2.24 problem 25
Internal
problem
ID
[8113]
Book
:
Second
order
enumerated
odes
Section
:
section
2
Problem
number
:
25
Date
solved
:
Monday, October 21, 2024 at 04:53:17 PM
CAS
classification
:
[[_2nd_order, _linear, _nonhomogeneous]]
Solve
\begin{align*} x y^{\prime \prime }-y^{\prime }+4 x^{3} y&=x^{5} \end{align*}
2.24.1 Solved as second order ode using change of variable on x method 2
Time used: 0.376 (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
\[
x y^{\prime \prime }-y^{\prime }+4 x^{3} y = 0
\]
In normal form the ode
\begin{align*} x y^{\prime \prime }-y^{\prime }+4 x^{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 {1}{x}\\ q \left (x \right )&=4 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 {1}{x}d x \right )}d x\\ &= \int e^{\ln \left (x \right )} \,dx\\ &= \int x d x\\ &= \frac {x^{2}}{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 {4 x^{2}}{x^{2}}\\ &= 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 )+4 y \left (\tau \right )&=0 \end{align*}
The above ode is now solved for \(y \left (\tau \right )\).This is second order with constant coefficients homogeneous
ODE. In standard form the ODE is
\[ A y''(\tau ) + B y'(\tau ) + C y(\tau ) = 0 \]
Where in the above \(A=1, B=0, C=4\). Let the solution be \(y \left (\tau \right )=e^{\lambda \tau }\). Substituting
this into the ODE gives
\[ \lambda ^{2} {\mathrm e}^{\tau \lambda }+4 \,{\mathrm e}^{\tau \lambda } = 0 \tag {1} \]
Since exponential function is never zero, then dividing Eq(2)
throughout by \(e^{\lambda \tau }\) gives
\[ \lambda ^{2}+4 = 0 \tag {2} \]
Equation (2) is the characteristic equation of the ODE. Its roots
determine the general solution form.Using the quadratic formula
\[ \lambda _{1,2} = \frac {-B}{2 A} \pm \frac {1}{2 A} \sqrt {B^2 - 4 A C} \]
Substituting \(A=1, B=0, C=4\) into the
above gives
\begin{align*} \lambda _{1,2} &= \frac {0}{(2) \left (1\right )} \pm \frac {1}{(2) \left (1\right )} \sqrt {0^2 - (4) \left (1\right )\left (4\right )}\\ &= \pm 2 i \end{align*}
Hence
\begin{align*} \lambda _1 &= + 2 i\\ \lambda _2 &= - 2 i \end{align*}
Which simplifies to
\begin{align*}
\lambda _1 &= 2 i \\
\lambda _2 &= -2 i \\
\end{align*}
Since roots are complex conjugate of each others, then let the
roots be
\[
\lambda _{1,2} = \alpha \pm i \beta
\]
Where \(\alpha =0\) and \(\beta =2\). Therefore the final solution, when using Euler relation,
can be written as
\[
y \left (\tau \right ) = e^{\alpha \tau } \left ( c_1 \cos (\beta \tau ) + c_2 \sin (\beta \tau ) \right )
\]
Which becomes
\[
y \left (\tau \right ) = e^{0}\left (c_1 \cos \left (2 \tau \right )+c_2 \sin \left (2 \tau \right )\right )
\]
Or
\[
y \left (\tau \right ) = c_1 \cos \left (2 \tau \right )+c_2 \sin \left (2 \tau \right )
\]
Will add steps showing solving for IC
soon.
