Question
The solution of $$\frac{{dy}}{{dx}} = \left| x \right|$$ is :
(Where $$c$$ is an arbitrary constant)
A.
$$y = \frac{{x\left| x \right|}}{2} + c$$
B.
$$y = \frac{{\left| x \right|}}{2} + c$$
C.
$$y = \frac{{{x^2}}}{2} + c$$
D.
$$y = \frac{{{x^3}}}{2} + c$$
Answer :
$$y = \frac{{x\left| x \right|}}{2} + c$$
Solution :
$$\eqalign{
& \frac{{dy}}{{dx}} = \left| x \right| \cr
& \frac{{dy}}{{dx}} = x{\text{ for }}x \geqslant 0\,;\frac{{dy}}{{dx}} = - x{\text{ for }}x < 0\,;\,\int {dy = } \int {x\,dx} \cr
& y = \frac{{{x^2}}}{2} + {C_1}......\left( {{\text{i}}} \right) \cr
& \int {dy = } - 1\,x\,dx \cr
& y = - \frac{{{x^2}}}{2} + {C_1}......\left( {{\text{ii}}} \right) \cr
& {\text{From }}\left( {\text{i}} \right){\text{ and }}\left( {{\text{ii}}} \right) \cr
& y = \frac{{x\left| x \right|}}{2} + C \cr} $$