W = x^{2} \] Which simplifies to \[ W = x^{2} \] Therefore Eq. (2) becomes \[ u_1 = -\int \frac {x^{7} \ln \left (x \right )}{x^{4}}\,dx \] Which simplifies to \[ u_1 = - \int x^{3} \ln \left (x \right )d x \] Hence \[ u_1 = -\frac {\ln \left (x \right ) x^{4}}{4}+\frac {x^{4}}{16} \] And Eq. (3) becomes \[ u_2 = \int \frac {x^{6} \ln \left (x \right )}{x^{4}}\,dx \] Which simplifies to \[ u_2 = \int \ln \left (x \right ) x^{2}d x \] Hence \[ u_2 = \frac {x^{3} \ln \left (x \right )}{3}-\frac {x^{3}}{9} \] Which simplifies to \begin{align*} u_1 &= -\frac {x^{4} \left (-1+4 \ln \left (x \right )\right )}{16} \\ u_2 &= \frac {x^{3} \left (-1+3 \ln \left (x \right )\right )}{9} \\ \end{align*} Therefore the particular solution, from equation (1) is \[ y_p(x) = -\frac {x^{5} \left (-1+4 \ln \left (x \right )\right )}{16}+\frac {x^{5} \left (-1+3 \ln \left (x \right )\right )}{9} \] Which simplifies to \[ y_p(x) = \frac {x^{5} \left (-7+12 \ln \left (x \right )\right )}{144} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (\left (-\frac {c_{1}}{x}+c_{2} \right ) x^{2}\right ) + \left (\frac {x^{5} \left (-7+12 \ln \left (x \right )\right )}{144}\right ) \\ &= \frac {x^{5} \left (-7+12 \ln \left (x \right )\right )}{144}+\left (-\frac {c_{1}}{x}+c_{2} \right ) x^{2} \\ \end{align*} Which simplifies to \[ y = \frac {x^{5} \left (-7+12 \ln \left (x \right )\right )}{144}+\left (-\frac {c_{1}}{x}+c_{2} \right ) x^{2} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {x^{5} \left (-7+12 \ln \left (x \right )\right )}{144}+\left (-\frac {c_{1}}{x}+c_{2} \right ) x^{2} \\ \end{align*}
Verification of solutions
\[ y = \frac {x^{5} \left (-7+12 \ln \left (x \right )\right )}{144}+\left (-\frac {c_{1}}{x}+c_{2} \right ) x^{2} \] Verified OK.
Given an ode of the form \begin {align*} A y^{\prime \prime } + B y^{\prime } + C y &= F(x) \end {align*}
This method reduces the order ode the ODE by one by applying the transformation \begin {align*} y&= B v \end {align*}
This results in \begin {align*} y' &=B' v+ v' B \\ y'' &=B'' v+ B' v' +v'' B + v' B' \\ &=v'' B+2 v'+ B'+B'' v \end {align*}
And now the original ode becomes \begin {align*} A\left ( v'' B+2v'B'+ B'' v\right )+B\left ( B'v+ v' B\right ) +CBv & =0\\ ABv'' +\left ( 2AB'+B^{2}\right ) v'+\left (AB''+BB'+CB\right ) v & =0 \tag {1} \end {align*}
If the term \(AB''+BB'+CB\) is zero, then this method works and can be used to solve \[ ABv''+\left ( 2AB' +B^{2}\right ) v'=0 \] By Using \(u=v'\) which reduces the order of the above ode to one. The new ode is \[ ABu'+\left ( 2AB'+B^{2}\right ) u=0 \] The above ode is first order ode which is solved for \(u\). Now a new ode \(v'=u\) is solved for \(v\) as first order ode. Then the final solution is obtain from \(y=Bv\).