The above solution is now transformed back to \(y\) using (6) which results in
\[
y = c_1 \cos \left (x^{2}\right )+c_2 \sin \left (x^{2}\right )
\]
Therefore the
homogeneous solution \(y_h\) is
\[
y_h = c_1 \cos \left (x^{2}\right )+c_2 \sin \left (x^{2}\right )
\]
The particular solution is now found using the method of
undetermined coefficients. Looking at the RHS of the ode, which is
\[ x^{5} \]
Shows that the
corresponding undetermined set of the basis functions (UC_set) for the trial solution is
\[ [\{1, x, x^{2}, x^{3}, x^{4}, x^{5}\}] \]
While the set of the basis functions for the homogeneous solution found earlier is
\[ \{\cos \left (x^{2}\right ), \sin \left (x^{2}\right )\} \]
Since there
is no duplication between the basis function in the UC_set and the basis functions of the
homogeneous solution, the trial solution is a linear combination of all the basis in the
UC_set.
\[
y_p = A_{6} x^{5}+A_{5} x^{4}+A_{4} x^{3}+A_{3} x^{2}+A_{2} x +A_{1}
\]
The unknowns \(\{A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}\}\) are found by substituting the above trial solution \(y_p\) into the
ODE and comparing coefficients. Substituting the trial solution into the ODE
and simplifying gives
\[
x \left (20 x^{3} A_{6}+12 x^{2} A_{5}+6 x A_{4}+2 A_{3}\right )-5 x^{4} A_{6}-4 x^{3} A_{5}-3 x^{2} A_{4}-2 x A_{3}-A_{2}+4 x^{3} \left (A_{6} x^{5}+A_{5} x^{4}+A_{4} x^{3}+A_{3} x^{2}+A_{2} x +A_{1}\right ) = x^{5}
\]
Solving for the unknowns by comparing coefficients results
in
\[ \left [A_{1} = 0, A_{2} = 0, A_{3} = {\frac {1}{4}}, A_{4} = 0, A_{5} = 0, A_{6} = 0\right ] \]
Substituting the above back in the above trial solution \(y_p\), gives the particular
solution
\[
y_p = \frac {x^{2}}{4}
\]
Therefore the general solution is
\begin{align*}
y &= y_h + y_p \\
&= \left (c_1 \cos \left (x^{2}\right )+c_2 \sin \left (x^{2}\right )\right ) + \left (\frac {x^{2}}{4}\right ) \\
\end{align*}
Will add steps showing solving for IC
soon.
2.24.2 Solved as second order ode using change of variable on x method 1
Time used: 0.160 (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 \(A=x, B=-1, C=4 x^{3}, f(x)=x^{5}\). 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
\[ x y^{\prime \prime }-y^{\prime }+4 x^{3} y = 0 \]
In normal form the ode
\begin{align*} x y^{\prime \prime }-y^{\prime }+4 x^{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 {1}{x}\\ q \left (x \right )&=4 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 {2 \sqrt {x^{2}}}{c}\tag {6} \\ \tau '' &= \frac {2 x}{c \sqrt {x^{2}}} \end{align*}
Substituting the above into (4) results in
\begin{align*} p_{1} \left (\tau \right ) &=\frac {\tau ^{\prime \prime }\left (x \right )+p \left (x \right ) \tau ^{\prime }\left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\\ &=\frac {\frac {2 x}{c \sqrt {x^{2}}}-\frac {1}{x}\frac {2 \sqrt {x^{2}}}{c}}{\left (\frac {2 \sqrt {x^{2}}}{c}\right )^2} \\ &=0 \end{align*}
Therefore ode (3) now becomes
\begin{align*} y \left (\tau \right )'' + p_1 y \left (\tau \right )' + q_1 y \left (\tau \right ) &= 0 \\ \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+c^{2} y \left (\tau \right ) &= 0 \tag {7} \end{align*}
The above ode is now solved for \(y \left (\tau \right )\). Since the ode is now constant coefficients, it can be easily
solved to give
\begin{align*} y \left (\tau \right ) &= c_1 \cos \left (c \tau \right )+c_2 \sin \left (c \tau \right ) \end{align*}
Now from (6)
\begin{align*} \tau &= \int \frac {1}{c} \sqrt q \,dx \\ &= \frac {\int 2 \sqrt {x^{2}}d x}{c}\\ &= \frac {x \sqrt {x^{2}}}{c} \end{align*}
Substituting the above into the solution obtained gives
\[
y = c_1 \cos \left (x \sqrt {x^{2}}\right )+c_2 \sin \left (x \sqrt {x^{2}}\right )
\]
The particular solution is now found
using the method of undetermined coefficients. Looking at the RHS of the ode, which is
\[ x^{5} \]
Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial
solution is
\[ [\{1, x, x^{2}, x^{3}, x^{4}, x^{5}\}] \]
While the set of the basis functions for the homogeneous solution found earlier is
\[ \left \{\cos \left (x \sqrt {x^{2}}\right ), \sin \left (x \sqrt {x^{2}}\right )\right \} \]
Since there is no duplication between the basis function in the UC_set and the basis
functions of the homogeneous solution, the trial solution is a linear combination of all the
basis in the UC_set.
