Question
What is the solution of $$\frac{{dy}}{{dx}} + 2y = 1$$ satisfying $$y\left( 0 \right) = 0\,?$$
A.
$$y = \frac{{1 - {e^{ - 2x}}}}{2}$$
B.
$$y = \frac{{1 + {e^{ - 2x}}}}{2}$$
C.
$$y = 1 + {e^x}$$
D.
$$y = \frac{{1 + {e^x}}}{2}$$
Answer :
$$y = \frac{{1 - {e^{ - 2x}}}}{2}$$
Solution :
$$\eqalign{
& \frac{{dy}}{{dx}} + 2y = 1\,;\,\frac{{dy}}{{dx}} = 1 - 2y \cr
& \int {\frac{{dy}}{{1 - 2y}}} = \int {dx} \cr
& - \frac{1}{2}\log \left| {1 - 2y} \right| = x + C \cr
& {\text{at }}x = 0,\,y = 0 \cr
& - \frac{1}{2}\log \,1 = 0 + C \Rightarrow C = 0 \cr
& 1 - 2y = {e^{ - 2x}}\,;\,y = \frac{{1 - {e^{ - 2x}}}}{2} \cr} $$