2.2.19 problem 20
Internal
problem
ID
[8818]
Book
:
Second
order
enumerated
odes
Section
:
section
2
Problem
number
:
20
Date
solved
:
Thursday, December 12, 2024 at 09:52:03 AM
CAS
classification
:
[[_2nd_order, _with_linear_symmetries]]
Solve
\begin{align*} x^{2} y^{\prime \prime }-3 x y^{\prime }+3 y&=2 x^{3}-x^{2} \end{align*}
Solved as second order Euler type ode
Time used: 0.178 (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 \(A=x^{2}, B=-3 x, C=3, f(x)=2 x^{3}-x^{2}\). Let the
solution be
\[ y = y_h + y_p \]
Where \(y_h\) is the solution to the homogeneous ODE \( A y''(x) + B y'(x) + C y(x) = 0\), and \(y_p\) is a particular solution
to the non-homogeneous ODE \(A y''(x) + B y'(x) + C y(x) = f(x)\). Solving for \(y_h\) from
\[ x^{2} y^{\prime \prime }-3 x y^{\prime }+3 y = 0 \]
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
\[ x^{2}(r(r-1))x^{r-2}-3 x r x^{r-1}+3 x^{r} = 0 \]
Simplifying gives
\[ r \left (r -1\right )x^{r}-3 r\,x^{r}+3 x^{r} = 0 \]
Since \(x^{r}\neq 0\) then dividing throughout by \(x^{r}\) gives
\[ r \left (r -1\right )-3 r+3 = 0 \]
Or
\[ r^{2}-4 r +3 = 0 \tag {1} \]
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 &= 3 \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
\[ y = c_2 \,x^{3}+c_1 x \]
Next,
we find the particular solution to the ODE
\[ x^{2} y^{\prime \prime }-3 x y^{\prime }+3 y = 2 x^{3}-x^{2} \]
The particular solution \(y_p\) can be found using
either the method of undetermined coefficients, or the method of variation of parameters.
The method of variation of parameters will be used as it is more general and can be used
when the coefficients of the ODE depend on \(x\) as well. Let
\begin{equation}
\tag{1} y_p(x) = u_1 y_1 + u_2 y_2
\end{equation}
Where \(u_1,u_2\) to be determined, and \(y_1,y_2\) are
the two basis solutions (the two linearly independent solutions of the homogeneous ODE)
found earlier when solving the homogeneous ODE as
\begin{align*}
y_1 &= x \\
y_2 &= x^{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
\[ W = \begin {vmatrix} x & x^{3} \\ \frac {d}{dx}\left (x\right ) & \frac {d}{dx}\left (x^{3}\right ) \end {vmatrix} \]
Which gives
\[ W = \begin {vmatrix} x & x^{3} \\ 1 & 3 x^{2} \end {vmatrix} \]
Therefore
\[
W = \left (x\right )\left (3 x^{2}\right ) - \left (x^{3}\right )\left (1\right )
\]
Which
simplifies to
\[
W = 2 x^{3}
\]
Which simplifies to
\[
W = 2 x^{3}
\]
Therefore Eq. (2) becomes
\[
u_1 = -\int \frac {x^{3} \left (2 x^{3}-x^{2}\right )}{2 x^{5}}\,dx
\]
Which simplifies to
\[
u_1 = - \int \left (x -\frac {1}{2}\right )d x
\]
Hence
\[
u_1 = -\frac {1}{2} x^{2}+\frac {1}{2} x
\]
And Eq. (3) becomes
\[
u_2 = \int \frac {x \left (2 x^{3}-x^{2}\right )}{2 x^{5}}\,dx
\]
Which simplifies to
\[
u_2 = \int \frac {2 x -1}{2 x^{2}}d x
\]
Hence
\[
u_2 = \ln \left (x \right )+\frac {1}{2 x}
\]
Therefore the particular
solution, from equation (1) is
\[
y_p(x) = \left (-\frac {1}{2} x^{2}+\frac {1}{2} x \right ) x +x^{3} \left (\ln \left (x \right )+\frac {1}{2 x}\right )
\]
Which simplifies to
\[
y_p(x) = x^{3} \ln \left (x \right )-\frac {x^{3}}{2}+x^{2}
\]
Therefore the general solution is
\begin{align*} y &= y_h + y_p \\ &= c_2 \,x^{3}+c_1 x +x^{3} \ln \left (x \right )-\frac {x^{3}}{2}+x^{2} \end{align*}
Will add steps showing solving for IC soon.
