Internal problem ID [9637]
Internal file name [OUTPUT/8579_Monday_June_06_2022_04_14_24_AM_20261698/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1310.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "kovacic", "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_2"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {x^{3} y^{\prime \prime }+3 x^{2} y^{\prime }+x y=1} \]
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 \[ x^{3} y^{\prime \prime }+3 x^{2} y^{\prime }+x y = 0 \] In normal form the ode \begin {align*} x^{3} y^{\prime \prime }+3 x^{2} y^{\prime }+x 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 {1}{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 {1}{x^{2}}}{\frac {1}{x^{6}}}\\ &= 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 )+x^{4} y \left (\tau \right )&=0 \\ \end {align*}
But in terms of \(\tau \) \begin {align*} x^{4}&=\frac {1}{4 \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 )}{4 \tau ^{2}}&=0 \end {align*}
The above ode is now solved for \(y \left (\tau \right )\). The ode can be written as \[ 4 \left (\frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )\right ) \tau ^{2}+y \left (\tau \right ) = 0 \] 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 \[ 4 \tau ^{2}(r(r-1))\tau ^{r-2}+0 r \tau ^{r-1}+\tau ^{r} = 0 \] Simplifying gives \[ 4 r \left (r -1\right )\tau ^{r}+0\,\tau ^{r}+\tau ^{r} = 0 \] Since \(\tau ^{r}\neq 0\) then dividing throughout by \(\tau ^{r}\) gives \[ 4 r \left (r -1\right )+0+1 = 0 \] Or \[ 4 r^{2}-4 r +1 = 0 \tag {1} \] 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}}\\ r_2 &= {\frac {1}{2}} \end {align*}
Since the roots are equal, then the general solution is \[ y \left (\tau \right )= c_{1} y_1 + c_{2} y_2 \] Where \(y_1 = \tau ^{r}\) and \(y_2 = \tau ^{r} \ln \left (\tau \right )\). Hence \[ y \left (\tau \right ) = c_{1} \sqrt {\tau }+c_{2} \sqrt {\tau }\, \ln \left (\tau \right ) \] The above solution is now transformed back to \(y\) using (6) which results in \begin {align*} y &= \frac {\sqrt {2}\, \sqrt {-\frac {1}{x^{2}}}\, \left (c_{1} -c_{2} \ln \left (2\right )+c_{2} \ln \left (-\frac {1}{x^{2}}\right )\right )}{2} \end {align*}
Therefore the homogeneous solution \(y_h\) is \[ y_h = \frac {\sqrt {2}\, \sqrt {-\frac {1}{x^{2}}}\, \left (c_{1} -c_{2} \ln \left (2\right )+c_{2} \ln \left (-\frac {1}{x^{2}}\right )\right )}{2} \] 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 &= \sqrt {-\frac {1}{x^{2}}} \\ y_2 &= -\frac {\sqrt {2}\, \sqrt {-\frac {1}{x^{2}}}\, \ln \left (2\right )}{2}+\frac {\sqrt {2}\, \sqrt {-\frac {1}{x^{2}}}\, \ln \left (-\frac {1}{x^{2}}\right )}{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 \[ W = \begin {vmatrix} \sqrt {-\frac {1}{x^{2}}} & -\frac {\sqrt {2}\, \sqrt {-\frac {1}{x^{2}}}\, \ln \left (2\right )}{2}+\frac {\sqrt {2}\, \sqrt {-\frac {1}{x^{2}}}\, \ln \left (-\frac {1}{x^{2}}\right )}{2} \\ \frac {d}{dx}\left (\sqrt {-\frac {1}{x^{2}}}\right ) & \frac {d}{dx}\left (-\frac {\sqrt {2}\, \sqrt {-\frac {1}{x^{2}}}\, \ln \left (2\right )}{2}+\frac {\sqrt {2}\, \sqrt {-\frac {1}{x^{2}}}\, \ln \left (-\frac {1}{x^{2}}\right )}{2}\right ) \end {vmatrix} \] Which gives \[ W = \begin {vmatrix} \sqrt {-\frac {1}{x^{2}}} & -\frac {\sqrt {2}\, \sqrt {-\frac {1}{x^{2}}}\, \ln \left (2\right )}{2}+\frac {\sqrt {2}\, \sqrt {-\frac {1}{x^{2}}}\, \ln \left (-\frac {1}{x^{2}}\right )}{2} \\ \frac {1}{\sqrt {-\frac {1}{x^{2}}}\, x^{3}} & -\frac {\sqrt {2}\, \ln \left (2\right )}{2 \sqrt {-\frac {1}{x^{2}}}\, x^{3}}+\frac {\sqrt {2}\, \ln \left (-\frac {1}{x^{2}}\right )}{2 \sqrt {-\frac {1}{x^{2}}}\, x^{3}}-\frac {\sqrt {2}\, \sqrt {-\frac {1}{x^{2}}}}{x} \end {vmatrix} \] Therefore \[ W = \left (\sqrt {-\frac {1}{x^{2}}}\right )\left (-\frac {\sqrt {2}\, \ln \left (2\right )}{2 \sqrt {-\frac {1}{x^{2}}}\, x^{3}}+\frac {\sqrt {2}\, \ln \left (-\frac {1}{x^{2}}\right )}{2 \sqrt {-\frac {1}{x^{2}}}\, x^{3}}-\frac {\sqrt {2}\, \sqrt {-\frac {1}{x^{2}}}}{x}\right ) - \left (-\frac {\sqrt {2}\, \sqrt {-\frac {1}{x^{2}}}\, \ln \left (2\right )}{2}+\frac {\sqrt {2}\, \sqrt {-\frac {1}{x^{2}}}\, \ln \left (-\frac {1}{x^{2}}\right )}{2}\right )\left (\frac {1}{\sqrt {-\frac {1}{x^{2}}}\, x^{3}}\right ) \] Which simplifies to \[ W = \frac {\sqrt {2}}{x^{3}} \] Which simplifies to \[ W = \frac {\sqrt {2}}{x^{3}} \] Therefore Eq. (2) becomes \[ u_1 = -\int \frac {-\frac {\sqrt {2}\, \sqrt {-\frac {1}{x^{2}}}\, \ln \left (2\right )}{2}+\frac {\sqrt {2}\, \sqrt {-\frac {1}{x^{2}}}\, \ln \left (-\frac {1}{x^{2}}\right )}{2}}{\sqrt {2}}\,dx \] Which simplifies to \[ u_1 = - \int -\frac {\sqrt {-\frac {1}{x^{2}}}\, \left (\ln \left (2\right )-\ln \left (-\frac {1}{x^{2}}\right )\right )}{2}d x \] Hence \[ u_1 = \frac {x \sqrt {-\frac {1}{x^{2}}}\, \ln \left (2\right ) \ln \left (x \right )}{2}+\frac {x \sqrt {-\frac {1}{x^{2}}}\, \ln \left (-\frac {1}{x^{2}}\right )^{2}}{8} \] And Eq. (3) becomes \[ u_2 = \int \frac {\sqrt {-\frac {1}{x^{2}}}}{\sqrt {2}}\,dx \] Which simplifies to \[ u_2 = \int \frac {\sqrt {2}\, \sqrt {-\frac {1}{x^{2}}}}{2}d x \] Hence \[ u_2 = \frac {\sqrt {2}\, \sqrt {-\frac {1}{x^{2}}}\, x \ln \left (x \right )}{2} \] Which simplifies to \begin{align*} u_1 &= \frac {x \sqrt {-\frac {1}{x^{2}}}\, \left (4 \ln \left (2\right ) \ln \left (x \right )+\ln \left (-\frac {1}{x^{2}}\right )^{2}\right )}{8} \\ u_2 &= \frac {\sqrt {2}\, \sqrt {-\frac {1}{x^{2}}}\, x \ln \left (x \right )}{2} \\ \end{align*} Therefore the particular solution, from equation (1) is \[ y_p(x) = -\frac {4 \ln \left (2\right ) \ln \left (x \right )+\ln \left (-\frac {1}{x^{2}}\right )^{2}}{8 x}+\frac {\sqrt {2}\, \sqrt {-\frac {1}{x^{2}}}\, x \ln \left (x \right ) \left (-\frac {\sqrt {2}\, \sqrt {-\frac {1}{x^{2}}}\, \ln \left (2\right )}{2}+\frac {\sqrt {2}\, \sqrt {-\frac {1}{x^{2}}}\, \ln \left (-\frac {1}{x^{2}}\right )}{2}\right )}{2} \] Which simplifies to \[ y_p(x) = -\frac {\ln \left (-\frac {1}{x^{2}}\right ) \left (4 \ln \left (x \right )+\ln \left (-\frac {1}{x^{2}}\right )\right )}{8 x} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (\frac {\sqrt {2}\, \sqrt {-\frac {1}{x^{2}}}\, \left (c_{1} -c_{2} \ln \left (2\right )+c_{2} \ln \left (-\frac {1}{x^{2}}\right )\right )}{2}\right ) + \left (-\frac {\ln \left (-\frac {1}{x^{2}}\right ) \left (4 \ln \left (x \right )+\ln \left (-\frac {1}{x^{2}}\right )\right )}{8 x}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\sqrt {2}\, \sqrt {-\frac {1}{x^{2}}}\, \left (c_{1} -c_{2} \ln \left (2\right )+c_{2} \ln \left (-\frac {1}{x^{2}}\right )\right )}{2}-\frac {\ln \left (-\frac {1}{x^{2}}\right ) \left (4 \ln \left (x \right )+\ln \left (-\frac {1}{x^{2}}\right )\right )}{8 x} \\ \end{align*}
