dx/dt=(x-8)^2, x(0)=c

Question

differential equations initial value problem

Answer 1

Rewrite as:

dx/(x-8)²=dt, (x-8)⁻²dx=dt, integrate wrt t: -(x-8)⁻¹=t+a where a is a constant to be found.

This can be written 1/(8-x)=t+a. When t=0, x=c, so 1/(8-c)=a.

Therefore 1/(8-x)=t+1/(8-c). Multiply through by (8-x)(8-c):

8-c=t(8-x)(8-c)+8-x, -c=64t-8ct-8xt+cxt-x=t(64-8c)+x(-8t+ct-1).

Therefore x=((64-8c)t+c)/((8-c)t+1) or x=(8t(8-c)+c)/((8-c)t+1).