Consider the ODE
y'(t) = A\cdot y(t) +B.
Determine {\color{red}a}, {\color{blue}b} and {\color{teal}C},
such that the function y with
y(t) = {\color{teal}C} \cdot e^{{\color{red}a} \cdot t} + {\color{blue}b}
satisfies this ODE with y(0) = Y0.
\color{red}a
=
A
\color{blue}b
=
-B/A
\color{teal}C
=
Y0+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.
Then, 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, {\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) .
Since the equation must hold for all real numbers C, it must also be valid for C = 0.
Solve A {\color{blue}b}+ B=0 for {\color{blue}b}.
This gives {\color{blue}b} = fractionReduce(-B,A).
We now use this {\color{blue}b} in {\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)
and set 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 eventually {\color{red}a} = A.
To find {\color{teal}C}, use the condition y(0) = Y0.
y(0) = Y0 = {\color{teal}C} \cdot e^{{\color{red}A} \cdot 0} +
{\color{blue}fractionReduce(-B,A)}
, which leads to {\color{teal}C} = Y0 - {\color{blue}fractionReduce(-B,A)}
= fractionReduce(Y0*A+B, A).