Verification of solutions
\[ y = \frac {\sqrt {2}\, \sqrt {-\frac {1}{x^{2}}}\, \left (c_{1} -c_{2} \ln \left (2\right )+c_{2} \ln \left (-\frac {1}{x^{2}}\right )\right )}{2}-\frac {\ln \left (-\frac {1}{x^{2}}\right ) \left (4 \ln \left (x \right )+\ln \left (-\frac {1}{x^{2}}\right )\right )}{8 x} \] 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 \(A=x^{3}, B=3 x^{2}, C=x, f(x)=1\). 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 \[ x^{3} y^{\prime \prime }+3 x^{2} y^{\prime }+x y = 0 \] In normal form the ode \begin {align*} x^{3} y^{\prime \prime }+3 x^{2} y^{\prime }+x 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 {1}{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 {\frac {1}{x^{2}}}}{c}\tag {6} \\ \tau '' &= -\frac {1}{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 {1}{c \sqrt {\frac {1}{x^{2}}}\, x^{3}}+\frac {3}{x}\frac {\sqrt {\frac {1}{x^{2}}}}{c}}{\left (\frac {\sqrt {\frac {1}{x^{2}}}}{c}\right )^2} \\ &=2 c \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 )+2 c \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}^{-c \tau } c_{1} \end {align*}
Now from (6) \begin {align*} \tau &= \int \frac {1}{c} \sqrt q \,dx \\ &= \frac {\int \sqrt {\frac {1}{x^{2}}}d x}{c}\\ &= \frac {\sqrt {\frac {1}{x^{2}}}\, x \ln \left (x \right )}{c} \end {align*}
Substituting the above into the solution obtained gives \[ y = \frac {c_{1}}{x} \] Now the particular solution to this ODE is found \[ x^{3} y^{\prime \prime }+3 x^{2} y^{\prime }+x y = 1 \] 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 &= \sqrt {-\frac {1}{x^{2}}} \\ y_2 &= -\frac {\sqrt {2}\, \sqrt {-\frac {1}{x^{2}}}\, \ln \left (2\right )}{2}+\frac {\sqrt {2}\, \sqrt {-\frac {1}{x^{2}}}\, \ln \left (-\frac {1}{x^{2}}\right )}{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 \[ W = \begin {vmatrix} \sqrt {-\frac {1}{x^{2}}} & -\frac {\sqrt {2}\, \sqrt {-\frac {1}{x^{2}}}\, \ln \left (2\right )}{2}+\frac {\sqrt {2}\, \sqrt {-\frac {1}{x^{2}}}\, \ln \left (-\frac {1}{x^{2}}\right )}{2} \\ \frac {d}{dx}\left (\sqrt {-\frac {1}{x^{2}}}\right ) & \frac {d}{dx}\left (-\frac {\sqrt {2}\, \sqrt {-\frac {1}{x^{2}}}\, \ln \left (2\right )}{2}+\frac {\sqrt {2}\, \sqrt {-\frac {1}{x^{2}}}\, \ln \left (-\frac {1}{x^{2}}\right )}{2}\right ) \end {vmatrix} \] Which gives \[ W = \begin {vmatrix} \sqrt {-\frac {1}{x^{2}}} & -\frac {\sqrt {2}\, \sqrt {-\frac {1}{x^{2}}}\, \ln \left (2\right )}{2}+\frac {\sqrt {2}\, \sqrt {-\frac {1}{x^{2}}}\, \ln \left (-\frac {1}{x^{2}}\right )}{2} \\ \frac {1}{\sqrt {-\frac {1}{x^{2}}}\, x^{3}} & -\frac {\sqrt {2}\, \ln \left (2\right )}{2 \sqrt {-\frac {1}{x^{2}}}\, x^{3}}+\frac {\sqrt {2}\, \ln \left (-\frac {1}{x^{2}}\right )}{2 \sqrt {-\frac {1}{x^{2}}}\, x^{3}}-\frac {\sqrt {2}\, \sqrt {-\frac {1}{x^{2}}}}{x} \end {vmatrix} \] Therefore \[ W = \left (\sqrt {-\frac {1}{x^{2}}}\right )\left (-\frac {\sqrt {2}\, \ln \left (2\right )}{2 \sqrt {-\frac {1}{x^{2}}}\, x^{3}}+\frac {\sqrt {2}\, \ln \left (-\frac {1}{x^{2}}\right )}{2 \sqrt {-\frac {1}{x^{2}}}\, x^{3}}-\frac {\sqrt {2}\, \sqrt {-\frac {1}{x^{2}}}}{x}\right ) - \left (-\frac {\sqrt {2}\, \sqrt {-\frac {1}{x^{2}}}\, \ln \left (2\right )}{2}+\frac {\sqrt {2}\, \sqrt {-\frac {1}{x^{2}}}\, \ln \left (-\frac {1}{x^{2}}\right )}{2}\right )\left (\frac {1}{\sqrt {-\frac {1}{x^{2}}}\, x^{3}}\right ) \] Which simplifies to \[ W = \frac {\sqrt {2}}{x^{3}} \] Which simplifies to \[ W = \frac {\sqrt {2}}{x^{3}} \] Therefore Eq. (2) becomes \[ u_1 = -\int \frac {-\frac {\sqrt {2}\, \sqrt {-\frac {1}{x^{2}}}\, \ln \left (2\right )}{2}+\frac {\sqrt {2}\, \sqrt {-\frac {1}{x^{2}}}\, \ln \left (-\frac {1}{x^{2}}\right )}{2}}{\sqrt {2}}\,dx \] Which simplifies to \[ u_1 = - \int -\frac {\sqrt {-\frac {1}{x^{2}}}\, \left (\ln \left (2\right )-\ln \left (-\frac {1}{x^{2}}\right )\right )}{2}d x \] Hence \[ u_1 = \frac {x \sqrt {-\frac {1}{x^{2}}}\, \ln \left (2\right ) \ln \left (x \right )}{2}+\frac {x \sqrt {-\frac {1}{x^{2}}}\, \ln \left (-\frac {1}{x^{2}}\right )^{2}}{8} \] And Eq. (3) becomes \[ u_2 = \int \frac {\sqrt {-\frac {1}{x^{2}}}}{\sqrt {2}}\,dx \] Which simplifies to \[ u_2 = \int \frac {\sqrt {2}\, \sqrt {-\frac {1}{x^{2}}}}{2}d x \] Hence \[ u_2 = \frac {\sqrt {2}\, \sqrt {-\frac {1}{x^{2}}}\, x \ln \left (x \right )}{2} \] Which simplifies to \begin{align*} u_1 &= \frac {x \sqrt {-\frac {1}{x^{2}}}\, \left (4 \ln \left (2\right ) \ln \left (x \right )+\ln \left (-\frac {1}{x^{2}}\right )^{2}\right )}{8} \\ u_2 &= \frac {\sqrt {2}\, \sqrt {-\frac {1}{x^{2}}}\, x \ln \left (x \right )}{2} \\ \end{align*} Therefore the particular solution, from equation (1) is \[ y_p(x) = -\frac {4 \ln \left (2\right ) \ln \left (x \right )+\ln \left (-\frac {1}{x^{2}}\right )^{2}}{8 x}+\frac {\sqrt {2}\, \sqrt {-\frac {1}{x^{2}}}\, x \ln \left (x \right ) \left (-\frac {\sqrt {2}\, \sqrt {-\frac {1}{x^{2}}}\, \ln \left (2\right )}{2}+\frac {\sqrt {2}\, \sqrt {-\frac {1}{x^{2}}}\, \ln \left (-\frac {1}{x^{2}}\right )}{2}\right )}{2} \] Which simplifies to \[ y_p(x) = -\frac {\ln \left (-\frac {1}{x^{2}}\right ) \left (4 \ln \left (x \right )+\ln \left (-\frac {1}{x^{2}}\right )\right )}{8 x} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (\frac {c_{1}}{x}\right ) + \left (-\frac {\ln \left (-\frac {1}{x^{2}}\right ) \left (4 \ln \left (x \right )+\ln \left (-\frac {1}{x^{2}}\right )\right )}{8 x}\right ) \\ &= -\frac {\ln \left (-\frac {1}{x^{2}}\right ) \left (4 \ln \left (x \right )+\ln \left (-\frac {1}{x^{2}}\right )\right )}{8 x}+\frac {c_{1}}{x} \\ \end{align*} Which simplifies to \[ y = \frac {-\frac {\ln \left (-\frac {1}{x^{2}}\right ) \left (4 \ln \left (x \right )+\ln \left (-\frac {1}{x^{2}}\right )\right )}{8}+c_{1}}{x} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {-\frac {\ln \left (-\frac {1}{x^{2}}\right ) \left (4 \ln \left (x \right )+\ln \left (-\frac {1}{x^{2}}\right )\right )}{8}+c_{1}}{x} \\ \end{align*}
