Therefore, the constant of integration is: C=f (x)-F (x) =f (2)-F (2) =1-F (2) This is a simple answer, however for many students, it is very difficult to this this abstractly. So, let's look at a concrete
Solve NowIt flows out of the second container at a constant rate. Let x (t) = volume in container 1; y (t) = volume in container 2. EQ 1: dx/dt = -ax EQ 2: dy/dt = ax - b, where a, b are constants.
Technically when we integrate we should get, ∫ cos(1 +2x) +sin(1 +2x)dx = 1 2(sinu −cosu+c) ∫ cos ( 1 + 2 x) + sin ( 1 + 2 x) d x = 1 2 ( sin u − cos u + c) Since the whole
is called the constant of integration. It is easily determined that all of these functions are indeed antiderivatives of cos ( x ) {\displaystyle \cos(x)} : d d x [ sin ( x ) + C ] = d d x sin ( x ) + d d x
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the differential equation with s replacing x gives dy ds = 3s2. Integrating this with respect to s from 2 to x : Z x 2 dy ds ds = Z x 2 3s2 ds H⇒ y(x) − y(2) = s3 x 2 = x3 − 23. Solving for y(x) (and
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