2.2.11 Internal
problem
ID
[18200]Book
:
Elementary
Differential
Equations.
By
R.L.E.
Schwarzenberger.
Chapman
and
Hall.
London.
First
Edition
(1969) Section
:
Chapter
4.
Autonomous
systems.
Exercises
at
page
69 Problem
number
:
6
(iv)Date
solved
:
Thursday, December 19, 2024 at 06:18:00 PMCAS
classification
:
[[_2nd_order, _missing_x]]
Solve
\begin{align*} x^{\prime \prime }-2 x^{\prime }+2 x&=0 \end{align*}
With initial conditions
\begin{align*} x \left (0\right )&=0\\ x^{\prime }\left (0\right )&=1 \end{align*}
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) &=2\\ F &=0 \end{align*}
Hence the ode is
\begin{align*} x^{\prime \prime }-2 x^{\prime }+2 x = 0 \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 \(q(t)=2\) is
\[
\{-\infty <t <\infty \}
\]
And the point \(t_0 = 0\) is
also inside this domain. Hence solution exists and is unique.
Time used: 0.116 (sec)
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 \(A=1, B=-2, C=2\) . Let the solution be \(x=e^{\lambda t}\) . Substituting this into the ODE gives
\[ \lambda ^{2} {\mathrm e}^{t \lambda }-2 \lambda \,{\mathrm e}^{t \lambda }+2 \,{\mathrm e}^{t \lambda } = 0 \tag {1} \]
Since exponential function is never zero, then dividing Eq(2) throughout by \(e^{\lambda t}\) gives
\[ \lambda ^{2}-2 \lambda +2 = 0 \tag {2} \]
Equation (2) is the characteristic equation of the ODE. Its roots determine the
general solution form.Using the quadratic formula
\[ \lambda _{1,2} = \frac {-B}{2 A} \pm \frac {1}{2 A} \sqrt {B^2 - 4 A C} \]
Substituting \(A=1, B=-2, C=2\) into the above gives
\begin{align*} \lambda _{1,2} &= \frac {2}{(2) \left (1\right )} \pm \frac {1}{(2) \left (1\right )} \sqrt {-2^2 - (4) \left (1\right )\left (2\right )}\\ &= 1 \pm i \end{align*}
Hence
\begin{align*} \lambda _1 &= 1 + i\\ \lambda _2 &= 1 - i \end{align*}
Which simplifies to
\begin{align*}
\lambda _1 &= 1+i \\
\lambda _2 &= 1-i \\
\end{align*}
Since roots are complex conjugate of each others, then let the roots be
\[
\lambda _{1,2} = \alpha \pm i \beta
\]
Where \(\alpha =1\) and \(\beta =1\) . Therefore the final solution, when using Euler relation, can be written as
\[
x = e^{\alpha t} \left ( c_1 \cos (\beta t) + c_2 \sin (\beta t) \right )
\]
Which becomes
\[
x = e^{t}\left (c_1 \cos \left (t \right )+c_2 \sin \left (t \right )\right )
\]
Will add steps showing solving for IC soon.
Summary of solutions found
\begin{align*}
x &= {\mathrm e}^{t} \sin \left (t \right ) \\
\end{align*}
Time used: 0.067 (sec)
Writing the ode as
\begin{align*} x^{\prime \prime }-2 x^{\prime }+2 x &= 0 \tag {1} \\ A x^{\prime \prime } + B x^{\prime } + C x &= 0 \tag {2} \end{align*}
Comparing (1) and (2) shows that
\begin{align*} A &= 1 \\ B &= -2\tag {3} \\ C &= 2 \end{align*}
Applying the Liouville transformation on the dependent variable gives
\begin{align*} z(t) &= x e^{\int \frac {B}{2 A} \,dt} \end{align*}
Then (2) becomes
\begin{align*} z''(t) = r z(t)\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 {-1}{1}\tag {6} \end{align*}
Comparing the above to (5) shows that
\begin{align*} s &= -1\\ t &= 1 \end{align*}
Therefore eq. (4) becomes
\begin{align*} z''(t) &= -z \left (t \right ) \tag {7} \end{align*}
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.