Summary of solutions found
\begin{align*}
y &= c_2 \,x^{3}+c_1 x +x^{3} \ln \left (x \right )-\frac {x^{3}}{2}+x^{2} \\
\end{align*}
Solved as second order ode using change of variable on x method 2
Time used: 0.663 (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
\[
x^{2} y^{\prime \prime }-3 x y^{\prime }+3 y = 0
\]
In normal form the ode
\begin{align*} x^{2} y^{\prime \prime }-3 x y^{\prime }+3 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 {3}{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}^{-\int p \left (x \right )d x}d x\\ &= \int {\mathrm e}^{-\int -\frac {3}{x}d x}d x\\ &= \int e^{3 \ln \left (x \right )} \,dx\\ &= \int x^{3}d x\\ &= \frac {x^{4}}{4}\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 {3}{x^{2}}}{x^{6}}\\ &= \frac {3}{x^{8}}\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 {3 y \left (\tau \right )}{x^{8}}&=0 \\ \end{align*}
But in terms of \(\tau \)
\begin{align*} \frac {3}{x^{8}}&=\frac {3}{16 \tau ^{2}} \end{align*}
Hence the above ode becomes
\begin{align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+\frac {3 y \left (\tau \right )}{16 \tau ^{2}}&=0 \end{align*}
The above ode is now solved for \(y \left (\tau \right )\). Writing the ode as
\begin{align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+\frac {3 y \left (\tau \right )}{16 \tau ^{2}} &= 0 \tag {1} \\ A \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right ) + B \frac {d}{d \tau }y \left (\tau \right ) + C y \left (\tau \right ) &= 0 \tag {2} \end{align*}
Comparing (1) and (2) shows that
\begin{align*} A &= 1 \\ B &= 0\tag {3} \\ C &= \frac {3}{16 \tau ^{2}} \end{align*}
Applying the Liouville transformation on the dependent variable gives
\begin{align*} z(\tau ) &= y \left (\tau \right ) e^{\int \frac {B}{2 A} \,d\tau } \end{align*}
Then (2) becomes
\begin{align*} z''(\tau ) = r z(\tau )\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 {-3}{16 \tau ^{2}}\tag {6} \end{align*}
Comparing the above to (5) shows that
\begin{align*} s &= -3\\ t &= 16 \tau ^{2} \end{align*}
Therefore eq. (4) becomes
\begin{align*} z''(\tau ) &= \left ( -\frac {3}{16 \tau ^{2}}\right ) z(\tau )\tag {7} \end{align*}
Equation (7) is now solved. After finding \(z(\tau )\) then \(y \left (\tau \right )\) is found using the inverse transformation
\begin{align*} y \left (\tau \right ) &= z \left (\tau \right ) e^{-\int \frac {B}{2 A} \,d\tau } \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 2.37: 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 \(t=16 \tau ^{2}\).
There is a pole at \(\tau =0\) 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
\[
r = -\frac {3}{16 \tau ^{2}}
\]
For the pole at \(\tau =0\) let \(b\)
be the coefficient of \(\frac {1}{ \tau ^{2}}\) in the partial fractions decomposition of \(r\) given above. Therefore \(b=-{\frac {3}{16}}\). Hence
\begin{alignat*}{2} [\sqrt r]_c &= 0 \\ \alpha _c^{+} &= \frac {1}{2} + \sqrt {1+4 b} &&= {\frac {3}{4}}\\ \alpha _c^{-} &= \frac {1}{2} - \sqrt {1+4 b} &&= {\frac {1}{4}} \end{alignat*}
Since the order of \(r\) at \(\infty \) is 2 then \([\sqrt r]_\infty =0\). Let \(b\) be the coefficient of \(\frac {1}{\tau ^{2}}\) 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 {3}{16 \tau ^{2}} \end{alignat*}
Since the \(\text {gcd}(s,t)=1\). This gives \(b=-{\frac {3}{16}}\). Hence
\begin{alignat*}{2} [\sqrt r]_\infty &= 0 \\ \alpha _{\infty }^{+} &= \frac {1}{2} + \sqrt {1+4 b} &&= {\frac {3}{4}}\\ \alpha _{\infty }^{-} &= \frac {1}{2} - \sqrt {1+4 b} &&= {\frac {1}{4}} \end{alignat*}
The following table summarizes the findings so far for poles and for the order of \(r\) at \(\infty \) where \(r\)
is
\[ r=-\frac {3}{16 \tau ^{2}} \]
| | | | |
| pole \(c\) location |
pole order |
\([\sqrt r]_c\) |
\(\alpha _c^{+}\) |
\(\alpha _c^{-}\) |
| | | | |
| \(0\) | \(2\) | \(0\) | \(\frac {3}{4}\) | \(\frac {1}{4}\) |
| | | | |
| | | |
| Order of \(r\) at \(\infty \) |
\([\sqrt r]_\infty \) |
\(\alpha _\infty ^{+}\) |
\(\alpha _\infty ^{-}\) |
| | | |
| \(2\) |
\(0\) | \(\frac {3}{4}\) | \(\frac {1}{4}\) |
| | | |
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 \(\alpha _\infty ^{-} = {\frac {1}{4}}\) then
\begin{align*} d &= \alpha _\infty ^{-} - \left ( \alpha _{c_1}^{-} \right ) \\ &= {\frac {1}{4}} - \left ( {\frac {1}{4}} \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)}}{\tau -c} \right ) + s(\infty ) [\sqrt r]_\infty \end{align*}
The above gives
