6.3.1.3 Example 3

This ode is a standard second order representing the oscillating harmonics ode with constant coefficients and does not require Kovacic algorithm to solve it as it can be readily solved using standard method by finding the roots of the characteristic equation. It is included here in order to illustrate the Kovacic algorithm.

\begin{align*} y''+y'+y&=0\\ A y''+ B y' + C y&=0 \end{align*}

Converting it to \(z''=r z\) as shown before gives

\begin{align*} z''&=\frac {s}{t} z\\ &=\frac {-3}{4} z \end{align*}

Hence \(r=\frac {-3}{4}\). There are no poles therefore \(\Gamma =\{\}\), and \(\mathcal {O}(\infty )=\deg (t)-\deg (s)=0\). Table 6.1 shows that the necessary conditions for case one are only satisfied. Since the set \(\Gamma \) is empty, then only the quantities related to \(\mathcal {O}(\infty )\) need to be calculated. The order of \(r\) at \(\infty \) is \(O_r(\infty ) = 0\) therefore \(v=0\). \(r\) has no \(x\) in it, hence the Laurent series of \(\sqrt r\) at \(\infty \) is itself

\begin{align*} \sqrt r =\frac {i \sqrt {3}}{2} \end{align*}

Therefore

\begin{align*} a = \frac {i \sqrt {3}}{2} \end{align*}

And since \(r\) is constant then \(b=0\). Hence

\begin{alignat*}{3} [\sqrt r]_\infty &= \frac {i \sqrt {3}}{2}\\ \alpha _{\infty }^{+} &= \frac {1}{2} \left ( \frac {b}{a} - v \right ) &&= \frac {1}{2} \left ( \frac {0}{\frac {i \sqrt {3}}{2}} - 0 \right ) &&= 0\\ \alpha _{\infty }^{-} &= \frac {1}{2} \left ( -\frac {b}{a} - v \right ) &&= \frac {1}{2} \left ( -\frac {0}{\frac {i \sqrt {3}}{2}} - 0 \right ) &&= 0 \end{alignat*}

This completes step 1 for case one. Step 2 searches for non-negative integer \(d\) using

\begin{align*} d&=\alpha _{\infty }^{\pm }-\sum _{c\in \Gamma }\alpha _{c}^{\pm } \end{align*}

Since there are no poles then

\begin{align*} d &= \alpha _\infty ^{-} \\ &= 0 \end{align*}

Since \(d\) is non-negative integer integer it can be used to find \(\omega \) using

\begin{align*} \omega &= \sum _{c \in \Gamma } \left ( s(c) [\sqrt r]_c + \frac {\alpha _c^{s(c)}}{x-c} \right ) + s(\infty ) [\sqrt r]_\infty \end{align*}

The above reduces to

\begin{align*} \omega &= (-) [\sqrt r]_\infty \\ &= 0 + (-) \left ( \frac {i \sqrt {3}}{2} \right ) \\ &= -\frac {i \sqrt {3}}{2} \end{align*}

Now that \(\omega \) is determined, the next step is find a corresponding minimal polynomial \(p(x)\) of degree \(d=0\) to solve the ode. The polynomial \(p(x)\) needs to satisfy the equation

\begin{align*} p'' + 2 \omega p' + \left ( \omega ' +\omega ^2 -r\right ) p &= 0\tag {1} \end{align*}

Since \(d=0\) then let \(p(x) = 1\). Substituting this in the above gives

\begin{align*} \left (0\right ) + 2 \left (-\frac {i \sqrt {3}}{2}\right ) \left (0\right ) + \left ( \left (0\right ) + \left (-\frac {i \sqrt {3}}{2}\right )^2 - \left (-{\frac {3}{4}}\right ) \right ) &= 0\\ 0 &= 0 \end{align*}

The equation is satisfied. Therefore the first solution to the ode \(z'' = r z\) is

\begin{align*} z(x) &= p e^{ \int \omega \,dx} \\ &= {\mathrm e}^{\int -\frac {i \sqrt {3}}{2}d x}\\ &= {\mathrm e}^{-\frac {i \sqrt {3}\, x}{2}} \end{align*}

The first solution to the original ode in \(y\) is now found from (using \(A=1,B=1\))

\begin{align*} y_1 &= z e^{ \int -\frac {1}{2} \frac {B}{A} \,dx}\\ &= z e^{ -\int \frac {1}{2} \,dx}\\ &= z e^{-\frac {x}{2}}\\ &= {\mathrm e}^{-\frac {i \sqrt {3}\, x}{2}} \left ({\mathrm e}^{-\frac {x}{2}}\right )\\ &={\mathrm e}^{-\frac {x \left (1+\mathrm {I} \sqrt {3}\right )}{2}} \end{align*}

The second solution \(y_2\) to the original ode is found using reduction of order

\begin{align*} y_2 &= y_1 \int \frac { e^{\int -\frac {B}{A} \,dx}}{y_1^2} \,dx \\ &= y_1 \int \frac { e^{-\int \,dx}}{\left (y_1\right )^2} \,dx\\ &= y_1 \int \frac { e^{-x}}{\left (y_1\right )^2} \,dx\\ &= \left ({\mathrm e}^{-\frac {x \left (1+\mathrm {I} \sqrt {3}\right )}{2}}\right ) \left (-\frac {i \sqrt {3}\, {\mathrm e}^{i \sqrt {3}\, x}}{3}\right )\\ &=-\frac {\mathrm {I}}{3} {\mathrm e}^{\frac {x \left (\mathrm {I} \sqrt {3}-1\right )}{2}} \sqrt {3} \end{align*}

Therefore the general solution is

\begin{align*} y &= c_1 y_1 + c_2 y_2\\ &= c_1 \left ({\mathrm e}^{-\frac {x \left (1+i \sqrt {3}\right )}{2}}\right ) + c_2 \left (-\frac {i {\mathrm e}^{\frac {x \left (i \sqrt {3}-1\right )}{2}} \sqrt {3}}{3}\right ) \end{align*}

Using Euler’s formula the above can be simplified to the standard looking solution

\begin{align*} y(x) &= {\mathrm e}^{-\frac {x}{2}} \left ( C_1 \sin \left (\frac {\sqrt {3}\, x}{2}\right ) + C_2 \cos \left (\frac {\sqrt {3}\, x}{2}\right ) \right ) \end{align*}