Question

can someone double check that i did these right please, diff eq. of growth models

1) dy/dx ky = ce^(kx)
2) dy/dx ky(m-y) = m/(1-ce^(-mkx))
3) dy/dx k(m-y) = m-ce^(-kt)
4) y=m(1-ce^(-kx) is derived to mcke^kx

Answer 1

what you wrote is wrong and nonsensical... if for (1) you meant \(dy/dx=ky\) and \(y=Ce^{kx}\) then yes you're correct. for (2) we have:$$\frac{dy}{dx}=ky(m-y)\\\frac1{y(m-y)}dy=k\,dx\\\left(\frac{A}y+\frac{B}{m-y}\right)dy=k\,dx$$usign partial fractions we have:$$A(m-y)+By=1\\(-A+B)y+Am=1$$hence \(A=B=1/m\)$$\frac1m\left(\frac1y+\frac1{m-y}\right)dy=k\,dx\\\frac1m\int\left(\frac1y+\frac1{m-y}\right)dy=k\int\,dx\\\log y-\log(m-y)=mkx+C\\\log(m-y)-\log y=-mkx+C\\\log\frac{m-y}{y}=-mkx+C\\\frac{m}y-1=Ce^{-mkx}\\\frac{m}y=1+Ce^{-mkx}\\y=\frac{m}{1+Ce^{-mkx}}$$

Answer 2

for (3) we have:$$\frac{dy}{dx}=k(m-y)\\\frac1{m-y}dy=k\,dx\\\int\frac1{m-y}dy=k\int\,dx\\-\log(m-y)=kx+C\\\log(m-y)=-kx+C\\m-y=Ce^{-kx}\\y=m+Ce^{-kx}$$

Answer 3

$$y=m(1-Ce^{-kx})\\y'=Cmke^{-kx}$$

Answer 4

check 3, m-y+ce^(-kx)
-y=ce^(-kx)-m
y=m-ce^(-kx)
or I'm a wrong

Answer 5

@freddy_eighty7 \(C\) and \(-C\) are equally arbitrary constants