TITLE: Cauchy differential equation
QUESTION [5 upvotes]: I'm trying to resolve this cauchy problem:
$ y'=2y+1$ such as $y(0)=1$
the general integral for the differential equation is $\frac{1}{2}(e^{2x+2c_1}-1)$
for $y(0)=1$ :
$y(0)=\frac{1}{2}(e^{2c_1}-1)=1$
my doubt is about the fact that i don't know how to "get down" that $c_1$ and i can't resolve the problem, can you help me with this one? 
Thank you for your time.

REPLY [2 votes]: $$y(x)=\frac{1}{2}(e^{2x+2c_1}-1)$$
$$t(x)=\frac{1}{2}(e^{2x}e^{2c_1}-1)$$
$e^{2c_1}$ is just a constant 
$$y(x)=\frac{1}{2}(Ke^{2x}-1) $$
And therefore
$$y(0)=1 \implies K=3 \implies y(x)=\frac{1}{2}(3e^{2x}-1)$$
edit
$$y(0)=\frac{1}{2}(e^{2c_1}-1)=1$$
$$\implies e^{2c_1}=3 $$
so substitute to $e^{2c_1}$ its value 3...you don't need to evaluate $c_1$
$$\frac{1}{2}(e^{2c_1}-1)=1$$
$$(e^{2c_1}-1)=2$$
$$e^{2c_1}=3$$
$$e^{2c_1}=e^{\ln 3}$$
$$(c_1= \frac {\ln 3}2)$$