\begin{align*} \omega &= \left ( (-)[\sqrt r]_{c_1} + \frac { \alpha _{c_1}^{-} }{\tau - c_1}\right ) + (-) [\sqrt r]_\infty \\ &= \frac {1}{4 \tau } + (-) \left ( 0 \right ) \\ &= \frac {1}{4 \tau }\\ &= \frac {1}{4 \tau } \end{align*}
Now that \(\omega \) is determined, the next step is find a corresponding minimal polynomial \(p(\tau )\) of degree
\(d=0\) to solve the ode. The polynomial \(p(\tau )\) 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(\tau ) &= 1\tag {2A} \end{align*}
Substituting the above in eq. (1A) gives
\begin{align*} \left (0\right ) + 2 \left (\frac {1}{4 \tau }\right ) \left (0\right ) + \left ( \left (-\frac {1}{4 \tau ^{2}}\right ) + \left (\frac {1}{4 \tau }\right )^2 - \left (-\frac {3}{16 \tau ^{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(\tau ) &= p e^{ \int \omega \,d\tau } \\ &= {\mathrm e}^{\int \frac {1}{4 \tau }d \tau }\\ &= \tau ^{{1}/{4}} \end{align*}
The first solution to the original ode in \(y \left (\tau \right )\) is found from
\[
y_1 = z_1 e^{ \int -\frac {1}{2} \frac {B}{A} \,d\tau }
\]
Since \(B=0\) then the above reduces to
\begin{align*}
y_1 &= z_1 \\
&= \tau ^{{1}/{4}} \\
\end{align*}
Which simplifies to
\[
y_1 = \tau ^{{1}/{4}}
\]
The second solution \(y_2\) to the original
ode is found using reduction of order
\[ y_2 = y_1 \int \frac { e^{\int -\frac {B}{A} \,d\tau }}{y_1^2} \,d\tau \]
Since \(B=0\) then the above becomes
\begin{align*}
y_2 &= y_1 \int \frac {1}{y_1^2} \,d\tau \\
&= \tau ^{{1}/{4}}\int \frac {1}{\sqrt {\tau }} \,d\tau \\
&= \tau ^{{1}/{4}}\left (2 \sqrt {\tau }\right ) \\
\end{align*}
Therefore the solution
is
\begin{align*}
y \left (\tau \right ) &= c_1 y_1 + c_2 y_2 \\
&= c_1 \left (\tau ^{{1}/{4}}\right ) + c_2 \left (\tau ^{{1}/{4}}\left (2 \sqrt {\tau }\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
\[
y = \frac {c_1 4^{{3}/{4}} \left (x^{4}\right )^{{1}/{4}}}{4}+\frac {c_2 4^{{1}/{4}} \left (x^{4}\right )^{{3}/{4}}}{2}
\]
Therefore the
homogeneous solution \(y_h\) is
\[
y_h = \frac {c_1 4^{{3}/{4}} \left (x^{4}\right )^{{1}/{4}}}{4}+\frac {c_2 4^{{1}/{4}} \left (x^{4}\right )^{{3}/{4}}}{2}
\]
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 &= \left (x^{4}\right )^{{1}/{4}} \\
y_2 &= \left (x^{4}\right )^{{3}/{4}} \\
\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
\[ W = \begin {vmatrix} \left (x^{4}\right )^{{1}/{4}} & \left (x^{4}\right )^{{3}/{4}} \\ \frac {d}{dx}\left (\left (x^{4}\right )^{{1}/{4}}\right ) & \frac {d}{dx}\left (\left (x^{4}\right )^{{3}/{4}}\right ) \end {vmatrix} \]
Which gives
\[ W = \begin {vmatrix} \left (x^{4}\right )^{{1}/{4}} & \left (x^{4}\right )^{{3}/{4}} \\ \frac {x^{3}}{\left (x^{4}\right )^{{3}/{4}}} & \frac {3 x^{3}}{\left (x^{4}\right )^{{1}/{4}}} \end {vmatrix} \]
Therefore
\[
W = \left (\left (x^{4}\right )^{{1}/{4}}\right )\left (\frac {3 x^{3}}{\left (x^{4}\right )^{{1}/{4}}}\right ) - \left (\left (x^{4}\right )^{{3}/{4}}\right )\left (\frac {x^{3}}{\left (x^{4}\right )^{{3}/{4}}}\right )
\]
Which simplifies to
\[
W = 2 x^{3}
\]
Which simplifies to
\[
W = 2 x^{3}
\]
Therefore Eq. (2) becomes
\[
u_1 = -\int \frac {\left (x^{4}\right )^{{3}/{4}} \left (2 x^{3}-x^{2}\right )}{2 x^{5}}\,dx
\]
Which simplifies to
\[
u_1 = - \int \frac {\left (x^{4}\right )^{{3}/{4}} \left (2 x -1\right )}{2 x^{3}}d x
\]
Hence
\[
u_1 = -\frac {\left (x -1\right ) \left (x^{4}\right )^{{3}/{4}}}{2 x^{2}}
\]
And Eq. (3) becomes
\[
u_2 = \int \frac {\left (x^{4}\right )^{{1}/{4}} \left (2 x^{3}-x^{2}\right )}{2 x^{5}}\,dx
\]
Which
simplifies to
\[
u_2 = \int \frac {\left (x^{4}\right )^{{1}/{4}} \left (2 x -1\right )}{2 x^{3}}d x
\]
Hence
\[
u_2 = \frac {\left (x^{4}\right )^{{1}/{4}}}{2 x^{2}}+\frac {\left (x^{4}\right )^{{1}/{4}} \ln \left (x \right )}{x}
\]
Therefore the particular solution, from equation (1) is
\[
y_p(x) = -\frac {x^{2} \left (x -1\right )}{2}+\left (x^{4}\right )^{{3}/{4}} \left (\frac {\left (x^{4}\right )^{{1}/{4}}}{2 x^{2}}+\frac {\left (x^{4}\right )^{{1}/{4}} \ln \left (x \right )}{x}\right )
\]
Which
simplifies to
\[
y_p(x) = x^{3} \ln \left (x \right )-\frac {x^{3}}{2}+x^{2}
\]
Therefore the general solution is
\begin{align*}
y &= y_h + y_p \\
&= \left (\frac {c_1 4^{{3}/{4}} \left (x^{4}\right )^{{1}/{4}}}{4}+\frac {c_2 4^{{1}/{4}} \left (x^{4}\right )^{{3}/{4}}}{2}\right ) + \left (x^{3} \ln \left (x \right )-\frac {x^{3}}{2}+x^{2}\right ) \\
\end{align*}
Will add steps showing solving for IC
soon.