\[
y_p = A_{6} x^{5}+A_{5} x^{4}+A_{4} x^{3}+A_{3} x^{2}+A_{2} x +A_{1}
\]
The unknowns \(\{A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}\}\) are found by substituting the above trial solution \(y_p\) into
the ODE and comparing coefficients. Substituting the trial solution into the ODE
and simplifying gives
\[
x \left (20 x^{3} A_{6}+12 x^{2} A_{5}+6 x A_{4}+2 A_{3}\right )-5 x^{4} A_{6}-4 x^{3} A_{5}-3 x^{2} A_{4}-2 x A_{3}-A_{2}+4 x^{3} \left (A_{6} x^{5}+A_{5} x^{4}+A_{4} x^{3}+A_{3} x^{2}+A_{2} x +A_{1}\right ) = x^{5}
\]
Solving for the unknowns by comparing coefficients results
in
\[ \left [A_{1} = 0, A_{2} = 0, A_{3} = {\frac {1}{4}}, A_{4} = 0, A_{5} = 0, A_{6} = 0\right ] \]
Substituting the above back in the above trial solution \(y_p\), gives the particular
solution
\[
y_p = \frac {x^{2}}{4}
\]
Therefore the general solution is
\begin{align*}
y &= y_h + y_p \\
&= \left (c_1 \cos \left (x \sqrt {x^{2}}\right )+c_2 \sin \left (x \sqrt {x^{2}}\right )\right ) + \left (\frac {x^{2}}{4}\right ) \\
\end{align*}
Will add steps showing solving for IC
soon.
2.24.3 Solved as second order Bessel ode
Time used: 0.428 (sec)
Writing the ode as
\begin{align*} x^{2} y^{\prime \prime }-y^{\prime } x +4 x^{4} y = x^{6}\tag {1} \end{align*}
Let the solution be
\begin{align*} y &= y_h + y_p \end{align*}
Where \(y_h\) is the solution to the homogeneous ODE and \(y_p\) is a particular solution to the
non-homogeneous ODE. Bessel ode has the form
\begin{align*} x^{2} y^{\prime \prime }+y^{\prime } x +\left (-n^{2}+x^{2}\right ) y = 0\tag {2} \end{align*}
The generalized form of Bessel ode is given by Bowman (1958) as the following
\begin{align*} x^{2} y^{\prime \prime }+\left (1-2 \alpha \right ) x y^{\prime }+\left (\beta ^{2} \gamma ^{2} x^{2 \gamma }-n^{2} \gamma ^{2}+\alpha ^{2}\right ) y = 0\tag {3} \end{align*}
With the standard solution
\begin{align*} y&=x^{\alpha } \left (c_1 \operatorname {BesselJ}\left (n , \beta \,x^{\gamma }\right )+c_2 \operatorname {BesselY}\left (n , \beta \,x^{\gamma }\right )\right )\tag {4} \end{align*}
Comparing (3) to (1) and solving for \(\alpha ,\beta ,n,\gamma \) gives
\begin{align*} \alpha &= 1\\ \beta &= 1\\ n &= {\frac {1}{2}}\\ \gamma &= 2 \end{align*}
