i've a question, i'm doing an exercise of differential equations, but my result is wrong due to a step that compared with wolfram alpha i don't understood.
You can check the screenshot, how the $C$ arbitrary constant is came out from exponential? I know that i can do (for power property): $$e^{\frac{x^2}{2}}*e^C$$ but how wolfram simplified the $C$ for came out from exponential?
Why c came out of e (simplified)?
And how wolfram come out: $$sqrt(e^{1/2(x^2-2)})$$
Why when substitute $C=1/e$ into $y(x)$ this output again $e^{1/2}$?
Waiting for answer, Thanks you very much.
To address the first part of your question, the thing is that wolfram alpha simply just made $e^C$ into a constant called $c_1$. So yes you were right in that $e^{\frac{x^2}{2} + c_1} = e^{\frac{x^2}{2}} * e^c$. But if you think about it, no matter what the value of $c$ is, $e^c$ will always come out to be some constant because $c$ is not variable. I know it seems a bit confusing, but all wolfram is doing is making the equation look simpler by reducing the constant term to one single variable, $c_1$. Basically, $c_1$ = $e^c$ for the first screenshot.
To address the second part of your question, simply plug in $\frac{1}{e}$ for $c_1$
$\sqrt{c_1*e^\frac{x^2}{2} - 1}$
$\sqrt{\frac{1}{e} * e^\frac{x^2}{2} - 1}$
$\sqrt{e^{-1} * e^\frac{x^2}{2} - 1}$
$\sqrt{e^{\frac{x^2}{2} - 1} - 1}$
$\sqrt{e^{\frac{x^2}{2} - \frac{2}{2}} - 1}$
$\sqrt{e^\frac{x^2 - 2}{2} - 1}$