Verification of solutions
\[ y = \frac {-\frac {\ln \left (-\frac {1}{x^{2}}\right ) \left (4 \ln \left (x \right )+\ln \left (-\frac {1}{x^{2}}\right )\right )}{8}+c_{1}}{x} \] 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 \(A=x^{3}, B=3 x^{2}, C=x, f(x)=1\). 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 \[ x^{3} y^{\prime \prime }+3 x^{2} y^{\prime }+x y = 0 \] In normal form the ode \begin {align*} x^{3} y^{\prime \prime }+3 x^{2} y^{\prime }+x 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 {1}{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 {1}{x^{2}}&=0 \tag {5} \end {align*}
Solving (5) for \(n\) gives \begin {align*} n&=-1 \tag {6} \end {align*}
Substituting this value in (3) gives \begin {align*} v^{\prime \prime }\left (x \right )+\frac {v^{\prime }\left (x \right )}{x}&=0 \\ v^{\prime \prime }\left (x \right )+\frac {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 {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 {u}{x} \end {align*}
Where \(f(x)=-\frac {1}{x}\) and \(g(u)=u\). Integrating both sides gives \begin {align*} \frac {1}{u} \,du &= -\frac {1}{x} \,d x\\ \int { \frac {1}{u} \,du} &= \int {-\frac {1}{x} \,d x}\\ \ln \left (u \right )&=-\ln \left (x \right )+c_{1}\\ u&={\mathrm e}^{-\ln \left (x \right )+c_{1}}\\ &=\frac {c_{1}}{x} \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}\\ &= c_{1} \ln \left (x \right )+c_{2} \end {align*}
Hence \begin {align*} y&= v \left (x \right ) x^{n}\\ &= \frac {c_{1} \ln \left (x \right )+c_{2}}{x}\\ &= \frac {c_{1} \ln \left (x \right )+c_{2}}{x}\\ \end {align*}
Now the particular solution to this ODE is found \[ x^{3} y^{\prime \prime }+3 x^{2} y^{\prime }+x y = 1 \] 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 {1}{x} \\ y_2 &= \frac {\ln \left (x \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 \[ W = \begin {vmatrix} \frac {1}{x} & \frac {\ln \left (x \right )}{x} \\ \frac {d}{dx}\left (\frac {1}{x}\right ) & \frac {d}{dx}\left (\frac {\ln \left (x \right )}{x}\right ) \end {vmatrix} \] Which gives \[ W = \begin {vmatrix} \frac {1}{x} & \frac {\ln \left (x \right )}{x} \\ -\frac {1}{x^{2}} & \frac {1}{x^{2}}-\frac {\ln \left (x \right )}{x^{2}} \end {vmatrix} \] Therefore \[ W = \left (\frac {1}{x}\right )\left (\frac {1}{x^{2}}-\frac {\ln \left (x \right )}{x^{2}}\right ) - \left (\frac {\ln \left (x \right )}{x}\right )\left (-\frac {1}{x^{2}}\right ) \] Which simplifies to \[ W = \frac {1}{x^{3}} \] Which simplifies to \[ W = \frac {1}{x^{3}} \] Therefore Eq. (2) becomes \[ u_1 = -\int \frac {\frac {\ln \left (x \right )}{x}}{1}\,dx \] Which simplifies to \[ u_1 = - \int \frac {\ln \left (x \right )}{x}d x \] Hence \[ u_1 = -\frac {\ln \left (x \right )^{2}}{2} \] And Eq. (3) becomes \[ u_2 = \int \frac {\frac {1}{x}}{1}\,dx \] Which simplifies to \[ u_2 = \int \frac {1}{x}d x \] Hence \[ u_2 = \ln \left (x \right ) \] Therefore the particular solution, from equation (1) is \[ y_p(x) = \frac {\ln \left (x \right )^{2}}{2 x} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (\frac {c_{1} \ln \left (x \right )+c_{2}}{x}\right ) + \left (\frac {\ln \left (x \right )^{2}}{2 x}\right ) \\ &= \frac {\ln \left (x \right )^{2}}{2 x}+\frac {c_{1} \ln \left (x \right )+c_{2}}{x} \\ \end{align*} Which simplifies to \[ y = \frac {\frac {\ln \left (x \right )^{2}}{2}+c_{1} \ln \left (x \right )+c_{2}}{x} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\frac {\ln \left (x \right )^{2}}{2}+c_{1} \ln \left (x \right )+c_{2}}{x} \\ \end{align*}