This method works only if the term \(AB''+BB'+CB\) is zero. The given ODE shows that \begin {align*} A &= x^{2}\\ B &= -2 x\\ C &= 2\\ F &= x^{5} \ln \left (x \right ) \end {align*}
The above shows that for this ode \begin {align*} AB''+BB'+CB &= \left (x^{2}\right ) \left (0\right ) + \left (-2 x\right ) \left (-2\right ) + \left (2\right ) \left (-2 x\right ) \\ &=0 \end {align*}
Hence the ode in \(v\) given in (1) now simplifies to \begin {align*} -2 x^{3} v'' +\left ( 0\right ) v' & =0 \end {align*}
Now by applying \(v'=u\) the above becomes \begin {align*} -2 x^{3} u^{\prime }\left (x \right ) = 0 \end {align*}
Which is now solved for \(u\). Integrating both sides gives \begin {align*} u \left (x \right ) &= \int { 0\,\mathop {\mathrm {d}x}}\\ &= c_{1} \end {align*}
The ode for \(v\) now becomes \begin {align*} v' &= u\\ &=c_{1} \end {align*}
Which is now solved for \(v\). Integrating both sides gives \begin {align*} v \left (x \right ) &= \int { c_{1}\,\mathop {\mathrm {d}x}}\\ &= c_{1} x +c_{2} \end {align*}
Therefore the homogeneous solution is \begin {align*} y_h(x) &= B v\\ &= \left (-2 x\right ) \left (c_{1} x +c_{2}\right ) \\ &= -2 x \left (c_{1} x +c_{2} \right ) \end {align*}
And now the particular solution \(y_p(x)\) will be found. 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 \\ y_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 \[ W = \begin {vmatrix} x & x^{2} \\ \frac {d}{dx}\left (x\right ) & \frac {d}{dx}\left (x^{2}\right ) \end {vmatrix} \] Which gives \[ W = \begin {vmatrix} x & x^{2} \\ 1 & 2 x \end {vmatrix} \] Therefore \[ W = \left (x\right )\left (2 x\right ) - \left (x^{2}\right )\left (1\right ) \] Which simplifies to \[ W = x^{2} \] Which simplifies to \[ W = x^{2} \] Therefore Eq. (2) becomes \[ u_1 = -\int \frac {x^{7} \ln \left (x \right )}{x^{4}}\,dx \] Which simplifies to \[ u_1 = - \int x^{3} \ln \left (x \right )d x \] Hence \[ u_1 = -\frac {\ln \left (x \right ) x^{4}}{4}+\frac {x^{4}}{16} \] And Eq. (3) becomes \[ u_2 = \int \frac {x^{6} \ln \left (x \right )}{x^{4}}\,dx \] Which simplifies to \[ u_2 = \int \ln \left (x \right ) x^{2}d x \] Hence \[ u_2 = \frac {x^{3} \ln \left (x \right )}{3}-\frac {x^{3}}{9} \] Which simplifies to \begin{align*} u_1 &= -\frac {x^{4} \left (-1+4 \ln \left (x \right )\right )}{16} \\ u_2 &= \frac {x^{3} \left (-1+3 \ln \left (x \right )\right )}{9} \\ \end{align*} Therefore the particular solution, from equation (1) is \[ y_p(x) = -\frac {x^{5} \left (-1+4 \ln \left (x \right )\right )}{16}+\frac {x^{5} \left (-1+3 \ln \left (x \right )\right )}{9} \] Which simplifies to \[ y_p(x) = \frac {x^{5} \left (-7+12 \ln \left (x \right )\right )}{144} \] Hence the complete solution is \begin {align*} y(x) &= y_h + y_p \\ &= \left (-2 x \left (c_{1} x +c_{2} \right )\right ) + \left (\frac {x^{5} \left (-7+12 \ln \left (x \right )\right )}{144}\right )\\ &= \frac {x^{5} \ln \left (x \right )}{12}-\frac {7 x^{5}}{144}-2 c_{1} x^{2}-2 c_{2} x \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {x^{5} \ln \left (x \right )}{12}-\frac {7 x^{5}}{144}-2 c_{1} x^{2}-2 c_{2} x \\ \end{align*}
Verification of solutions
\[ y = \frac {x^{5} \ln \left (x \right )}{12}-\frac {7 x^{5}}{144}-2 c_{1} x^{2}-2 c_{2} x \] Verified OK.