Case
Allowed pole order for \(r\)
Allowed value for \(\mathcal {O}(\infty )\)
1
\(\left \{ 0,1,2,4,6,8,\cdots \right \} \)
\(\left \{ \cdots ,-6,-4,-2,0,2,3,4,5,6,\cdots \right \} \)
2
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 2.12: Necessary conditions for each Kovacic case
The order of \(r\) at \(\infty \) is the degree of \(t\) minus the degree of \(s\) . Therefore
\begin{align*} O\left (\infty \right ) &= \text {deg}(t) - \text {deg}(s) \\ &= 0 - 0 \\ &= 0 \end{align*}
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 \(r = -1\) 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
\[ z_1(t) = \cos \left (t \right ) \]
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
\begin{align*}
x_1 &= z_1 e^{ \int -\frac {1}{2} \frac {B}{A} \,dt} \\
&= z_1 e^{ -\int \frac {1}{2} \frac {-2}{1} \,dt} \\
&= z_1 e^{t} \\
&= z_1 \left ({\mathrm e}^{t}\right ) \\
\end{align*}
Which simplifies to
\[
x_1 = {\mathrm e}^{t} \cos \left (t \right )
\]
The second solution \(x_2\) to the original ode is found using reduction of order
\[ x_2 = x_1 \int \frac { e^{\int -\frac {B}{A} \,dt}}{x_1^2} \,dt \]
Substituting gives
\begin{align*}
x_2 &= x_1 \int \frac { e^{\int -\frac {-2}{1} \,dt}}{\left (x_1\right )^2} \,dt \\
&= x_1 \int \frac { e^{2 t}}{\left (x_1\right )^2} \,dt \\
&= x_1 \left (\tan \left (t \right )\right ) \\
\end{align*}
Therefore the solution is
\begin{align*}
x &= c_1 x_1 + c_2 x_2 \\
&= c_1 \left ({\mathrm e}^{t} \cos \left (t \right )\right ) + c_2 \left ({\mathrm e}^{t} \cos \left (t \right )\left (\tan \left (t \right )\right )\right ) \\
\end{align*}
Will add steps showing solving for IC soon.
Summary of solutions found
\begin{align*}
x &= {\mathrm e}^{t} \sin \left (t \right ) \\
\end{align*}
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [x^{\prime \prime }-2 x^{\prime }+2 x=0, x \left (0\right )=0, x^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & x^{\prime \prime } \\ \bullet & {} & \textrm {Characteristic polynomial of ODE}\hspace {3pt} \\ {} & {} & r^{2}-2 r +2=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {2\pm \left (\sqrt {-4}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (1-\mathrm {I}, 1+\mathrm {I}\right ) \\ \bullet & {} & \textrm {1st solution of the ODE}\hspace {3pt} \\ {} & {} & x_{1}\left (t \right )={\mathrm e}^{t} \cos \left (t \right ) \\ \bullet & {} & \textrm {2nd solution of the ODE}\hspace {3pt} \\ {} & {} & x_{2}\left (t \right )={\mathrm e}^{t} \sin \left (t \right ) \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & x=\mathit {C1} x_{1}\left (t \right )+\mathit {C2} x_{2}\left (t \right ) \\ \bullet & {} & \textrm {Substitute in solutions}\hspace {3pt} \\ {} & {} & x=\mathit {C1} \,{\mathrm e}^{t} \cos \left (t \right )+\mathit {C2} \,{\mathrm e}^{t} \sin \left (t \right ) \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} x=\textit {\_C1} {\mathrm e}^{t} \cos \left (t \right )+\textit {\_C2} {\mathrm e}^{t} \sin \left (t \right ) \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} x \left (0\right )=0 \\ {} & {} & 0=\textit {\_C1} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & x^{\prime }=\textit {\_C1} \,{\mathrm e}^{t} \cos \left (t \right )-\textit {\_C1} \,{\mathrm e}^{t} \sin \left (t \right )+\textit {\_C2} \,{\mathrm e}^{t} \sin \left (t \right )+\textit {\_C2} \,{\mathrm e}^{t} \cos \left (t \right ) \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} x^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=1 \\ {} & {} & 1=\textit {\_C1} +\textit {\_C2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} \textit {\_C1} \hspace {3pt}\textrm {and}\hspace {3pt} \textit {\_C2} \\ {} & {} & \left \{\textit {\_C1} =0, \textit {\_C2} =1\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & x={\mathrm e}^{t} \sin \left (t \right ) \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & x={\mathrm e}^{t} \sin \left (t \right ) \end {array} \]
` Methods for second order ODEs:
--- Trying classification methods ---
trying a quadrature
checking if the LODE has constant coefficients
<- constant coefficients successful `
Solving time : 0.008
dsolve ([ diff ( diff ( x ( t ), t ), t )-2* diff ( x ( t ), t )+2* x ( t ) = 0,
op ([ x (0) = 0, D ( x )(0) = 1])],x(t),singsol=all)
\[
x = {\mathrm e}^{t} \sin \left (t \right )
\]
Solving time : 0.012
DSolve [{ D [ x [ t ],{ t ,2}]-2* D [ x [ t ], t ]+2* x [ t ]==0,{ x [0]==0, Derivative [1][ x ][0] == 1}},
x[t],t,IncludeSingularSolutions-> True ]
\[
x(t)\to e^t \sin (t)
\]