Summary of solutions found
\begin{align*}
y &= x^{3} \ln \left (x \right )-\frac {x^{3}}{2}+x^{2}+\frac {c_1 4^{{3}/{4}} \left (x^{4}\right )^{{1}/{4}}}{4}+\frac {c_2 4^{{1}/{4}} \left (x^{4}\right )^{{3}/{4}}}{2} \\
\end{align*}
Solved as second order ode using change of variable on x method 1
Time used: 0.244 (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 \(A=x^{2}, B=-3 x, C=3, f(x)=2 x^{3}-x^{2}\). Let the
solution be
\[ y = y_h + y_p \]
Where \(y_h\) is the solution to the homogeneous ODE \( A y''(x) + B y'(x) + C y(x) = 0\), and \(y_p\) is a particular
solution to the non-homogeneous ODE \(A y''(x) + B y'(x) + C y(x) = f(x)\). Solving for \(y_h\) from
\[ x^{2} y^{\prime \prime }-3 x y^{\prime }+3 y = 0 \]
In normal form the ode
\begin{align*} x^{2} y^{\prime \prime }-3 x y^{\prime }+3 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 {3}{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 {3}\, \sqrt {\frac {1}{x^{2}}}}{c}\tag {6} \\ \tau '' &= -\frac {\sqrt {3}}{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 {3}}{c \sqrt {\frac {1}{x^{2}}}\, x^{3}}-\frac {3}{x}\frac {\sqrt {3}\, \sqrt {\frac {1}{x^{2}}}}{c}}{\left (\frac {\sqrt {3}\, \sqrt {\frac {1}{x^{2}}}}{c}\right )^2} \\ &=-\frac {4 c \sqrt {3}}{3} \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 {4 c \sqrt {3}\, \left (\frac {d}{d \tau }y \left (\tau \right )\right )}{3}+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 {2 \sqrt {3}\, c \tau }{3}} \left (c_1 \cosh \left (\frac {\sqrt {3}\, c \tau }{3}\right )+i c_2 \sinh \left (\frac {\sqrt {3}\, c \tau }{3}\right )\right ) \end{align*}
Now from (6)
\begin{align*} \tau &= \int \frac {1}{c} \sqrt q \,dx \\ &= \frac {\int \sqrt {3}\, \sqrt {\frac {1}{x^{2}}}d x}{c}\\ &= \frac {\sqrt {3}\, \ln \left (x \right )}{c} \end{align*}
Substituting the above into the solution obtained gives
\[
y = x^{2} \left (c_1 \left (\frac {x}{2}+\frac {1}{2 x}\right )+i c_2 \left (\frac {x}{2}-\frac {1}{2 x}\right )\right )
\]
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^{2} \left (\frac {x}{2}+\frac {1}{2 x}\right ) \\
y_2 &= i x^{2} \left (\frac {x}{2}-\frac {1}{2 x}\right ) \\
\end{align*}
In the Variation of parameters \(u_1,u_2\) are found using
\begin{align*}
\tag{2} u_1 &= -\int \frac {y_2 f(x)}{a W(x)} \\
\tag{3} u_2 &= \int \frac {y_1 f(x)}{a W(x)} \\
\end{align*}
Where \(W(x)\) is the
Wronskian and \(a\) is the coefficient in front of \(y''\) in the given ODE. The Wronskian is given
by \(W= \begin {vmatrix} y_1 & y_{2} \\ y_{1}^{\prime } & y_{2}^{\prime } \end {vmatrix} \). Hence
\[ W = \begin {vmatrix} x^{2} \left (\frac {x}{2}+\frac {1}{2 x}\right ) & i x^{2} \left (\frac {x}{2}-\frac {1}{2 x}\right ) \\ \frac {d}{dx}\left (x^{2} \left (\frac {x}{2}+\frac {1}{2 x}\right )\right ) & \frac {d}{dx}\left (i x^{2} \left (\frac {x}{2}-\frac {1}{2 x}\right )\right ) \end {vmatrix} \]
Which gives
\[ W = \begin {vmatrix} x^{2} \left (\frac {x}{2}+\frac {1}{2 x}\right ) & i x^{2} \left (\frac {x}{2}-\frac {1}{2 x}\right ) \\ 2 x \left (\frac {x}{2}+\frac {1}{2 x}\right )+x^{2} \left (\frac {1}{2}-\frac {1}{2 x^{2}}\right ) & 2 i x \left (\frac {x}{2}-\frac {1}{2 x}\right )+i x^{2} \left (\frac {1}{2}+\frac {1}{2 x^{2}}\right ) \end {vmatrix} \]
Therefore
\[
W = \left (x^{2} \left (\frac {x}{2}+\frac {1}{2 x}\right )\right )\left (2 i x \left (\frac {x}{2}-\frac {1}{2 x}\right )+i x^{2} \left (\frac {1}{2}+\frac {1}{2 x^{2}}\right )\right ) - \left (i x^{2} \left (\frac {x}{2}-\frac {1}{2 x}\right )\right )\left (2 x \left (\frac {x}{2}+\frac {1}{2 x}\right )+x^{2} \left (\frac {1}{2}-\frac {1}{2 x^{2}}\right )\right )
\]
Which simplifies to
\[
W = i x^{3}
\]
Which simplifies to
\[
W = i x^{3}
\]
Therefore Eq. (2) becomes
\[
u_1 = -\int \frac {i x^{2} \left (\frac {x}{2}-\frac {1}{2 x}\right ) \left (2 x^{3}-x^{2}\right )}{i x^{5}}\,dx
\]
Which simplifies to
\[
u_1 = - \int \frac {\left (x^{2}-1\right ) \left (2 x -1\right )}{2 x^{2}}d x
\]
Hence
\[
u_1 = -\frac {x^{2}}{2}+\frac {x}{2}+\ln \left (x \right )+\frac {1}{2 x}
\]
And Eq. (3) becomes
\[
u_2 = \int \frac {x^{2} \left (\frac {x}{2}+\frac {1}{2 x}\right ) \left (2 x^{3}-x^{2}\right )}{i x^{5}}\,dx
\]
Which
simplifies to
\[
u_2 = \int -\frac {i \left (x^{2}+1\right ) \left (2 x -1\right )}{2 x^{2}}d x
\]
Hence
\[
u_2 = -\frac {i \left (x^{2}-x +2 \ln \left (x \right )+\frac {1}{x}\right )}{2}
\]
Therefore the particular solution, from equation (1) is
\[
y_p(x) = \left (-\frac {x^{2}}{2}+\frac {x}{2}+\ln \left (x \right )+\frac {1}{2 x}\right ) x^{2} \left (\frac {x}{2}+\frac {1}{2 x}\right )+\frac {x^{2} \left (\frac {x}{2}-\frac {1}{2 x}\right ) \left (x^{2}-x +2 \ln \left (x \right )+\frac {1}{x}\right )}{2}
\]
Which
simplifies to
\[
y_p(x) = x^{3} \ln \left (x \right )-\frac {x^{3}}{2}+x^{2}
\]
Therefore the general solution is
\begin{align*}
y &= y_h + y_p \\
&= \left (x^{2} \left (c_1 \left (\frac {x}{2}+\frac {1}{2 x}\right )+i c_2 \left (\frac {x}{2}-\frac {1}{2 x}\right )\right )\right ) + \left (x^{3} \ln \left (x \right )-\frac {x^{3}}{2}+x^{2}\right ) \\
\end{align*}
Will add steps showing solving for IC
soon.
