The general solution of the logistic equation is y = 14 / [1 - C · tⁿ], where a = - 14² / 3 and C is an integration constant. The particular solution for y(0) = 10 is y = 14 / [1 - (4 / 10) · tⁿ], where n = - 14² / 3.
How to find the solution of an ordinary differential equation with separable variables
Herein we have a kind of ordinary differential equation with separable variables, that is, that variables t and y can be separated at each side of the expression prior solving the expression:
dy / dt = 3 · y · (1 - y / 14)
dy / [3 · y · (1 - y / 14)] = dt
dy / [- (3 / 14) · y · (y - 14)] = dt
By partial fractions we find the following expression:
- (1 / 14) ∫ dy / y + (1 / 14) ∫ dy / (y - 14) = - (14 / 3) ∫ dt
- (1 / 14) · ln |y| + (1 / 14) · ln |y - 14| = - (14 / 3) · ln |t| + C, where C is the integration constant.
y = 14 / [1 - C · tⁿ], where n = - 14² / 3.
If y(0) = 10, then the particular solution is:
y = 14 / [1 - (4 / 10) · tⁿ], where n = - 14² / 3.
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