Internal
problem
ID
[8068] Book
:
Second
order
enumerated
odes Section
:
section
1 Problem
number
:
31 Date
solved
:
Monday, October 21, 2024 at 04:47:15 PM CAS
classification
:
[[_2nd_order, _missing_x]]
1.31.1 Solved as second order linear constant coeff ode
Time used: 0.102 (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 \(A=1, B=1, C=0, f(x)=1\).
Let the solution be
\[ y = y_h + y_p \]
Where \(y_h\) is the solution to the homogeneous ODE \( A y''(x) + B y'(x) + C y(x) = 0\), and \(y_p\) is a
particular solution to the non-homogeneous ODE \(A y''(x) + B y'(x) + C y(x) = f(x)\). \(y_h\) is the solution to
\[ y^{\prime \prime }+y^{\prime } = 0 \]
This is second
order with constant coefficients homogeneous ODE. In standard form the ODE is
\[ A y''(x) + B y'(x) + C y(x) = 0 \]
Where in the above \(A=1, B=1, C=0\). Let the solution be \(y=e^{\lambda x}\). Substituting this into the ODE gives
The particular solution is now found using the
method of undetermined coefficients. Looking at the RHS of the ode, which is
\[ 1 \]
Shows
that the corresponding undetermined set of the basis functions (UC_set) for the
trial solution is
\[ [\{1\}] \]
While the set of the basis functions for the homogeneous solution
found earlier is
\[ \{1, {\mathrm e}^{-x}\} \]
Since \(1\) is duplicated in the UC_set, then this basis is multiplied
by extra \(x\). The UC_set becomes
\[ [\{x\}] \]
Since there was duplication between the basis
functions in the UC_set and the basis functions of the homogeneous solution, the trial
solution is a linear combination of all the basis function in the above updated
UC_set.
\[
y_p = A_{1} x
\]
The unknowns \(\{A_{1}\}\) 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
\[
A_{1} = 1
\]
Solving for the unknowns by comparing coefficients results
in
\[ [A_{1} = 1] \]
Substituting the above back in the above trial solution \(y_p\), gives the particular
solution
Which is now solve for \(p(x)\) as first order ode. Integrating gives
\begin{align*} \int \frac {1}{1-p}d p &= dx\\ -\ln \left (-1+p \right )&= x +c_1 \end{align*}
Singular solutions are found by solving
\begin{align*} 1-p&= 0 \end{align*}
for \(p \left (x \right )\). This is because we had to divide by this in the above step. This gives the following
singular solution(s), which also have to satisfy the given ODE.
\begin{align*} p \left (x \right ) = 1 \end{align*}
Solving for \(p \left (x \right )\) from the above solution(s) gives (after possible removing of solutions that do not
verify)
\begin{align*} p \left (x \right )&=1\\ p \left (x \right )&={\mathrm e}^{-x -c_1}+1 \end{align*}
For solution (1) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is
\begin{align*} y^{\prime } = 1 \end{align*}
Since the ode has the form \(y^{\prime }=f(x)\), then we only need to integrate \(f(x)\).
\begin{align*} \int {dy} &= \int {1\, dx}\\ y &= x + c_2 \end{align*}
For solution (2) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is
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 16: Necessary conditions for each Kovacic case
The order of \(r\) at \(\infty \) is the degree of \(t\) minus the degree of \(s\). Therefore
There are no poles in \(r\). Therefore the set of poles \(\Gamma \) is empty. Since there is no odd order pole
larger than \(2\) and the order at \(\infty \) is \(0\) then the necessary conditions for case one are met.
Therefore
\begin{align*} L &= [1] \end{align*}
Since \(r = {\frac {1}{4}}\) is not a function of \(x\), then there is no need run Kovacic algorithm to obtain a solution
for transformed ode \(z''=r z\) as one solution is
\[ z_1(x) = {\mathrm e}^{-\frac {x}{2}} \]
Using the above, the solution for the original ode can
now be found. The first solution to the original ode in \(y\) is found from
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
\[
y^{\prime \prime }+y^{\prime } = 0
\]
The homogeneous solution is found using the Kovacic algorithm which results in
\[
y_h = c_1 \,{\mathrm e}^{-x}+c_2
\]
The particular solution is now found using the method of undetermined coefficients. Looking
at the RHS of the ode, which is
\[ 1 \]
Shows that the corresponding undetermined set of the basis
functions (UC_set) for the trial solution is
\[ [\{1\}] \]
While the set of the basis functions for the
homogeneous solution found earlier is
\[ \{1, {\mathrm e}^{-x}\} \]
Since \(1\) is duplicated in the UC_set, then this basis is
multiplied by extra \(x\). The UC_set becomes
\[ [\{x\}] \]
Since there was duplication between the basis
functions in the UC_set and the basis functions of the homogeneous solution, the trial
solution is a linear combination of all the basis function in the above updated
UC_set.
\[
y_p = A_{1} x
\]
The unknowns \(\{A_{1}\}\) 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
\[
A_{1} = 1
\]
Solving for the unknowns by comparing coefficients results
in
\[ [A_{1} = 1] \]
Substituting the above back in the above trial solution \(y_p\), gives the particular
solution