Verification of solutions
\[ y = \frac {\frac {\ln \left (x \right )^{2}}{2}+c_{1} \ln \left (x \right )+c_{2}}{x} \] Verified OK.
Writing the ode as \begin {align*} x^{3} y^{\prime \prime }+3 x^{2} y^{\prime }+x 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^{3} \\ B &= 3 x^{2}\tag {3} \\ C &= x \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 {-1}{4 x^{2}}\tag {6} \end {align*}
Comparing the above to (5) shows that \begin {align*} s &= -1\\ t &= 4 x^{2} \end {align*}
Therefore eq. (4) becomes \begin {align*} z''(x) &= \left ( -\frac {1}{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 |
\(\left \{ 1,2\right \} \) |
\(\left \{ 2,3,4,5,6,7,\cdots \right \} \) |
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 \[ r = -\frac {1}{4 x^{2}} \] 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 \(b=-{\frac {1}{4}}\). Hence \begin {alignat*} {2} [\sqrt r]_c &= 0 \\ \alpha _c^{+} &= \frac {1}{2} + \sqrt {1+4 b} &&= {\frac {1}{2}}\\ \alpha _c^{-} &= \frac {1}{2} - \sqrt {1+4 b} &&= {\frac {1}{2}} \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 {1}{4 x^{2}} \end {alignat*}
Since the \(\text {gcd}(s,t)=1\). This gives \(b=-{\frac {1}{4}}\). Hence \begin {alignat*} {2} [\sqrt r]_\infty &= 0 \\ \alpha _{\infty }^{+} &= \frac {1}{2} + \sqrt {1+4 b} &&= {\frac {1}{2}}\\ \alpha _{\infty }^{-} &= \frac {1}{2} - \sqrt {1+4 b} &&= {\frac {1}{2}} \end {alignat*}
The following table summarizes the findings so far for poles and for the order of \(r\) at \(\infty \) where \(r\) is \[ r=-\frac {1}{4 x^{2}} \]
| pole \(c\) location | pole order | \([\sqrt r]_c\) | \(\alpha _c^{+}\) | \(\alpha _c^{-}\) |
| \(0\) | \(2\) | \(0\) | \(\frac {1}{2}\) | \(\frac {1}{2}\) |
| Order of \(r\) at \(\infty \) | \([\sqrt r]_\infty \) | \(\alpha _\infty ^{+}\) | \(\alpha _\infty ^{-}\) |
| \(2\) | \(0\) | \(\frac {1}{2}\) | \(\frac {1}{2}\) |
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 \(\alpha _\infty ^{-} = {\frac {1}{2}}\) then \begin {align*} d &= \alpha _\infty ^{-} - \left ( \alpha _{c_1}^{-} \right ) \\ &= {\frac {1}{2}} - \left ( {\frac {1}{2}} \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 {1}{2 x} + (-) \left ( 0 \right ) \\ &= \frac {1}{2 x}\\ &= \frac {1}{2 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 {1}{2 x}\right ) \left (0\right ) + \left ( \left (-\frac {1}{2 x^{2}}\right ) + \left (\frac {1}{2 x}\right )^2 - \left (-\frac {1}{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 {1}{2 x}d x}\\ &= \sqrt {x} \end {align*}
