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Answer 1 · 2017-06-03T01:40:11

#y=x^2/4+2#

First, integrate both sides with respect to #x#.

#intdy/dxdx = int1/2xdx#

#intdy = 1/2intxdx#

#y + C_1 = x^2/4 + C_2#

#y = x^2/4 + C_2 - C_1#

Let #C = C_2 - C_1#

#y = x^2/4 + C#

Now, to figure out what C is, plug in known #x# and #y# values and solve for #C#.

#3 = 2^2/4 + C#

#3 = 1+C#

#2 = C#

So we can therefore write #y=f(x)# like this:

#y=x^2/4+2#

Final Answer