Summary of solutions found
\begin{align*}
y &= x^{2} \left (c_1 \left (\frac {x}{2}+\frac {1}{2 x}\right )+i c_2 \left (\frac {x}{2}-\frac {1}{2 x}\right )\right )+x^{3} \ln \left (x \right )-\frac {x^{3}}{2}+x^{2} \\
\end{align*}
Solved as second order ode using change of variable on y method 2
Time used: 0.253 (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 \(A=x^{2}, B=-3 x, C=3, f(x)=2 x^{3}-x^{2}\). Let the
solution be
\[
y = y_h + y_p
\]
Where \(y_h\) is the solution to the homogeneous ODE \( A y''(x) + B y'(x) + C y(x) = 0\), and \(y_p\) is a particular
solution to the non-homogeneous ODE \(A y''(x) + B y'(x) + C y(x) = f(x)\). Solving for \(y_h\) from
\[
x^{2} y^{\prime \prime }-3 x y^{\prime }+3 y = 0
\]
In normal form the ode
\begin{align*} x^{2} y^{\prime \prime }-3 x y^{\prime }+3 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 {3}{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 {3}{x^{2}}&=0 \tag {5} \end{align*}
Solving (5) for \(n\) gives
\begin{align*} n&=3 \tag {6} \end{align*}
Substituting this value in (3) gives
\begin{align*} v^{\prime \prime }\left (x \right )+\frac {3 v^{\prime }\left (x \right )}{x}&=0 \\ v^{\prime \prime }\left (x \right )+\frac {3 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 {3 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 {3}{x}\\ p(x) &=0 \end{align*}
The integrating factor \(\mu \) is
\begin{align*} \mu &= e^{\int {q\,dx}}\\ &= {\mathrm e}^{\int \frac {3}{x}d x}\\ &= x^{3} \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^{3}\right ) &= 0 \end{align*}
Integrating gives
\begin{align*} u \,x^{3}&= \int {0 \,dx} + c_1 \\ &=c_1 \end{align*}
Dividing throughout by the integrating factor \(x^{3}\) gives the final solution
\[ u \left (x \right ) = \frac {c_1}{x^{3}} \]
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}{2 x^{2}}+c_2 \end{align*}
Hence
\begin{align*} y&= v \left (x \right ) x^{n}\\ &= \left (-\frac {c_1}{2 x^{2}}+c_2 \right ) x^{3}\\ &= c_2 \,x^{3}-\frac {1}{2} c_1 x\\ \end{align*}
Now the particular solution to this ODE is found
\[
x^{2} y^{\prime \prime }-3 x y^{\prime }+3 y = 2 x^{3}-x^{2}
\]
The particular solution \(y_p\) can be found
using either the method of undetermined coefficients, or the method of variation
of parameters. The method of variation of parameters will be used as it is more
general and can be used when the coefficients of the ODE depend on \(x\) as well.
Let
\begin{equation}
\tag{1} y_p(x) = u_1 y_1 + u_2 y_2
\end{equation}
Where \(u_1,u_2\) to be determined, and \(y_1,y_2\) are the two basis solutions (the two linearly
independent solutions of the homogeneous ODE) found earlier when solving the
homogeneous ODE as
\begin{align*}
y_1 &= x \\
y_2 &= x^{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
\[ W = \begin {vmatrix} x & x^{3} \\ \frac {d}{dx}\left (x\right ) & \frac {d}{dx}\left (x^{3}\right ) \end {vmatrix} \]
Which gives
\[ W = \begin {vmatrix} x & x^{3} \\ 1 & 3 x^{2} \end {vmatrix} \]
Therefore
\[
W = \left (x\right )\left (3 x^{2}\right ) - \left (x^{3}\right )\left (1\right )
\]
Which simplifies to
\[
W = 2 x^{3}
\]
Which simplifies to
\[
W = 2 x^{3}
\]
Therefore Eq. (2) becomes
\[
u_1 = -\int \frac {x^{3} \left (2 x^{3}-x^{2}\right )}{2 x^{5}}\,dx
\]
Which simplifies to
\[
u_1 = - \int \left (x -\frac {1}{2}\right )d x
\]
Hence
\[
u_1 = -\frac {1}{2} x^{2}+\frac {1}{2} x
\]
And Eq. (3) becomes
\[
u_2 = \int \frac {x \left (2 x^{3}-x^{2}\right )}{2 x^{5}}\,dx
\]
Which
simplifies to
\[
u_2 = \int \frac {2 x -1}{2 x^{2}}d x
\]
Hence
\[
u_2 = \ln \left (x \right )+\frac {1}{2 x}
\]
Therefore the particular solution, from equation (1) is
\[
y_p(x) = \left (-\frac {1}{2} x^{2}+\frac {1}{2} x \right ) x +x^{3} \left (\ln \left (x \right )+\frac {1}{2 x}\right )
\]
Which
simplifies to
\[
y_p(x) = x^{3} \ln \left (x \right )-\frac {x^{3}}{2}+x^{2}
\]
Therefore the general solution is
\begin{align*}
y &= y_h + y_p \\
&= \left (\left (-\frac {c_1}{2 x^{2}}+c_2 \right ) x^{3}\right ) + \left (x^{3} \ln \left (x \right )-\frac {x^{3}}{2}+x^{2}\right ) \\
\end{align*}
Will add steps showing solving for IC
soon.
