Consider the ODE
y'(t) = A\cdot y(t) +B.
Determine {\color{red}a} and {\color{blue}b},
such that for all real numbers C the function y with
y(t) = C \cdot e^{{\color{red}a} \cdot t} + {\color{blue}b} is a solution of the ODE.
\color{red}a
=
A
\color{blue}b
=
-B/A
Substitute the function
y(t) = C \cdot e^{{\color{red}a} \cdot t} + {\color{blue}b}
into the ODE
y'(t) = A\cdot y(t) +B.
The derivative reads y'(t) = \left (C \cdot e^{{\color{red}a} \cdot t} + {\color{blue}b}\right)' =
{\color{red}a} \cdot C \cdot e^{{\color{red}a} \cdot t}.
Thus it follows {\color{red}a} \cdot C \cdot e^{{\color{red}a} \cdot t} =
A\cdot \left(C \cdot e^{{\color{red}a} \cdot t} + {\color{blue}b} \right) +B
= A \cdot C \cdot e^{{\color{red}a} \cdot t} +\left( A {\color{blue}b}+ B \right) .
As this should hold for all C, it must hold in particular if C = 0.
We thus solve A {\color{blue}b}+ B=0 for {\color{blue}b}.
This gives {\color{blue}b} = fractionReduce(-B,A).
Substitute {\color{blue}b} into {\color{red}a} \cdot C \cdot e^{{\color{red}a} \cdot t}
= A \cdot C \cdot e^{{\color{red}a} \cdot t} +\left( A {\color{blue}b}+ B \right)
with t = 0.
This results in {\color{red}a} \cdot C \cdot 1
= A \cdot C \cdot 1+\left( A {\color{blue}b}+ B \right)
= A \cdot C +0 and finally {\color{red}a} = A .