Substituting all the above into (4) gives the solution as
\begin{align*} y = \frac {c_1 x \sqrt {2}\, \sin \left (x^{2}\right )}{\sqrt {\pi }\, \sqrt {x^{2}}}-\frac {c_2 x \sqrt {2}\, \cos \left (x^{2}\right )}{\sqrt {\pi }\, \sqrt {x^{2}}} \end{align*}
Therefore the homogeneous solution \(y_h\) is
\[
y_h = \frac {c_1 x \sqrt {2}\, \sin \left (x^{2}\right )}{\sqrt {\pi }\, \sqrt {x^{2}}}-\frac {c_2 x \sqrt {2}\, \cos \left (x^{2}\right )}{\sqrt {\pi }\, \sqrt {x^{2}}}
\]
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 &= \frac {x \sqrt {2}\, \sin \left (x^{2}\right )}{\sqrt {\pi }\, \sqrt {x^{2}}} \\
y_2 &= -\frac {x \sqrt {2}\, \cos \left (x^{2}\right )}{\sqrt {\pi }\, \sqrt {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
\[ W = \begin {vmatrix} \frac {x \sqrt {2}\, \sin \left (x^{2}\right )}{\sqrt {\pi }\, \sqrt {x^{2}}} & -\frac {x \sqrt {2}\, \cos \left (x^{2}\right )}{\sqrt {\pi }\, \sqrt {x^{2}}} \\ \frac {d}{dx}\left (\frac {x \sqrt {2}\, \sin \left (x^{2}\right )}{\sqrt {\pi }\, \sqrt {x^{2}}}\right ) & \frac {d}{dx}\left (-\frac {x \sqrt {2}\, \cos \left (x^{2}\right )}{\sqrt {\pi }\, \sqrt {x^{2}}}\right ) \end {vmatrix} \]
Which gives
\[ W = \begin {vmatrix} \frac {x \sqrt {2}\, \sin \left (x^{2}\right )}{\sqrt {\pi }\, \sqrt {x^{2}}} & -\frac {x \sqrt {2}\, \cos \left (x^{2}\right )}{\sqrt {\pi }\, \sqrt {x^{2}}} \\ \frac {\sqrt {2}\, \sin \left (x^{2}\right )}{\sqrt {\pi }\, \sqrt {x^{2}}}-\frac {x^{2} \sqrt {2}\, \sin \left (x^{2}\right )}{\sqrt {\pi }\, \left (x^{2}\right )^{{3}/{2}}}+\frac {2 x^{2} \sqrt {2}\, \cos \left (x^{2}\right )}{\sqrt {\pi }\, \sqrt {x^{2}}} & -\frac {\sqrt {2}\, \cos \left (x^{2}\right )}{\sqrt {\pi }\, \sqrt {x^{2}}}+\frac {x^{2} \sqrt {2}\, \cos \left (x^{2}\right )}{\sqrt {\pi }\, \left (x^{2}\right )^{{3}/{2}}}+\frac {2 x^{2} \sqrt {2}\, \sin \left (x^{2}\right )}{\sqrt {\pi }\, \sqrt {x^{2}}} \end {vmatrix} \]
Therefore
\[