Summary of solutions found
\begin{align*}
y &= \left (-\frac {c_1}{2 x^{2}}+c_2 \right ) x^{3}+x^{3} \ln \left (x \right )-\frac {x^{3}}{2}+x^{2} \\
\end{align*}
Time used: 0.118 (sec)
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 &= -3 x\\ C &= 3\\ F &= 2 x^{3}-x^{2} \end{align*}
The above shows that for this ode
\begin{align*} AB''+BB'+CB &= \left (x^{2}\right ) \left (0\right ) + \left (-3 x\right ) \left (-3\right ) + \left (3\right ) \left (-3 x\right ) \\ &=0 \end{align*}
Hence the ode in \(v\) given in (1) now simplifies to
\begin{align*} -3 x^{3} v'' +\left ( 3 x^{2}\right ) v' & =0 \end{align*}
Now by applying \(v'=u\) the above becomes
\begin{align*} -3 x^{2} \left (u^{\prime }\left (x \right ) x -u \left (x \right )\right ) = 0 \end{align*}
Which is now solved for \(u\). 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}{x}\\ p(x) &=0 \end{align*}
The integrating factor \(\mu \) is
\begin{align*} \mu &= e^{\int {q\,dx}}\\ &= {\mathrm e}^{\int -\frac {1}{x}d x}\\ &= \frac {1}{x} \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 (\frac {u}{x}\right ) &= 0 \end{align*}
Integrating gives
\begin{align*} \frac {u}{x}&= \int {0 \,dx} + c_1 \\ &=c_1 \end{align*}
Dividing throughout by the integrating factor \(\frac {1}{x}\) gives the final solution
\[ u \left (x \right ) = c_1 x \]
The ode for \(v\) now
becomes
\[
v^{\prime }\left (x \right ) = c_1 x
\]
Which is now solved for \(v\). Since the ode has the form \(v^{\prime }\left (x \right )=f(x)\), then we only need to
integrate \(f(x)\).
\begin{align*} \int {dv} &= \int {c_1 x\, dx}\\ v \left (x \right ) &= \frac {c_1 \,x^{2}}{2} + c_2 \end{align*}
Replacing \(v \left (x \right )\) above by \(-\frac {y}{3 x}\), then the homogeneous solution is
\begin{align*} y_h(x) &= B v\\ &= -\frac {3 \left (c_1 \,x^{2}+2 c_2 \right ) x}{2} \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^{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
\[ W = \begin {vmatrix} x & x^{3} \\ \frac {d}{dx}\left (x\right ) & \frac {d}{dx}\left (x^{3}\right ) \end {vmatrix} \]
Which gives
\[ W = \begin {vmatrix} x & x^{3} \\ 1 & 3 x^{2} \end {vmatrix} \]
Therefore
\[
W = \left (x\right )\left (3 x^{2}\right ) - \left (x^{3}\right )\left (1\right )
\]
Which
simplifies to
\[
W = 2 x^{3}
\]
Which simplifies to
\[
W = 2 x^{3}
\]
Therefore Eq. (2) becomes
\[
u_1 = -\int \frac {x^{3} \left (2 x^{3}-x^{2}\right )}{2 x^{5}}\,dx
\]
Which simplifies to
\[
u_1 = - \int \left (x -\frac {1}{2}\right )d x
\]
Hence
\[
u_1 = -\frac {1}{2} x^{2}+\frac {1}{2} x
\]
And Eq. (3) becomes
\[
u_2 = \int \frac {x \left (2 x^{3}-x^{2}\right )}{2 x^{5}}\,dx
\]
Which simplifies to
\[
u_2 = \int \frac {2 x -1}{2 x^{2}}d x
\]
Hence
\[
u_2 = \ln \left (x \right )+\frac {1}{2 x}
\]
Therefore the particular
solution, from equation (1) is
\[
y_p(x) = \left (-\frac {1}{2} x^{2}+\frac {1}{2} x \right ) x +x^{3} \left (\ln \left (x \right )+\frac {1}{2 x}\right )
\]
Which simplifies to
\[
y_p(x) = x^{3} \ln \left (x \right )-\frac {x^{3}}{2}+x^{2}
\]
Hence the complete solution is
\begin{align*} y(x) &= y_h + y_p \\ &= \left (-\frac {3 \left (c_1 \,x^{2}+2 c_2 \right ) x}{2}\right ) + \left (x^{3} \ln \left (x \right )-\frac {x^{3}}{2}+x^{2}\right )\\ &= x^{3} \ln \left (x \right )+\frac {\left (-3 c_1 -1\right ) x^{3}}{2}+x^{2}-3 c_2 x \end{align*}
Will add steps showing solving for IC soon.