The first solution to the original ode in \(y\) is found from \begin{align*} y_1 &= z_1 e^{ \int -\frac {1}{2} \frac {B}{A} \,dx} \\ &= z_1 e^{ -\int \frac {1}{2} \frac {3 x^{2}}{x^{3}} \,dx} \\ &= z_1 e^{-\frac {3 \ln \left (x \right )}{2}} \\ &= z_1 \left (\frac {1}{x^{\frac {3}{2}}}\right ) \\ \end{align*} Which simplifies to \[ y_1 = \frac {1}{x} \] 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 {3 x^{2}}{x^{3}} \,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 (\ln \left (x \right )\right ) \\ \end{align*} Therefore the solution is
\begin{align*} y &= c_{1} y_1 + c_{2} y_2 \\ &= c_{1} \left (\frac {1}{x}\right ) + c_{2} \left (\frac {1}{x}\left (\ln \left (x \right )\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 \[ x^{3} y^{\prime \prime }+3 x^{2} y^{\prime }+x y = 0 \] The homogeneous solution is found using the Kovacic algorithm which results in \[ y_h = \frac {c_{1}}{x}+\frac {c_{2} \ln \left (x \right )}{x} \] 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 &= \frac {1}{x} \\ y_2 &= \frac {\ln \left (x \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 \[ W = \begin {vmatrix} \frac {1}{x} & \frac {\ln \left (x \right )}{x} \\ \frac {d}{dx}\left (\frac {1}{x}\right ) & \frac {d}{dx}\left (\frac {\ln \left (x \right )}{x}\right ) \end {vmatrix} \] Which gives \[ W = \begin {vmatrix} \frac {1}{x} & \frac {\ln \left (x \right )}{x} \\ -\frac {1}{x^{2}} & \frac {1}{x^{2}}-\frac {\ln \left (x \right )}{x^{2}} \end {vmatrix} \] Therefore \[ W = \left (\frac {1}{x}\right )\left (\frac {1}{x^{2}}-\frac {\ln \left (x \right )}{x^{2}}\right ) - \left (\frac {\ln \left (x \right )}{x}\right )\left (-\frac {1}{x^{2}}\right ) \] Which simplifies to \[ W = \frac {1}{x^{3}} \] Which simplifies to \[ W = \frac {1}{x^{3}} \] Therefore Eq. (2) becomes \[ u_1 = -\int \frac {\frac {\ln \left (x \right )}{x}}{1}\,dx \] Which simplifies to \[ u_1 = - \int \frac {\ln \left (x \right )}{x}d x \] Hence \[ u_1 = -\frac {\ln \left (x \right )^{2}}{2} \] And Eq. (3) becomes \[ u_2 = \int \frac {\frac {1}{x}}{1}\,dx \] Which simplifies to \[ u_2 = \int \frac {1}{x}d x \] Hence \[ u_2 = \ln \left (x \right ) \] Therefore the particular solution, from equation (1) is \[ y_p(x) = \frac {\ln \left (x \right )^{2}}{2 x} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (\frac {c_{1}}{x}+\frac {c_{2} \ln \left (x \right )}{x}\right ) + \left (\frac {\ln \left (x \right )^{2}}{2 x}\right ) \\ \end{align*} Which simplifies to \[ y = \frac {c_{2} \ln \left (x \right )+c_{1}}{x}+\frac {\ln \left (x \right )^{2}}{2 x} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {c_{2} \ln \left (x \right )+c_{1}}{x}+\frac {\ln \left (x \right )^{2}}{2 x} \\ \end{align*}
Verification of solutions
\[ y = \frac {c_{2} \ln \left (x \right )+c_{1}}{x}+\frac {\ln \left (x \right )^{2}}{2 x} \] 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)] <- double symmetry of the form [xi=0, eta=F(x)] successful`
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 20
dsolve(x^3*diff(diff(y(x),x),x)+3*x^2*diff(y(x),x)+x*y(x)-1=0,y(x), singsol=all)
\[ y \left (x \right ) = \frac {c_{1} \ln \left (x \right )+\frac {\ln \left (x \right )^{2}}{2}+c_{2}}{x} \]
✓ Solution by Mathematica
Time used: 0.021 (sec). Leaf size: 27
DSolve[-1 + x*y[x] + 3*x^2*y'[x] + x^3*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {\log ^2(x)+2 c_2 \log (x)+2 c_1}{2 x} \]