Internal
problem
ID
[15386] Book
:
INTRODUCTORY
DIFFERENTIAL
EQUATIONS.
Martha
L.
Abell,
James
P.
Braselton.
Fourth
edition
2014.
ElScAe.
2014 Section
:
Chapter
4.
Higher
Order
Equations.
Exercises
4.7,
page
195 Problem
number
:
39 Date
solved
:
Friday, October 18, 2024 at 01:28:53 PM CAS
classification
:
[[_2nd_order, _exact, _linear, _nonhomogeneous]]
And the point \(x_0 = 1\) is inside this domain. The domain of \(q(x)=-\frac {1}{2 x^{2}}\) is
\[
\{x <0\boldsymbol {\lor }0<x\}
\]
And the point \(x_0 = 1\) is
also inside this domain. The domain of \(F =\frac {1}{2 x^{4}}\) is
\[
\{x <0\boldsymbol {\lor }0<x\}
\]
And the point \(x_0 = 1\) is also inside this domain. Hence
solution exists and is unique.
15.39.2 Solved as second order Euler type ode
Time used: 0.186 (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 \(A=2 x^{2}, B=3 x, C=-1, f(x)=\frac {1}{x^{2}}\). 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
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
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
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
Dividing throughout by the integrating factor \(\frac {1}{\sqrt {x}}\) gives the final solution
\[ y = \frac {15 c_2 \,x^{{5}/{2}}-5 c_1 x +3}{15 x^{2}} \]
Will add steps
showing solving for IC soon.
15.39.6 Solved as second order ode using change of variable on x method 2
Time used: 0.447 (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
The above ode is now solved for \(y \left (\tau \right )\). This is Euler second order ODE. Let the solution be \(y \left (\tau \right ) = \tau ^r\), then
\(y'=r \tau ^{r-1}\) and \(y''=r(r-1) \tau ^{r-2}\). Substituting these back into the given ODE gives
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
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
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
15.39.7 Solved as second order ode using change of variable on y method 2
Time used: 0.276 (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 \(A=2 x^{2}, B=3 x, C=-1, f(x)=\frac {1}{x^{2}}\). 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
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 \(y\).
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
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
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
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.
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 211: Necessary conditions for each Kovacic case
The order of \(r\) at \(\infty \) is the degree of \(t\) minus the degree of \(s\). Therefore
The poles of \(r\) in eq. (7) and the order of each pole are determined by solving for the roots of \(t=16 x^{2}\).
There is a pole at \(x=0\) of order \(2\). Since there is no odd order pole larger than \(2\) and the order at \(\infty \) is
\(2\) then the necessary conditions for case one are met. Since there is a pole of order \(2\) then
necessary conditions for case two are met. Since pole order is not larger than \(2\) and
the order at \(\infty \) is \(2\) then the necessary conditions for case three are met. Therefore
\begin{align*} L &= [1, 2, 4, 6, 12] \end{align*}
Attempting to find a solution using case \(n=1\).
Looking at poles of order 2. The partial fractions decomposition of \(r\) is
\[
r = \frac {5}{16 x^{2}}
\]
For the pole at \(x=0\) let \(b\)
be the coefficient of \(\frac {1}{ x^{2}}\) in the partial fractions decomposition of \(r\) given above. Therefore \(b={\frac {5}{16}}\). Hence
Since the order of \(r\) at \(\infty \) is 2 then \([\sqrt r]_\infty =0\). Let \(b\) be the coefficient of \(\frac {1}{x^{2}}\) in the Laurent series expansion of \(r\)
at \(\infty \). which can be found by dividing the leading coefficient of \(s\) by the leading coefficient of \(t\)
from
\begin{alignat*}{2} r &= \frac {s}{t} &&= \frac {5}{16 x^{2}} \end{alignat*}
Since the \(\text {gcd}(s,t)=1\). This gives \(b={\frac {5}{16}}\). Hence
The following table summarizes the findings so far for poles and for the order of \(r\) at \(\infty \) where \(r\)
is
\[ r=\frac {5}{16 x^{2}} \]
pole \(c\) location
pole order
\([\sqrt r]_c\)
\(\alpha _c^{+}\)
\(\alpha _c^{-}\)
\(0\)
\(2\)
\(0\)
\(\frac {5}{4}\)
\(-{\frac {1}{4}}\)
Order of \(r\) at \(\infty \)
\([\sqrt r]_\infty \)
\(\alpha _\infty ^{+}\)
\(\alpha _\infty ^{-}\)
\(2\)
\(0\)
\(\frac {5}{4}\)
\(-{\frac {1}{4}}\)
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
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
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
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
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
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
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
Which is solved for \(\xi (x)\). This is Euler second order ODE. Let the solution be \(\xi = x^r\), then \(\xi '=r x^{r-1}\) and \(\xi ''=r(r-1) x^{r-2}\).
Substituting these back into the given ODE gives
\[ x^{2}(r(r-1))x^{r-2}-\frac {3 x}{2} r x^{r-1}+x^{r} = 0 \]
Dividing throughout by the integrating factor \(\frac {x}{x^{{3}/{2}} c_1 +c_2}\) gives the final solution
\[ y = \frac {5 x^{{5}/{2}} c_1 c_3 +5 c_3 c_2 x +1}{5 x^{2}} \]
Hence, the solution
found using Lagrange adjoint equation method is
\begin{align*}
y &= \frac {5 x^{{5}/{2}} c_1 c_3 +5 c_3 c_2 x +1}{5 x^{2}} \\
\end{align*}
Will add steps showing solving for IC
soon.
15.39.10 Maple step by step solution
15.39.11 Maple trace
`Methodsfor second order ODEs:---Trying classification methods ---tryinga quadraturetryinghigh order exact linear fully integrable<-high order exact linear fully integrable successful`