Summary of solutions found
\begin{align*}
y &= x^{3} \ln \left (x \right )+\frac {\left (-3 c_1 -1\right ) x^{3}}{2}+x^{2}-3 c_2 x \\
\end{align*}
Solved as second order ode using Kovacic algorithm
Time used: 0.230 (sec)
Writing the ode as
\begin{align*} x^{2} y^{\prime \prime }-3 x y^{\prime }+3 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 &= 3 \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 {3}{4 x^{2}}\tag {6} \end{align*}
Comparing the above to (5) shows that
\begin{align*} s &= 3\\ t &= 4 x^{2} \end{align*}
Therefore eq. (4) becomes
\begin{align*} z''(x) &= \left ( \frac {3}{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 2.38: 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
\[
r = \frac {3}{4 x^{2}}
\]
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 \(b={\frac {3}{4}}\). Hence
\begin{alignat*}{2} [\sqrt r]_c &= 0 \\ \alpha _c^{+} &= \frac {1}{2} + \sqrt {1+4 b} &&= {\frac {3}{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 {3}{4 x^{2}} \end{alignat*}
Since the \(\text {gcd}(s,t)=1\). This gives \(b={\frac {3}{4}}\). Hence
\begin{alignat*}{2} [\sqrt r]_\infty &= 0 \\ \alpha _{\infty }^{+} &= \frac {1}{2} + \sqrt {1+4 b} &&= {\frac {3}{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
\[ r=\frac {3}{4 x^{2}} \]
| | | | |
| pole \(c\) location |
pole order |
\([\sqrt r]_c\) |
\(\alpha _c^{+}\) |
\(\alpha _c^{-}\) |
| | | | |
| \(0\) | \(2\) | \(0\) | \(\frac {3}{2}\) | \(-{\frac {1}{2}}\) |
| | | | |
| | | |
| Order of \(r\) at \(\infty \) |
\([\sqrt r]_\infty \) |
\(\alpha _\infty ^{+}\) |
\(\alpha _\infty ^{-}\) |
| | | |
| \(2\) |
\(0\) | \(\frac {3}{2}\) | \(-{\frac {1}{2}}\) |
| | | |
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 {3}{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}\\ &= \frac {1}{\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}{x^{2}} \,dx} \\
&= z_1 e^{\frac {3 \ln \left (x \right )}{2}} \\
&= z_1 \left (x^{{3}/{2}}\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 {-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 {x^{2}}{2}\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 (\frac {x^{2}}{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
\[
x^{2} y^{\prime \prime }-3 x y^{\prime }+3 y = 0
\]
The homogeneous solution is found using the Kovacic algorithm which results in
\[
y_h = c_1 x +\frac {1}{2} c_2 \,x^{3}
\]
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 &= x \\
y_2 &= \frac {x^{3}}{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
\[ W = \begin {vmatrix} x & \frac {x^{3}}{2} \\ \frac {d}{dx}\left (x\right ) & \frac {d}{dx}\left (\frac {x^{3}}{2}\right ) \end {vmatrix} \]
Which gives
\[ W = \begin {vmatrix} x & \frac {x^{3}}{2} \\ 1 & \frac {3 x^{2}}{2} \end {vmatrix} \]
Therefore
\[
W = \left (x\right )\left (\frac {3 x^{2}}{2}\right ) - \left (\frac {x^{3}}{2}\right )\left (1\right )
\]
Which simplifies to
\[
W = x^{3}
\]
Which simplifies to
\[
W = x^{3}
\]
Therefore Eq. (2) becomes
\[
u_1 = -\int \frac {\frac {x^{3} \left (2 x^{3}-x^{2}\right )}{2}}{x^{5}}\,dx
\]
Which simplifies to
\[
u_1 = - \int \left (x -\frac {1}{2}\right )d x
\]
Hence
\[
u_1 = -\frac {1}{2} x^{2}+\frac {1}{2} x
\]
And Eq. (3) becomes
\[
u_2 = \int \frac {x \left (2 x^{3}-x^{2}\right )}{x^{5}}\,dx
\]
Which
simplifies to
\[
u_2 = \int \frac {2 x -1}{x^{2}}d x
\]
Hence
\[
u_2 = 2 \ln \left (x \right )+\frac {1}{x}
\]
Therefore the particular solution, from equation (1) is
\[
y_p(x) = \left (-\frac {1}{2} x^{2}+\frac {1}{2} x \right ) x +\frac {x^{3} \left (2 \ln \left (x \right )+\frac {1}{x}\right )}{2}
\]
Which
simplifies to
\[
y_p(x) = x^{3} \ln \left (x \right )-\frac {x^{3}}{2}+x^{2}
\]
Therefore the general solution is
\begin{align*}
y &= y_h + y_p \\
&= \left (c_1 x +\frac {1}{2} c_2 \,x^{3}\right ) + \left (x^{3} \ln \left (x \right )-\frac {x^{3}}{2}+x^{2}\right ) \\
\end{align*}
Will add steps showing solving for IC
soon.
Summary of solutions found
\begin{align*}
y &= c_1 x +\frac {c_2 \,x^{3}}{2}+x^{3} \ln \left (x \right )-\frac {x^{3}}{2}+x^{2} \\
\end{align*}
Solved as second order ode adjoint method
Time used: 0.209 (sec)
In normal form the ode
\begin{align*} x^{2} y^{\prime \prime }-3 x y^{\prime }+3 y = 2 x^{3}-x^{2} \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 {3}{x^{2}}\\ r \left (x \right )&=2 x -1 \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 {3 \xi \left (x \right )}{x^{2}}\right ) &= 0\\ \xi ^{\prime \prime }\left (x \right )+\frac {3 \xi ^{\prime }\left (x \right )}{x}&= 0 \end{align*}
Which is solved for \(\xi (x)\). This is second order ode with missing dependent variable \(\xi \). Let
\begin{align*} p(x) &= \xi ^{\prime } \end{align*}
Then
\begin{align*} p'(x) &= \xi ^{\prime \prime } \end{align*}
Hence the ode becomes
\begin{align*} p^{\prime }\left (x \right )+\frac {3 p \left (x \right )}{x} = 0 \end{align*}
Which is now solve for \(p(x)\) as first order ode. In canonical form a linear first order is
\begin{align*} p^{\prime }\left (x \right ) + q(x)p \left (x \right ) &= p(x) \end{align*}
Comparing the above to the given ode shows that
\begin{align*} q(x) &=\frac {3}{x}\\ p(x) &=0 \end{align*}
The integrating factor \(\mu \) is
\begin{align*} \mu &= e^{\int {q\,dx}}\\ &= {\mathrm e}^{\int \frac {3}{x}d x}\\ &= x^{3} \end{align*}
The ode becomes
\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \mu p &= 0 \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (p \,x^{3}\right ) &= 0 \end{align*}
Integrating gives
\begin{align*} p \,x^{3}&= \int {0 \,dx} + c_1 \\ &=c_1 \end{align*}
Dividing throughout by the integrating factor \(x^{3}\) gives the final solution
\[ p \left (x \right ) = \frac {c_1}{x^{3}} \]
For solution
(1) found earlier, since \(p=\xi ^{\prime }\) then we now have a new first order ode to solve which is
\begin{align*} \xi ^{\prime } = \frac {c_1}{x^{3}} \end{align*}
Since the ode has the form \(\xi ^{\prime }=f(x)\), then we only need to integrate \(f(x)\).
\begin{align*} \int {d\xi } &= \int {\frac {c_1}{x^{3}}\, dx}\\ \xi &= -\frac {c_1}{2 x^{2}} + c_2 \end{align*}
Will add steps showing solving for IC soon.