W = \left (\frac {x \sqrt {2}\, \sin \left (x^{2}\right )}{\sqrt {\pi }\, \sqrt {x^{2}}}\right )\left (-\frac {\sqrt {2}\, \cos \left (x^{2}\right )}{\sqrt {\pi }\, \sqrt {x^{2}}}+\frac {x^{2} \sqrt {2}\, \cos \left (x^{2}\right )}{\sqrt {\pi }\, \left (x^{2}\right )^{{3}/{2}}}+\frac {2 x^{2} \sqrt {2}\, \sin \left (x^{2}\right )}{\sqrt {\pi }\, \sqrt {x^{2}}}\right ) - \left (-\frac {x \sqrt {2}\, \cos \left (x^{2}\right )}{\sqrt {\pi }\, \sqrt {x^{2}}}\right )\left (\frac {\sqrt {2}\, \sin \left (x^{2}\right )}{\sqrt {\pi }\, \sqrt {x^{2}}}-\frac {x^{2} \sqrt {2}\, \sin \left (x^{2}\right )}{\sqrt {\pi }\, \left (x^{2}\right )^{{3}/{2}}}+\frac {2 x^{2} \sqrt {2}\, \cos \left (x^{2}\right )}{\sqrt {\pi }\, \sqrt {x^{2}}}\right )
\]
Which simplifies to
\[
W = \frac {4 x \left (\sin \left (x^{2}\right )^{2}+\cos \left (x^{2}\right )^{2}\right )}{\pi }
\]
Which simplifies to
\[
W = \frac {4 x}{\pi }
\]
Therefore Eq. (2)
becomes
\[
u_1 = -\int \frac {-\frac {x^{7} \sqrt {2}\, \cos \left (x^{2}\right )}{\sqrt {\pi }\, \sqrt {x^{2}}}}{\frac {4 x^{3}}{\pi }}\,dx
\]
Which simplifies for \(0<x\) to
\[
u_1 = - \int -\frac {x^{3} \sqrt {2}\, \sqrt {\pi }\, \cos \left (x^{2}\right )}{4}d x
\]
Hence
\[
u_1 = \frac {\sqrt {2}\, \sqrt {\pi }\, \left (\frac {\cos \left (x^{2}\right )}{2}+\frac {x^{2} \sin \left (x^{2}\right )}{2}\right )}{4}
\]
And Eq. (3) becomes
\[
u_2 = \int \frac {\frac {x^{7} \sqrt {2}\, \sin \left (x^{2}\right )}{\sqrt {\pi }\, \sqrt {x^{2}}}}{\frac {4 x^{3}}{\pi }}\,dx
\]
Which simplifies
for \(0<x\) to
\[
u_2 = \int \frac {x^{3} \sqrt {2}\, \sqrt {\pi }\, \sin \left (x^{2}\right )}{4}d x
\]
Hence
\[
u_2 = \frac {\sqrt {2}\, \sqrt {\pi }\, \left (\frac {\sin \left (x^{2}\right )}{2}-\frac {x^{2} \cos \left (x^{2}\right )}{2}\right )}{4}
\]
Therefore the particular solution, from equation (1) is
\[
y_p(x) = \frac {\left (\frac {\cos \left (x^{2}\right )}{2}+\frac {x^{2} \sin \left (x^{2}\right )}{2}\right ) x \sin \left (x^{2}\right )}{2 \sqrt {x^{2}}}-\frac {x \cos \left (x^{2}\right ) \left (\frac {\sin \left (x^{2}\right )}{2}-\frac {x^{2} \cos \left (x^{2}\right )}{2}\right )}{2 \sqrt {x^{2}}}
\]
Which
simplifies to
\[
y_p(x) = \frac {x \sqrt {x^{2}}}{4}
\]
Therefore the general solution is
\begin{align*}
y &= y_h + y_p \\
&= \left (\frac {c_1 x \sqrt {2}\, \sin \left (x^{2}\right )}{\sqrt {\pi }\, \sqrt {x^{2}}}-\frac {c_2 x \sqrt {2}\, \cos \left (x^{2}\right )}{\sqrt {\pi }\, \sqrt {x^{2}}}\right ) + \left (\frac {x \sqrt {x^{2}}}{4}\right ) \\
\end{align*}
Will add steps showing solving for IC
soon.
