Internal
problem
ID
[894] Book
:
Differential
equations
and
linear
algebra,
3rd
ed.,
Edwards
and
Penney Section
:
Section
5.5,
Nonhomogeneous
equations
and
undetermined
coefficients
Page
351 Problem
number
:
49 Date
solved
:
Thursday, October 17, 2024 at 01:20:36 AM CAS
classification
:
[[_2nd_order, _with_linear_symmetries]]
11.26.1 Solved as second order linear constant coeff ode
Time used: 0.119 (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 \(A=1, B=-4, C=4, f(x)=2 \,{\mathrm e}^{2 x}\).
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
\[ y^{\prime \prime }-4 y^{\prime }+4 y = 0 \]
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 \(A=1, B=-4, C=4\). Let the solution be \(y=e^{\lambda x}\). Substituting this into the ODE gives
Since \({\mathrm e}^{2 x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The
UC_set becomes
\[ [\{x \,{\mathrm e}^{2 x}\}] \]
Since \(x \,{\mathrm e}^{2 x}\) is duplicated in the UC_set, then this basis is multiplied
by extra \(x\). The UC_set becomes
\[ [\{x^{2} {\mathrm e}^{2 x}\}] \]
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.
\[
y_p = A_{1} x^{2} {\mathrm e}^{2 x}
\]
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
\[
y = c_1 x \,{\mathrm e}^{2 x}+x^{2} {\mathrm e}^{2 x}+c_2 \,{\mathrm e}^{2 x}
\]
Will add steps showing solving for IC soon.
11.26.3 Solved as second order ode using Kovacic algorithm
Time used: 0.085 (sec)
Writing the ode as
\begin{align*} y^{\prime \prime }-4 y^{\prime }+4 y &= 0 \tag {1} \\ A y^{\prime \prime } + B y^{\prime } + C y &= 0 \tag {2} \end{align*}
Comparing (1) and (2) shows that
\begin{align*} A &= 1 \\ B &= -4\tag {3} \\ C &= 4 \end{align*}
Applying the Liouville transformation on the dependent variable gives
\begin{align*} z(x) &= y e^{\int \frac {B}{2 A} \,dx} \end{align*}
Then (2) becomes
\begin{align*} z''(x) = r z(x)\tag {4} \end{align*}
Where \(r\) is given by
\begin{align*} r &= \frac {s}{t}\tag {5} \\ &= \frac {2 A B' - 2 B A' + B^2 - 4 A C}{4 A^2} \end{align*}
Substituting the values of \(A,B,C\) from (3) in the above and simplifying gives
\begin{align*} r &= \frac {0}{1}\tag {6} \end{align*}
Comparing the above to (5) shows that
\begin{align*} s &= 0\\ t &= 1 \end{align*}
Therefore eq. (4) becomes
\begin{align*} z''(x) &= 0 \tag {7} \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 88: 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 \(infinity\) then the necessary conditions for case one are met.
Therefore
\begin{align*} L &= [1] \end{align*}
Since \(r = 0\) 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
\[ z_1(x) = 1 \]
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
\[
y^{\prime \prime }-4 y^{\prime }+4 y = 0
\]
The homogeneous solution is found using the Kovacic
algorithm which results in
Since \({\mathrm e}^{2 x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The
UC_set becomes
\[ [\{x \,{\mathrm e}^{2 x}\}] \]
Since \(x \,{\mathrm e}^{2 x}\) is duplicated in the UC_set, then this basis is multiplied
by extra \(x\). The UC_set becomes
\[ [\{x^{2} {\mathrm e}^{2 x}\}] \]
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.
\[
y_p = A_{1} x^{2} {\mathrm e}^{2 x}
\]
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
`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`