Internal
problem
ID
[18203] Book
:
Elementary
Differential
Equations.
By
R.L.E.
Schwarzenberger.
Chapman
and
Hall.
London.
First
Edition
(1969) Section
:
Chapter
5.
Linear
equations.
Exercises
at
page
85 Problem
number
:
7
(iii) Date
solved
:
Thursday, December 19, 2024 at 06:18:04 PM CAS
classification
:
[[_2nd_order, _linear, _nonhomogeneous]]
\begin{align*} x \left (0\right )&=0\\ x^{\prime }\left (0\right )&=1 \end{align*}
Existence and uniqueness analysis
This is a linear ODE. In canonical form it is written as
\begin{align*} x^{\prime \prime } + p(t)x^{\prime } + q(t) x &= F \end{align*}
Where here
\begin{align*} p(t) &=2\\ q(t) &=4\\ F &={\mathrm e}^{t} \cos \left (2 t \right ) \end{align*}
Hence the ode is
\begin{align*} x^{\prime \prime }+2 x^{\prime }+4 x = {\mathrm e}^{t} \cos \left (2 t \right ) \end{align*}
The domain of \(p(t)=2\) is
\[
\{-\infty <t <\infty \}
\]
And the point \(t_0 = 0\) is inside this domain. The domain of \(q(t)=4\) is
\[
\{-\infty <t <\infty \}
\]
And the point \(t_0 = 0\) is
also inside this domain. The domain of \(F ={\mathrm e}^{t} \cos \left (2 t \right )\) is
\[
\{-\infty <t <\infty \}
\]
And the point \(t_0 = 0\) is also inside this domain. Hence
solution exists and is unique.
Solved as second order linear constant coeff ode
Time used: 0.376 (sec)
This is second order non-homogeneous ODE. In standard form the ODE is
\[ A x''(t) + B x'(t) + C x(t) = f(t) \]
Where \(A=1, B=2, C=4, f(t)={\mathrm e}^{t} \cos \left (2 t \right )\).
Let the solution be
\[ x = x_h + x_p \]
Where \(x_h\) is the solution to the homogeneous ODE \( A x''(t) + B x'(t) + C x(t) = 0\), and \(x_p\) is a
particular solution to the non-homogeneous ODE \(A x''(t) + B x'(t) + C x(t) = f(t)\). \(x_h\) is the solution to
\[ x^{\prime \prime }+2 x^{\prime }+4 x = 0 \]
This is second
order with constant coefficients homogeneous ODE. In standard form the ODE is
\[ A x''(t) + B x'(t) + C x(t) = 0 \]
Where in the above \(A=1, B=2, C=4\). Let the solution be \(x=e^{\lambda t}\). Substituting this into the ODE gives
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 \(x_p\) into the
ODE and comparing coefficients. Substituting the trial solution into the ODE
and simplifying gives
\[
3 A_{1} {\mathrm e}^{t} \cos \left (2 t \right )-8 A_{1} {\mathrm e}^{t} \sin \left (2 t \right )+3 A_{2} {\mathrm e}^{t} \sin \left (2 t \right )+8 A_{2} {\mathrm e}^{t} \cos \left (2 t \right ) = {\mathrm e}^{t} \cos \left (2 t \right )
\]
Solving for the unknowns by comparing coefficients results
in
Equation (7) is now solved. After finding \(z(t)\) then \(x\) is found using the inverse transformation
\begin{align*} x &= z \left (t \right ) 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 2.15: 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 \(r = -3\) 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
\[ z_1(t) = \cos \left (\sqrt {3}\, t \right ) \]
Using the above, the solution for the original ode can
now be found. The first solution to the original ode in \(x\) is found from
This is second order nonhomogeneous ODE. Let the solution be
\[
x = x_h + x_p
\]
Where \(x_h\) is the solution to
the homogeneous ODE \( A x''(t) + B x'(t) + C x(t) = 0\), and \(x_p\) is a particular solution to the nonhomogeneous ODE \(A x''(t) + B x'(t) + C x(t) = f(t)\). \(x_h\) is the
solution to
\[
x^{\prime \prime }+2 x^{\prime }+4 x = 0
\]
The homogeneous solution is found using the Kovacic algorithm which results in
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 \(x_p\) into the
ODE and comparing coefficients. Substituting the trial solution into the ODE
and simplifying gives
\[
3 A_{1} {\mathrm e}^{t} \cos \left (2 t \right )-8 A_{1} {\mathrm e}^{t} \sin \left (2 t \right )+3 A_{2} {\mathrm e}^{t} \sin \left (2 t \right )+8 A_{2} {\mathrm e}^{t} \cos \left (2 t \right ) = {\mathrm e}^{t} \cos \left (2 t \right )
\]
Solving for the unknowns by comparing coefficients results
in
`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)]->Try solving first the homogeneous part of the ODEchecking if the LODE has constant coefficients<- constant coefficients successful<-solving first the homogeneous part of the ODE successful`