Consider the ODE
y'(x) = p(x) \cdot y(x) with p(x) = 3x^2 - 2*(A+B+C)x + A*B+A*C + B*C
and y(0) = 1.
Determine the value \color{orange}\ln \left( y(L)
\right)
of the solution of the IVP.
\color{orange}\ln \left( y(L) \right)
=
pow(L,3) -(A+B+C)*L*L+(A*B+A*C+B*C)*L
The general solution of a linear homogeneous ODE takes the form y'(x) = p(x) \cdot y(x)
is y(x) = K \cdot e^{P(x)}.
Here, K is a constant and \displaystyle P(x) \in \int p(x) dx is an antiderivative of p.
We have \displaystyle P(x) \in \int p(x) dx =
\int \left( 3x^2 - 2*(A+B+C)x + A*B+A*C + B*C
\right) dx=
x^3 -(A+B+C)x^2+A*B+A*C+B*Cx+C.
The integration constant
C is determined by the initial condition y(0) = 1.
Thus C = 0 and
y(x) = e^{x^3 -(A+B+C)x^2+A*B+A*C+B*Cx}.
Substitution yields {\color{orange}\ln \left( y(L)
\right)} =
\ln \left(e^{L^3 -(A+B+C)\cdot L^2+A*B+A*C+B*C\cdot L}
\right) =
pow(L,3) -(A+B+C)*L*L+(A*B+A*C+B*C)*L.