I want to solve the following separable ODE to find an expression for the shock location $x=s(t)$. The initial condition I prescribe to this problem will be that $s(2)=3$.
$$\frac{ds}{dt}=\frac{s+1}{2t}$$
Here is my working for the problem:
As this is a separable ODE, we consider,
$$\int\frac{1}{s+1}ds=\int\frac{1}{2t}dt$$
$$\iff\ln(s+1)=\ln(2t)+c$$
$$\iff s+1=2t+d$$
Where we take $d:=\exp(c)$. Then, applying our initial condition, we get that $d=0$. So that our expression for the location of the shock is given by,
$$s(t)=2t-1$$
However, I should be getting $s(t)=2\sqrt {2t}-1$. What have I done wrong in the above? This should be incredibly easy and I know I'm missing something terribly simple.
HINT:
$$\frac{d(s+1)}{dt}=\frac{s+1}{2t}, \quad \frac{2 d(s+1)}{s+1}=\frac{dt}{t} $$
Integrating
$$ \frac{(s+1)^2}{t}= c_1$$
simplify getting
$$ s = \sqrt{tc_1} -1 = c_2\sqrt{t} -1 $$
$c_2$ evaluates with $s(2)=3, \rightarrow 2 \sqrt 2 $