Internal
problem
ID
[10069] Book
:
Own
collection
of
miscellaneous
problems Section
:
section
1.0 Problem
number
:
82 Date
solved
:
Thursday, November 27, 2025 at 10:10:56 AM CAS
classification
:
[[_2nd_order, _linear, _nonhomogeneous]]
This is second order non-homogeneous ODE. In standard form the ODE is
\[ A z''(t) + B z'(t) + C z(t) = f(t) \]
Where \(A=1, B=3, C=2, f(t)=24 \,{\mathrm e}^{-3 t}-24 \,{\mathrm e}^{-4 t}\). Let the
solution be
\[ z = z_h + z_p \]
Where \(z_h\) is the solution to the homogeneous ODE \( A z''(t) + B z'(t) + C z(t) = 0\), and \(z_p\) is a particular solution to the
non-homogeneous ODE \(A z''(t) + B z'(t) + C z(t) = f(t)\). \(z_h\) is the solution to
\[ z^{\prime \prime }+3 z^{\prime }+2 z = 0 \]
This is second order with constant coefficients homogeneous ODE. In standard form the
ODE is
\[ A z''(t) + B z'(t) + C z(t) = 0 \]
Where in the above \(A=1, B=3, C=2\). Let the solution be \(z=e^{\lambda t}\). Substituting this into the ODE
gives
While the set of the basis
functions for the homogeneous solution found earlier is
\[ \{{\mathrm e}^{-2 t}, {\mathrm e}^{-t}\} \]
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.
The
unknowns \(\{A_{1}, A_{2}\}\) are found by substituting the above trial solution \(z_p\) into the ODE and comparing
coefficients. Substituting the trial solution into the ODE and simplifying gives
Equation (7) is now solved. After finding \(z(t)\) then \(z\) is found using the inverse transformation
\begin{align*} z &= z e^{-\int \frac {B}{2 A} \,dt} \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 2.26: 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 \(t\), then there is no need run Kovacic algorithm to obtain a solution for
transformed ode \(z''=r z\) as one solution is
\[ z_1(t) = {\mathrm e}^{-\frac {t}{2}} \]
Using the above, the solution for the original ode can now be
found. The first solution to the original ode in \(z\) is found from
This is second order nonhomogeneous ODE. Let the solution be
\[
z = z_h + z_p
\]
Where \(z_h\) is the solution to the
homogeneous ODE \( A z''(t) + B z'(t) + C z(t) = 0\), and \(z_p\) is a particular solution to the nonhomogeneous ODE \(A z''(t) + B z'(t) + C z(t) = f(t)\). \(z_h\) is the solution
to
\[
z^{\prime \prime }+3 z^{\prime }+2 z = 0
\]
The homogeneous solution is found using the Kovacic algorithm which results in
While the set of the basis
functions for the homogeneous solution found earlier is
\[ \{{\mathrm e}^{-2 t}, {\mathrm e}^{-t}\} \]
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.
The
unknowns \(\{A_{1}, A_{2}\}\) are found by substituting the above trial solution \(z_p\) into the ODE and comparing
coefficients. Substituting the trial solution into the ODE and simplifying gives
Methodsfor second order ODEs:---Trying classification methods ---tryinga quadraturetryinghigh order exact linear fully integrabletryingdifferential order: 2; linear nonhomogeneous with symmetry [0,1]tryinga double symmetry of the form [xi=0, eta=F(x)]<-double symmetry of the form [xi=0, eta=F(x)] successful