2.3.1.3 second order kovacic

\[\begin {aligned} x^{\prime \prime }-x&=t^{2}\\ x \left (0\right ) &= 0\\ x^{\prime }\left (0\right ) &= 1\\ \end {aligned}\]

Entering kovacic solverWriting the ode as

\begin{align*} x^{\prime \prime }-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 &= 0\tag {3} \\ C &= -1 \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 2.18: 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

\[ z_1(t) = {\mathrm e}^{-t} \]

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} \end{align*}

Since \(B=0\) then the above reduces to

\begin{gather*} \begin {aligned} x_1 &= z_1 \\ &= {\mathrm e}^{-t} \end {aligned} \end{gather*}

Which simplifies to

\begin{align*} x_1 &= {\mathrm e}^{-t} \end{align*}

The second solution \(x_2\) to the original ode is found using reduction of order

\[ x_2 = x_1 \int \frac { e^{\int -\frac {B}{A} \,dt}}{x_1^2} \,dt \]

Since \(B=0\) then the above becomes

\begin{align*} x_2 &= x_1 \int \frac {1}{x_1^2} \,dt \\ &= {\mathrm e}^{-t}\int \frac {1}{{\mathrm e}^{-2 t}} \,dt\\ &= \frac {{\mathrm e}^{t}}{2} \end{align*}

Therefore the solution is

\begin{align*} x &= c_1 x_1 + c_2 x_2 \\ &= c_1\left ({\mathrm e}^{-t}\right ) + c_2\left (\frac {{\mathrm e}^{t}}{2}\right ) \end{align*}

This is second order nonhomogeneous ODE. Let the solution be

\begin{align*} x &= x_h + x_p \end{align*}

Where \(x_h\) is the solution to the homogeneous ODE

\begin{align*} A x''(t) + B x'(t) + C x(t) &= 0 \end{align*}

And \(x_p\) is a particular solution to the nonhomogeneous ODE

\begin{align*} A x''(t) + B x'(t) + C x(t) &= f(t) \end{align*}

\(x_h\) is the solution to

\begin{align*} x^{\prime \prime }-x = 0 \end{align*}

The homogeneous solution is found using the Kovacic algorithm which results in

\begin{align*} x_h &= c_1 \,{\mathrm e}^{-t}+\frac {c_2 \,{\mathrm e}^{t}}{2} \end{align*}

The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is

\[ t^{2} \]

Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is

\[ [\{1, t, t^{2}\}] \]

While the set of the basis functions for the homogeneous solution found earlier is

\[ \left \{\frac {{\mathrm e}^{t}}{2}, {\mathrm e}^{-t}\right \} \]

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.

\begin{align*} x_p &= A_{3} t^{2}+A_{2} t +A_{1} \end{align*}

The unknowns \(\{A_{1}, A_{2}, A_{3}\}\) 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

\begin{align*} -A_{3} t^{2}-A_{2} t -A_{1}+2 A_{3} = t^{2} \end{align*}

Solving for the unknowns by comparing coefficients results in

\[ [A_{1} = -2, A_{2} = 0, A_{3} = -1] \]

Substituting the above back in the above trial solution \(x_p\), gives the particular solution

\begin{align*} x_p &= -t^{2}-2 \end{align*}

Therefore the general solution is

\begin{gather*} \begin {aligned} x &= x_h + x_p \\ &= \left (c_1 \,{\mathrm e}^{-t}+\frac {c_2 \,{\mathrm e}^{t}}{2}\right ) + \left (-t^{2}-2\right ) \end {aligned} \end{gather*}

Applying the initial conditions \([x \left (0\right ) = 0, x^{\prime }\left (0\right ) = 1]\), the solution becomes

\begin{align*} x &= -t^{2}-2+\frac {3 \,{\mathrm e}^{t}}{2}+\frac {{\mathrm e}^{-t}}{2} \\ \end{align*}

\(x = -t^{2}-2+\frac {3 \,{\mathrm e}^{t}}{2}+\frac {{\mathrm e}^{-t}}{2}\)	Direction field