2.24.4 Solved as second order ode using Kovacic algorithm
Time used: 0.306 (sec)
Writing the ode as
\begin{align*} x y^{\prime \prime }-y^{\prime }+4 x^{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 \\ B &= -1\tag {3} \\ C &= 4 x^{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 {-16 x^{4}+3}{4 x^{2}}\tag {6} \end{align*}
Comparing the above to (5) shows that
\begin{align*} s &= -16 x^{4}+3\\ t &= 4 x^{2} \end{align*}
Therefore eq. (4) becomes
\begin{align*} z''(x) &= \left ( \frac {-16 x^{4}+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 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 - 4 \\ &= -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. Therefore
\begin{align*} L &= [1, 2] \end{align*}
Attempting to find a solution using case \(n=1\).
Looking at poles of order 2. The partial fractions decomposition of \(r\) is
\[
r = -4 x^{2}+\frac {3}{4 x^{2}}
\]
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 \(O_r(\infty ) = -2\) then
\begin{alignat*}{3} v &= \frac {-O_r(\infty )}{2} &&= \frac {2}{2} &&= 1 \end{alignat*}
\([\sqrt r]_\infty \) is the sum of terms involving \(x^i\) for \(0\leq i \leq v\) in the Laurent series for \(\sqrt r\) at \(\infty \). Therefore
\begin{align*} [\sqrt r]_\infty &= \sum _{i=0}^{v} a_i x^i \\ &= \sum _{i=0}^{1} a_i x^i \tag {8} \end{align*}
Let \(a\) be the coefficient of \(x^v=x^1\) in the above sum. The Laurent series of \(\sqrt r\) at \(\infty \) is
\[ \sqrt r \approx 2 i x -\frac {3 i}{16 x^{3}}-\frac {9 i}{1024 x^{7}}-\frac {27 i}{32768 x^{11}}-\frac {405 i}{4194304 x^{15}}-\frac {1701 i}{134217728 x^{19}}-\frac {15309 i}{8589934592 x^{23}}-\frac {72171 i}{274877906944 x^{27}} + \dots \tag {9} \]
Comparing Eq. (9)
with Eq. (8) shows that
\[ a = 2 i \]
From Eq. (9) the sum up to \(v=1\) gives
\begin{align*} [\sqrt r]_\infty &= \sum _{i=0}^{1} a_i x^i \\ &= 2 i x \tag {10} \end{align*}
Now we need to find \(b\), where \(b\) be the coefficient of \(x^{v-1} = x^{0}=1\) in \(r\) minus the coefficient of same term but
in \(\left ( [\sqrt r]_\infty \right )^2 \) where \([\sqrt r]_\infty \) was found above in Eq (10). Hence
\[ \left ( [\sqrt r]_\infty \right )^2 = -4 x^{2} \]
This shows that the coefficient of \(1\) in the
above is \(0\). Now we need to find the coefficient of \(1\) in \(r\). How this is done depends on if \(v=0\) or
not. Since \(v=1\) which is not zero, then starting \(r=\frac {s}{t}\), we do long division and write this
in the form
\[ r = Q + \frac {R}{t} \]
Where \(Q\) is the quotient and \(R\) is the remainder. Then the coefficient
of \(1\) in \(r\) will be the coefficient this term in the quotient. Doing long division gives
\begin{align*} r &= \frac {s}{t} \\ &= \frac {-16 x^{4}+3}{4 x^{2}} \\ &= Q + \frac {R}{4 x^{2}} \\ &= \left (-4 x^{2}\right ) + \left ( \frac {3}{4 x^{2}}\right ) \\ &= -4 x^{2}+\frac {3}{4 x^{2}} \end{align*}
We see that the coefficient of the term \(x\) in the quotient is \(0\). Now \(b\) can be found.
