Cauchy differential equation

Question

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.

Answer 1

$$y(x)=\frac{1}{2}(e^{2x+2c_1}-1)$$ $$t(x)=\frac{1}{2}(e^{2x}e^{2c_1}-1)$$ $e^{2c_1}$ 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 $c_1$

$$\frac{1}{2}(e^{2c_1}-1)=1$$ $$(e^{2c_1}-1)=2$$ $$e^{2c_1}=3$$ $$e^{2c_1}=e^{\ln 3}$$ $$(c_1= \frac {\ln 3}2)$$

Answer 2

You have $e^{2x+2x_1}=e^{2c_1}\,e^{2x}$. And, whenever you have a "function of a constant", you can think of it as a constant. So your solution is $$ y(x)=\frac12\,(d\,e^{2x}-1). $$ Then $$ 1=y(0)=\frac{d-1}2, $$ so $d=3$.