Answer:
No, and solutions is is not valid.
Step-by-step explanation:
Solving y'(x)= y(x)a|x| by parts yields y(x)=[tex]e^{\frac{a}{2}x^{2} }[/tex] is the solutions, for y(0)=0 (initial value), it yields [tex]e^{0}=0[/tex] which is not valid. Note a is assumed to be constant.
