<s>[INST] Finding reason for removal of the logarithm in ODE I'm given an ODE
$$y' = \frac{x + 2y + 1}{2x + 3},~y(5/2) = 1/4.$$
To solve this I'm doing the following:


*

*Solve system of equations
$$2x_0 + 2y_0 = 1,$$
$$2x_0  = 3,$$
this gives me solutions $x_0 = \frac{3}{2}$, $y_0 = -\frac{1}{4}$. Then, I'm transforming variables $s = x + \frac{3}{2}$, $t = y - \frac{1}{4}$.

*Transform the ODE using those new variables to the form
$$t' = \frac{s + 2t}{2s}$$
and using substitution of the form $t = s\cdot u(s)$ the equation is reduced to the
$$u' = \frac{1}{2s} \implies u' = \frac{1}{2}\ln|C\cdot s|.$$
Using initial conditions constant C can be found using $y(5/2) = 1/4$, $s(5/2) = 4$, $t(4) = 0$, $u(4) = 0$, hence $0 = \ln|4C| \implies C = \pm \frac{1}{4}$.

*Putting it all together yields me
$$y = \frac{1}{2}\left(x + \frac{3}{2}\right)\ln\left|\frac{x + 3/2}{4}\right| + 1/4$$
with two domains $I_1 = (-3/2, +\infty)$ and $I_2 = (-\infty, -3/2)$.
But my textbook states that the answer is 
$$y = \frac{1}{2}\left(x + \frac{3}{2}\right)\ln\frac{x + 3/2}{4} + 1/4,~I = (-3/2, +\infty)$$


Hence the following two questions:


*

*What was the reason for removal of absolute value in the logarithm?

*Why is mine domain of validity wrong?

 [/INST] Since the initial condition is given for $x=\frac52$ we are assuming the solution for the domain $x>-\frac32\implies \frac{x + 3/2}{4}>0$.
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