How to you find the general solution of #(dr)/(ds)=0.05s#?
1 Answer
Dec 3, 2016
Explanation:
We should first separate the variables. This means we want to place all terms with
To do this, we can treat the differential
#dr=0.05scolor(white).ds#
Now we integrate both sides to undo the differentials:
#intdr=int0.05scolor(white).ds#
#intdr=0.05intscolor(white).ds#
#r=0.05(s^2/2)+C#
#r=0.025s^2+C#