Internal
problem
ID
[9086] Book
:
Second
order
enumerated
odes Section
:
section
1 Problem
number
:
15 Date
solved
:
Wednesday, March 05, 2025 at 07:19:25 AM CAS
classification
:
[[_2nd_order, _missing_x]]
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 solved for \(u(x)\) as first order ode.
Integrating gives
\begin{align*} \int \frac {1}{-u +1}d u &= dx\\ -\ln \left (u -1\right )&= x +c_1 \end{align*}
Singular solutions are found by solving
\begin{align*} -u +1&= 0 \end{align*}
for \(u \left (x \right )\). 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*} u \left (x \right ) = 1 \end{align*}
In summary, these are the solution found for \(u(x)\)
\begin{align*}
-\ln \left (u \left (x \right )-1\right ) &= x +c_1 \\
u \left (x \right ) &= 1 \\
\end{align*}
For solution \(-\ln \left (u \left (x \right )-1\right ) = x +c_1\), since \(u=y^{\prime }\) then we now have a new first order ode to solve which is
\begin{align*} -\ln \left (y^{\prime }-1\right ) = x +c_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 {{\mathrm e}^{-x -c_1}+1\, dx}\\ y &= x -{\mathrm e}^{-x -c_1} + c_2 \end{align*}
For solution \(u \left (x \right ) = 1\), since \(u=y^{\prime }\) 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_3 \end{align*}
In summary, these are the solution found for \((y)\)
\begin{align*}
y &= x -{\mathrm e}^{-x -c_1}+c_2 \\
y &= x +c_3 \\
\end{align*}
Will add steps showing solving for IC soon.
Summary of solutions found
\begin{align*}
y &= x +c_3 \\
y &= x -{\mathrm e}^{-x -c_1}+c_2 \\
\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.
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 2.6: 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
`Methodsfor second order ODEs:---Trying classification methods ---tryinga quadraturetryinghigh order exact linear fully integrable<-high order exact linear fully integrable successful`