The original ode 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 )} \end{align*}
Or
\begin{align*} y^{\prime }+y \left (-\frac {3}{x}-\frac {c_1}{x^{3} \left (-\frac {c_1}{2 x^{2}}+c_2 \right )}\right )&=\frac {c_2 \,x^{2}-c_2 x -c_1 \ln \left (x \right )-\frac {c_1}{2 x}}{-\frac {c_1}{2 x^{2}}+c_2} \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 {-6 c_2 \,x^{2}+c_1}{x \left (-2 c_2 \,x^{2}+c_1 \right )}\\ p(x) &=\frac {x \left (-2 c_2 \,x^{3}+2 c_1 \ln \left (x \right ) x +2 c_2 \,x^{2}+c_1 \right )}{-2 c_2 \,x^{2}+c_1} \end{align*}
The integrating factor \(\mu \) is
\begin{align*} \mu &= e^{\int {q\,dx}}\\ &= {\mathrm e}^{\int -\frac {-6 c_2 \,x^{2}+c_1}{x \left (-2 c_2 \,x^{2}+c_1 \right )}d x}\\ &= -\frac {1}{x \left (-2 c_2 \,x^{2}+c_1 \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 {x \left (-2 c_2 \,x^{3}+2 c_1 \ln \left (x \right ) x +2 c_2 \,x^{2}+c_1 \right )}{-2 c_2 \,x^{2}+c_1}\right ) \\
\frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (-\frac {y}{x \left (-2 c_2 \,x^{2}+c_1 \right )}\right ) &= \left (-\frac {1}{x \left (-2 c_2 \,x^{2}+c_1 \right )}\right ) \left (\frac {x \left (-2 c_2 \,x^{3}+2 c_1 \ln \left (x \right ) x +2 c_2 \,x^{2}+c_1 \right )}{-2 c_2 \,x^{2}+c_1}\right ) \\
\mathrm {d} \left (-\frac {y}{x \left (-2 c_2 \,x^{2}+c_1 \right )}\right ) &= \left (-\frac {-2 c_2 \,x^{3}+2 c_1 \ln \left (x \right ) x +2 c_2 \,x^{2}+c_1}{\left (-2 c_2 \,x^{2}+c_1 \right )^{2}}\right )\, \mathrm {d} x \\
\end{align*}
Integrating gives
\begin{align*} -\frac {y}{x \left (-2 c_2 \,x^{2}+c_1 \right )}&= \int {-\frac {-2 c_2 \,x^{3}+2 c_1 \ln \left (x \right ) x +2 c_2 \,x^{2}+c_1}{\left (-2 c_2 \,x^{2}+c_1 \right )^{2}} \,dx} \\ &=\frac {\frac {x}{2}-\frac {c_1}{8 c_2}}{-\frac {c_1}{2}+c_2 \,x^{2}}+\frac {\ln \left (x \right ) x^{2}}{2 c_2 \,x^{2}-c_1} + c_3 \end{align*}
Dividing throughout by the integrating factor \(-\frac {1}{x \left (-2 c_2 \,x^{2}+c_1 \right )}\) gives the final solution
\[ y = \frac {x \left (8 c_2^{2} c_3 \,x^{2}+4 x^{2} \ln \left (x \right ) c_2 -4 c_1 c_2 c_3 +4 c_2 x -c_1 \right )}{4 c_2} \]
Hence, the solution
found using Lagrange adjoint equation method is
\begin{align*}
y &= \frac {x \left (8 c_2^{2} c_3 \,x^{2}+4 x^{2} \ln \left (x \right ) c_2 -4 c_1 c_2 c_3 +4 c_2 x -c_1 \right )}{4 c_2} \\
\end{align*}
The constants can be merged to give
\[
y = \frac {x \left (8 c_2^{2} x^{2}+4 x^{2} \ln \left (x \right ) c_2 -4 c_1 c_2 +4 c_2 x -c_1 \right )}{4 c_2}
\]
Will
add steps showing solving for IC soon.
Summary of solutions found
\begin{align*}
y &= \frac {x \left (8 c_2^{2} x^{2}+4 x^{2} \ln \left (x \right ) c_2 -4 c_1 c_2 +4 c_2 x -c_1 \right )}{4 c_2} \\
\end{align*}
Maple step by step solution
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`
Maple dsolve solution
Solving time : 0.005 (sec)
Leaf size : 28
dsolve(x^2*diff(diff(y(x),x),x)-3*diff(y(x),x)*x+3*y(x) = 2*x^3-x^2,
y(x),singsol=all)
\[
y = \frac {x \left (2 \ln \left (x \right ) x^{2}+\left (c_{1} -1\right ) x^{2}+2 x +2 c_{2} \right )}{2}
\]
Mathematica DSolve solution
Solving time : 0.026 (sec)
Leaf size : 27
DSolve[{x^2*D[y[x],{x,2}]-3*x*D[y[x],x]+3*y[x]==2*x^3-x^2,{}},
y[x],x,IncludeSingularSolutions->True]
\[
y(x)\to x \left (x^2 \log (x)+\left (-\frac {1}{2}+c_2\right ) x^2+x+c_1\right )
\]