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How to you find the general solution of #(dr)/(ds)=0.05s#?

Answer 1

#r=0.025s^2+C#

We should first separate the variables. This means we want to place all terms with #r# on one side of the equation and all terms with #s# on the other.

To do this, we can treat the differential #(dr)/(ds)# like division, meaning we can multiply both sides of the equation by #ds# to "move it" to the right side. That is, we can say that:

#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#

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Answer 2

Charles Arocha

To find the general solution of ( \frac{{dr}}{{ds}} = 0.05s ), integrate both sides with respect to ( s ).

[ \int \frac{{dr}}{{ds}} , ds = \int 0.05s , ds ]

[ r = \int 0.05s , ds ]

[ r = 0.05 \times \frac{{s^2}}{2} + C ]

where ( C ) is the constant of integration. Thus, the general solution is ( r = 0.025s^2 + C ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.