♥ thus y’/(yf-y) = k, where yf=6000;
♠ hence ln(yf-y) =-kx +C;
at x=0 y(0)=y0=1000, hence C= ln(yf-y0);
and we get:
ln((yf-y)/(yf-y0) =-kx; or;
yf-y =(yf-y0)*exp(-kx); at last
♦ y = 6000- 5100*exp(-0.0015*x);
dydt∝y=kyseparate the variablesdyy=kdtintegrate both sides:lny=kt+Cy(t)=Cektby plugging in t=0 we find that C the original populationy(0)=Ce0=C⟹C=y0so we gety=y0ekt