Determine the general solution of the ODE
y'(x) = \dfrac{2*Bx + C}{Bx^2 + Cx + D}(y(x) + A)^2
and the constant \color{red}C in case of the initial value y(0) = 0.
y(x)
=
-\frac{1}{\ln |Bx^2 + Cx +D| + C} - A
\color{red}C
=
-\lnabs(D) - 1/A
To get the general solution we apply the separation of variables.
We rewrite the ODE as: y'(x) = \dfrac{dy}{dx}
= \dfrac{2*Bx + C}{Bx^2 + Cx + D}(y(x) + A)^2
\implies \dfrac{1}{(y(x) + A)^2} \ dy
= \dfrac{2*Bx + C}{Bx^2 + Cx +D} \ dx.
Now we look for antiderivatives \displaystyle
\int \dfrac{1}{(y(x) + A)^2} \ dy = - \dfrac{1}{(y(x) + A)}
= \int\dfrac{2*Bx + C}{Bx^2 + Cx +D} \ dx
= \ln|Bx^2 + Cx +D| + C.
For the right-hand side use \displaystyle \int\dfrac{f'(x)}{f(x)} \ dx = \ln| f(x)| + C.
It remains to solve \displaystyle = - \dfrac{1}{(y(x) + A)}
= \ln |Bx^2 + Cx +D| + C
, i.e. to isolate y(x):
Taking the reciprocal and subtracting gives \displaystyle
y(x) = -\dfrac{1}{\ln |Bx^2 + Cx +D| + C} - A.
The constant C is obtained by setting x=0 and using the initial value y(0) = 0:
\displaystyle
y(0) = 0 = - \dfrac{1}{\ln |D| +C} - A = - \dfrac{1}{\ln (abs(D)) +C} - A
\implies {\color{red} C = - \ln(abs(D)) - fractionReduce(1,A)}.