Example 2. For the curve with equation y = f(x), it is known that f'(x) is proportional to x² + 1 and the curve passes through the point, (3, 0) and (0, 36). Find the equation of the curve.

Solution: Since f '(x) is proportional to x² + 1,

= k (x² + 1), where k is a constant.

So, y = ò k(x² + 1) dx

i.e. y = ò k x² dx + ò k dx

, where C is a constant.

Now, the curve passes through the point (3, 0). Thus,
           0 = 9k + 3k + C
           Þ  12k + C = 0

The curve also passes through the point (0, 36).

So, 36 = 0 + 0 + C
     Þ C = 36

Hence 12k = -36  Þ  =-3

Thus, the equation of the curve is

with k = -3 and c = 36

Thus, the equation of the curve is

y = -x³ - 3x + 36