# differential equation of first order - omitting absolute value operator

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: $$-\frac{1}{2}\ln\left(\left|1-2y\right|\right)=x+\ln\left(c\right)$$ Why Symbolab omits the absolute value operator and writes:

• 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.
– Moo
Nov 7 '20 at 12:40
• i.e, in the equation: $\left|y\right|=x^2$, you can also drop the absolute value operator ? Oh, think this is true. Nov 7 '20 at 13:09
• 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. Nov 7 '20 at 14:06

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.
$$y'+2y\:=1$$ 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.