2
$\begingroup$

I am trying to solve the following ODE with the given boundary condition.

$${Z(t)f(t)}-\frac{d Z(t)}{d t}=0$$

with the boundary condition

$$\lim_{t\rightarrow\infty}Z(t)=0$$

I have solved in the following way:

$$\int\frac{d Z(t)}{d t}\frac{1}{Z(t)}dt=\int f(t)dt$$

$$Z(t)=e^{\int f(t)dt-C}=C_{1}e^{\int f(t)dt}$$

where $e^{-c}=C_{1}$ is constant.

How can i use the boundary condition in the solution and to get rid of the constant?

  • 0
    I would say just take the limit of Z that you found in the first few steps since if you replace Z(t) in your equation you'd have to replace all of them not just some of them. Side note, if Z is only a function of a single variable no need for partial derivatives – 2017-01-28
  • 0
    Thanks for your comment. I think i didn't phrase my question properly, the purpose is to use the boundary condition in finding the solution at time (t). If I take the limit of the solution above it will not be the solution of the ODE at time (t) but at (infinity). It hope it makes better sense now. – 2017-01-28
  • 0
    @H.H. Rugh. $f(t)$ also approaches positive infinity as t approaches positive infinity. – 2017-01-28
  • 0
    Taking the limit is how you enforce the boundary condition, you already have your solution for any time t, but with an unknown constant that needs to meet the boundary condition. But if f(t) becomes arbitrarily large only $C_1=0$ satisfies this condition and you get the trivial solution – 2017-01-28

1 Answers 1

0

Either $Z\equiv 0$ or any solution (for any $C_1$) will solve the boundary condition. It depends upon the behavior of $\int f(t)dt$ at $+\infty$ (whether ot not it goes to $-\infty$).