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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.

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  • $\begingroup$ it should be correct! thank you for you help, i'll try to do as you said $\endgroup$ – Gabriele Sortino Apr 27 '18 at 0:41
  • $\begingroup$ If the solution you showed is correct then you simply have to solve: $\frac{1}{2}\left(e^{2C_{1}}-1\right) = 1$ for $C_{1}$. You should get $C_{1} = \frac{ln(3)}{2}$. Sorry for the confusion in the first comment. $\endgroup$ – Desperados Apr 27 '18 at 0:43
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$$y(x)=\frac{1}{2}(e^{2x+2c_1}-1)$$ $$t(x)=\frac{1}{2}(e^{2x}e^{2c_1}-1)$$ $e^{2c_1}$ 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 $c_1$

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

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  • $\begingroup$ how do I assign that 3 though?, why is $c_1=\frac{ln(3)}{2} $ ? $\endgroup$ – Gabriele Sortino Apr 27 '18 at 1:06
  • $\begingroup$ ok I derive $e^2x$ right? $\endgroup$ – Gabriele Sortino Apr 27 '18 at 1:08
  • $\begingroup$ for the initial condition remeber that $y(0)=1$ means that when $x=0$ then $y=1$ @GabrieleSortino so yes thats the formula you posted with ln $\endgroup$ – Isham Apr 27 '18 at 1:09
  • $\begingroup$ @GabrieleSortino I added some lines...and yes $c_1= \ln (3 )/2$ $\endgroup$ – Isham Apr 27 '18 at 1:14
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    $\begingroup$ You are the man, thank you $\endgroup$ – Gabriele Sortino Apr 27 '18 at 1:27
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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 $d=3$.

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