Consider the ODE
y''(t) - T\cdot y'(t) +D\cdot y(t) = 0
with the initial conditions y(0) = 0 und y'(0) =L2 - L1.
Determine for the solution of the IVP y the value
\color{orange}\ln \left(- \dfrac{y(K)}{y(-K)}\right)
=
K*(L1+L2)
First use the characteristic equation \lambda^2 - T\cdot \lambda +D = 0
to find the general solution of the ODE.
The two roots of the quadratic equation are \lambda_1 = L1
and \lambda_2 = L2.
Thus, the general solution is
y(t) = {\color{blue}C_1} \cdot e^{L1 \cdot t} + {\color{blue}C_2} \cdot e^{L2 \cdot t}.
The constants
{\color{blue}C_1} and {\color{blue}C_2} can be found with the initial conditions
y(0) = 0 und y'(0) =L2 - L1.
Using y(0) = 0 yields C_1 + C_2 = 0 and \color{teal}C_1 = - C_2.
With y'(0) =L2 - L1 we get the relation
\color{purple}L1 \cdot C_1 + L2 \cdot C_2 = L2 - L1.
Together with \color{teal}C_1 = - C_2 in
\color{purple}L1 \cdot C_1 + L2 \cdot C_2 = L2 - L1, we
deduce that \color{blue}C_2 = 1 and thus \color{blue}C_1= -1.
The particular solution is y(t) = e^{L2 \cdot t} - e^{L1 \cdot t} with values
y(K) = e^{L2 \cdot K} - e^{L1 \cdot K}
= e^{L2*K} - e^{L1*K} and
y(-K) = e^{L2 \cdot negParens(-K)} - e^{L1 \cdot negParens(-K)}
= e^{-L2*K} - e^{-L1*K} = \dfrac{e^{L1*K} - e^{L2*K}}{e^{L1*K + L2*K}}.
With these valued we first calculate - \dfrac{y(K)}{y(-K)}
=- \left(-e^{L1*K + L2*K} \right) = e^{L1*K + L2*K} = e^{(L1+L2)*K}
and thus arrive at \color{orange}\ln \left(- \dfrac{y(K)}{y(-K)}\right)
= K*(L1+L2).