Consdier the ODE
y'(t) = A\cdot y(t) +B.
Let y be the solution with
y(t) = {\color{teal}C} \cdot e^{{\color{red}a} \cdot t} + {\color{blue}b}
and y(0) = Y0.
Determine the value \color{orange}y \left(fractionReduce(L,A) \ln(Y1)
\right).
\color{orange}y \left(fractionReduce(L,A) \ln(Y1)\right)
=
C*pow(Y1,L)-Q
Calculate {\color{red}a}, {\color{blue}b}, {\color{teal}C} as in one of the other problems:
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}.
This implies that {\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},to find {\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 yields {\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.
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).
Therefore, the solution function is: y(t) = C \cdot e^{A \cdot t} - Q.
With t = fractionReduce(L,A) \ln(Y1)
this gives \color{orange}y \left(fractionReduce(L,A) \ln(Y1)
\right) =
C*pow(Y1,L)-Q.