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i have the equation: $$y'+2y\:=1$$

and i solve it the regular way for first order differential equation: $$y'\:=1-2y$$ $$\frac{dy}{dx}=1-2y$$ $$\int \:\frac{1}{1-2y}dy=\int \:1dx$$ $$-\frac{1}{2}\int \:-\frac{2}{1-2y}dy=\int \:1dx$$ and using the integral formula: enter image description here $$-\frac{1}{2}\ln\left(\left|1-2y\right|\right)=x+\ln\left(c\right)$$ Why Symbolab omits the absolute value operator and writes: enter image description here

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    $\begingroup$ When you take the exponential of of both sides, you have $$|\pm e^{-2(x+c)}| = e^{-2(x+c)}$$ Because of this, over the reals, the absolute value drops, that is, the exponential is always positive. $\endgroup$
    – Moo
    Nov 7 '20 at 12:40
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    $\begingroup$ i.e, in the equation: $\left|y\right|=x^2$, you can also drop the absolute value operator ? Oh, think this is true. $\endgroup$
    – Ilya.K.
    Nov 7 '20 at 13:09
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    $\begingroup$ Symbolab might take this solution over the complex numbers. Then the sign difference under the logarithm is a difference of $i\pi$ in the integration constant. $\endgroup$ Nov 7 '20 at 14:06
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Since $$\int\frac{1}{1-2y}dy=\int 1dx, \text{ or }-\int\frac{1}{2y-1}dy=\int1dx $$ one has $$ -\frac12\ln|2y-1|=x+C $$ or $$ \ln|2y-1|=-2x-2C$$ So $$ 2y-1=\pm e^{-2C}e^{-2x}$$ Let $k=\pm e^{-2C}$. Then $$ y=\frac12+\frac12ke^{-2x}. $$ Since $k$ absorbs the signs, it does not matter if you have absolute value for $2y-1$ or not.

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$$y'+2y\:=1$$ With integrating factor method: $$(ye^{2x})'=e^{2x}$$ $$ye^{2x}=\dfrac 12 e^{2x}+K$$ $$\boxed {y(x)=\dfrac 12 +Ke^{-2x}}$$ Then we can rewrite this as: $$\dfrac {2y-1}K=e^{-2x} \geq 0$$ $$\ln \left (\dfrac {2y-1}K\right )=-2x $$ It seems to me that the absolute value is needed.

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