Writing the ode as \begin {align*} x^{2} y^{\prime \prime }-2 y^{\prime } x +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 &= -2 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 {0}{1}\tag {6} \end {align*}
Comparing the above to (5) shows that \begin {align*} s &= 0\\ t &= 1 \end {align*}
Therefore eq. (4) becomes \begin {align*} z''(x) &= 0 \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) \\ &= 0 - -\infty \\ &= \infty \end {align*}
There are no poles in \(r\). Therefore the set of poles \(\Gamma \) is empty. Since there is no odd order pole larger than \(2\) and the order at \(\infty \) is \(infinity\) then the necessary conditions for case one are met. Therefore \begin {align*} L &= [1] \end {align*}
Since \(r = 0\) is not a function of \(x\), then there is no need run Kovacic algorithm to obtain a solution for transformed ode \(z''=r z\) as one solution is \[ z_1(x) = 1 \] Using the above, the solution for the original ode can now be found. 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 x}{x^{2}} \,dx} \\ &= z_1 e^{\ln \left (x \right )} \\ &= z_1 \left (x\right ) \\ \end{align*} Which simplifies to \[ y_1 = x \] 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 x}{x^{2}} \,dx}}{\left (y_1\right )^2} \,dx \\ &= y_1 \int \frac { e^{2 \ln \left (x \right )}}{\left (y_1\right )^2} \,dx \\ &= y_1 \left (x\right ) \\ \end{align*} Therefore the solution is
\begin{align*} y &= c_{1} y_1 + c_{2} y_2 \\ &= c_{1} \left (x\right ) + c_{2} \left (x\left (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 \[ x^{2} y^{\prime \prime }-2 y^{\prime } x +2 y = 0 \] The homogeneous solution is found using the Kovacic algorithm which results in \[ y_h = c_{2} x^{2}+c_{1} x \] 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 &= x \\ y_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 \[ W = \begin {vmatrix} x & x^{2} \\ \frac {d}{dx}\left (x\right ) & \frac {d}{dx}\left (x^{2}\right ) \end {vmatrix} \] Which gives \[ W = \begin {vmatrix} x & x^{2} \\ 1 & 2 x \end {vmatrix} \] Therefore \[ W = \left (x\right )\left (2 x\right ) - \left (x^{2}\right )\left (1\right ) \] Which simplifies to \[ W = x^{2} \] Which simplifies to \[ W = x^{2} \] Therefore Eq. (2) becomes \[ u_1 = -\int \frac {x^{7} \ln \left (x \right )}{x^{4}}\,dx \] Which simplifies to \[ u_1 = - \int x^{3} \ln \left (x \right )d x \] Hence \[ u_1 = -\frac {\ln \left (x \right ) x^{4}}{4}+\frac {x^{4}}{16} \] And Eq. (3) becomes \[ u_2 = \int \frac {x^{6} \ln \left (x \right )}{x^{4}}\,dx \] Which simplifies to \[ u_2 = \int \ln \left (x \right ) x^{2}d x \] Hence \[ u_2 = \frac {x^{3} \ln \left (x \right )}{3}-\frac {x^{3}}{9} \] Which simplifies to \begin{align*} u_1 &= -\frac {x^{4} \left (-1+4 \ln \left (x \right )\right )}{16} \\ u_2 &= \frac {x^{3} \left (-1+3 \ln \left (x \right )\right )}{9} \\ \end{align*} Therefore the particular solution, from equation (1) is \[ y_p(x) = -\frac {x^{5} \left (-1+4 \ln \left (x \right )\right )}{16}+\frac {x^{5} \left (-1+3 \ln \left (x \right )\right )}{9} \] Which simplifies to \[ y_p(x) = \frac {x^{5} \left (-7+12 \ln \left (x \right )\right )}{144} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{2} x^{2}+c_{1} x\right ) + \left (\frac {x^{5} \left (-7+12 \ln \left (x \right )\right )}{144}\right ) \\ \end{align*} Which simplifies to \[ y = x \left (c_{2} x +c_{1} \right )+\frac {x^{5} \left (-7+12 \ln \left (x \right )\right )}{144} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= x \left (c_{2} x +c_{1} \right )+\frac {x^{5} \left (-7+12 \ln \left (x \right )\right )}{144} \\ \end{align*}
Verification of solutions
\[ y = x \left (c_{2} x +c_{1} \right )+\frac {x^{5} \left (-7+12 \ln \left (x \right )\right )}{144} \] 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 <- LODE of Euler type successful <- solving first the homogeneous part of the ODE successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 25
dsolve(x^2*diff(diff(y(x),x),x)-2*x*diff(y(x),x)+2*y(x)-x^5*ln(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = \frac {x^{5} \ln \left (x \right )}{12}-\frac {7 x^{5}}{144}+c_{2} x^{2}+c_{1} x \]
✓ Solution by Mathematica
Time used: 0.02 (sec). Leaf size: 32
DSolve[-(x^5*Log[x]) + 2*y[x] - 2*x*y'[x] + x^2*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to -\frac {7 x^5}{144}+\frac {1}{12} x^5 \log (x)+c_2 x^2+c_1 x \]