\begin{align*} b &= \left (0\right )-\left (0\right )\\ &= 0 \end{align*}
Hence
\begin{alignat*}{3} [\sqrt r]_\infty &= 2 i x\\ \alpha _{\infty }^{+} &= \frac {1}{2} \left ( \frac {b}{a} - v \right ) &&= \frac {1}{2} \left ( \frac {0}{2 i} - 1 \right ) &&= -{\frac {1}{2}}\\ \alpha _{\infty }^{-} &= \frac {1}{2} \left ( -\frac {b}{a} - v \right ) &&= \frac {1}{2} \left ( -\frac {0}{2 i} - 1 \right ) &&= -{\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
\[ r=\frac {-16 x^{4}+3}{4 x^{2}} \]
| | | | |
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 \(\infty \) |
\([\sqrt r]_\infty \) |
\(\alpha _\infty ^{+}\) |
\(\alpha _\infty ^{-}\) |
| | | |
\(-2\) |
\(2 i x\) | \(-{\frac {1}{2}}\) | \(-{\frac {1}{2}}\) |
| | | |
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 ( 2 i x \right ) \\ &= -\frac {1}{2 x}-2 i x\\ &= -\frac {1}{2 x}-2 i x \end{align*}
Now that \(\omega \) is determined, the next step is find a corresponding minimal polynomial \(p(x)\) of degree
\(d=0\) to solve the ode. The polynomial \(p(x)\) needs to satisfy the equation
\begin{align*} p'' + 2 \omega p' + \left ( \omega ' +\omega ^2 -r\right ) p = 0 \tag {1A} \end{align*}
Let
\begin{align*} p(x) &= 1\tag {2A} \end{align*}
Substituting the above in eq. (1A) gives
\begin{align*} \left (0\right ) + 2 \left (-\frac {1}{2 x}-2 i x\right ) \left (0\right ) + \left ( \left (\frac {1}{2 x^{2}}-2 i\right ) + \left (-\frac {1}{2 x}-2 i x\right )^2 - \left (\frac {-16 x^{4}+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 \left (-\frac {1}{2 x}-2 i x \right )d x}\\ &= \frac {{\mathrm e}^{-i x^{2}}}{\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 {-1}{x} \,dx} \\
&= z_1 e^{\frac {\ln \left (x \right )}{2}} \\
&= z_1 \left (\sqrt {x}\right ) \\
\end{align*}
Which simplifies to
\[
y_1 = {\mathrm e}^{-i x^{2}}
\]
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 {-1}{x} \,dx}}{\left (y_1\right )^2} \,dx \\
&= y_1 \int \frac { e^{\ln \left (x \right )}}{\left (y_1\right )^2} \,dx \\
&= y_1 \left (-\frac {i {\mathrm e}^{2 i x^{2}}}{4}\right ) \\
\end{align*}
Therefore
the solution is
\begin{align*}
y &= c_1 y_1 + c_2 y_2 \\
&= c_1 \left ({\mathrm e}^{-i x^{2}}\right ) + c_2 \left ({\mathrm e}^{-i x^{2}}\left (-\frac {i {\mathrm e}^{2 i x^{2}}}{4}\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
\[
x y^{\prime \prime }-y^{\prime }+4 x^{3} y = 0
\]
The homogeneous solution is found using the Kovacic algorithm which results in
\[
y_h = c_1 \,{\mathrm e}^{-i x^{2}}-\frac {i c_2 \,{\mathrm e}^{i x^{2}}}{4}
\]
The particular solution \(y_p\) can be found using either the method of undetermined coefficients,
or the method of variation of parameters. The method of variation of parameters will be
used as it is more general and can be used when the coefficients of the ODE depend on \(x\) as
