I have the following simple DE:

$$\frac{dy}{dx} + f(x)y = 0$$

I am given that $y=1$ when $x=0$. I am also given that $\frac{dF}{dx} =f(x)$ and that $F(n) = nF(1)$.

But I am struggling to set the correct integral bounds after I have separated the variables (I am to show that $y(n) \rightarrow 0$ as $n \rightarrow \infty$):

$$\frac{dy}{y} = -f(x)dx$$

I am very confused about the integral bounds; I know that $y=1$ when $x=0$ should be used to get rid of the integration constant, but then what bounds should I use for the integral; upper bound on the right side should definitely be $n$ because we need to get $y$ in terms of $n$, but what upper bound is that on the $y$ side? this is what I mean:

$$\int_{y=?}^{y=?} f(y) dy = \int^{x=n}_{x=?}f(x) dx$$

I think I am super confused right now; because this for example makes sense to me at the moment (even though it is not correct):

$$\int^{y=1}_{y=?} f(y) dy = \int_{x=?}^{x=0} f(x) dx$$

i.e. I match the bounds. I really hope this makes sense and you understand what I mean. It is like if $x=n$ then what value does $y$ take? and then that should be $y$ upper bound. Again, I am pretty sure I am confusing my self right now.

EDIT: Ok I know how to solve that thing. I just use the indefinite integral. But my question still remains, for other cases in general, does one have to "match" the values of $y$ and $x$ in the bounds? and if not then that does not really make sense to me right now.


The solution should be $$ \int\frac{\mathrm{d}y}{y}=-\int f(x)\,\mathrm{d}x\tag{1} $$ when converted to definite integrals, this is, $$ \int_{\color{#C00000}{1}}^y\frac{\mathrm{d}t}{t}=-\int_{\color{#C00000}{0}}^x f(t)\,\mathrm{d}t\tag{2} $$ which becomes $$ \log\left(\frac y{\color{#C00000}{1}}\right)=-\int_{\color{#C00000}{0}}^x f(t)\,\mathrm{d}t\tag{3} $$ or $$ y=\color{#C00000}{1}\,e^{-\int_{\color{#C00000}{0}}^x f(t)\,\mathrm{d}t}\tag{4} $$ The red terms are those from the initial conditions.

Note that the derivative of $(2)$ is the original equation, if we simply apply the Fundamental Theorem of Calculus.

  • $\begingroup$ oh, so that is how I convert it into a definite integral! thanks $\endgroup$ – i squared - Keep it Real Apr 6 '16 at 7:18
  • $\begingroup$ I have solved it similarly just now; I used the indefinite integral, then used the initial condition and then used the value $x=n$ to find $y(n)$ $\endgroup$ – i squared - Keep it Real Apr 6 '16 at 7:19
  • $\begingroup$ Yes, there are different approaches to this. We could also get $$\log(y)+C=-\int f(x)\,\mathrm{d}x$$ and adjust the constant of integration $C$ accordingly. $\endgroup$ – robjohn Apr 6 '16 at 7:21

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