Question #a249f
1 Answer
Jun 3, 2017
Explanation:
First, integrate both sides with respect to
#intdy/dxdx = int1/2xdx#
#intdy = 1/2intxdx#
#y + C_1 = x^2/4 + C_2#
#y = x^2/4 + C_2 - C_1#
Let
#y = x^2/4 + C#
Now, to figure out what C is, plug in known
#3 = 2^2/4 + C#
#3 = 1+C#
#2 = C#
So we can therefore write
#y=x^2/4+2#
Final Answer