well. Let
\begin{equation}
\tag{1} y_p(x) = u_1 y_1 + u_2 y_2
\end{equation}
Where \(u_1,u_2\) to be determined, and \(y_1,y_2\) are the two basis solutions (the two linearly
independent solutions of the homogeneous ODE) found earlier when solving the
homogeneous ODE as
\begin{align*}
y_1 &= {\mathrm e}^{-i x^{2}} \\
y_2 &= -\frac {i {\mathrm e}^{i x^{2}}}{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
\[ W = \begin {vmatrix} {\mathrm e}^{-i x^{2}} & -\frac {i {\mathrm e}^{i x^{2}}}{4} \\ \frac {d}{dx}\left ({\mathrm e}^{-i x^{2}}\right ) & \frac {d}{dx}\left (-\frac {i {\mathrm e}^{i x^{2}}}{4}\right ) \end {vmatrix} \]
Which gives
\[ W = \begin {vmatrix} {\mathrm e}^{-i x^{2}} & -\frac {i {\mathrm e}^{i x^{2}}}{4} \\ -2 i x \,{\mathrm e}^{-i x^{2}} & \frac {x \,{\mathrm e}^{i x^{2}}}{2} \end {vmatrix} \]
Therefore
\[
W = \left ({\mathrm e}^{-i x^{2}}\right )\left (\frac {x \,{\mathrm e}^{i x^{2}}}{2}\right ) - \left (-\frac {i {\mathrm e}^{i x^{2}}}{4}\right )\left (-2 i x \,{\mathrm e}^{-i x^{2}}\right )
\]
Which simplifies to
\[
W = {\mathrm e}^{-i x^{2}} x \,{\mathrm e}^{i x^{2}}
\]
Which simplifies to
\[
W = x
\]
Therefore Eq. (2) becomes
\[
u_1 = -\int \frac {-\frac {i {\mathrm e}^{i x^{2}} x^{5}}{4}}{x^{2}}\,dx
\]
Which simplifies to
\[
u_1 = - \int -\frac {i {\mathrm e}^{i x^{2}} x^{3}}{4}d x
\]
Hence
\[
u_1 = -\frac {i \left (i x^{2}-1\right ) {\mathrm e}^{i x^{2}}}{8}
\]
And Eq. (3) becomes
\[
u_2 = \int \frac {{\mathrm e}^{-i x^{2}} x^{5}}{x^{2}}\,dx
\]
Which
simplifies to
\[
u_2 = \int {\mathrm e}^{-i x^{2}} x^{3}d x
\]
Hence
\[
u_2 = \frac {\left (i x^{2}+1\right ) {\mathrm e}^{-i x^{2}}}{2}
\]
Therefore the particular solution, from equation (1) is
\[
y_p(x) = -\frac {i \left (i x^{2}-1\right ) {\mathrm e}^{i x^{2}} {\mathrm e}^{-i x^{2}}}{8}-\frac {i {\mathrm e}^{i x^{2}} \left (i x^{2}+1\right ) {\mathrm e}^{-i x^{2}}}{8}
\]
Which
simplifies to
\[
y_p(x) = \frac {x^{2}}{4}
\]
Therefore the general solution is
\begin{align*}
y &= y_h + y_p \\
&= \left (c_1 \,{\mathrm e}^{-i x^{2}}-\frac {i c_2 \,{\mathrm e}^{i x^{2}}}{4}\right ) + \left (\frac {x^{2}}{4}\right ) \\
\end{align*}
Will add steps showing solving for IC
soon.
2.24.5 Maple step by step solution
2.24.6 Maple trace
Methods for second order ODEs:
2.24.7 Maple dsolve solution
Solving time : 0.004
(sec)
Leaf size : 22
dsolve(x*diff(diff(y(x),x),x)-diff(y(x),x)+4*x^3*y(x) = x^5,
y(x),singsol=all)
\[
y = \frac {x^{2}}{4}+c_1 \cos \left (x^{2}\right )+c_2 \sin \left (x^{2}\right )
\]
2.24.8 Mathematica DSolve solution
Solving time : 0.064
(sec)
Leaf size : 27
DSolve[{x*D[y[x],{x,2}]-D[y[x],x]+4*x^3*y[x]==x^5,{}},
y[x],x,IncludeSingularSolutions->True]
\[
y(x)\to \frac {x^2}{4}+c_1 \cos \left (x^2\right )+c_2 \sin \left (x^2\right )
\]