So I was reading my professor's differential equation book until I came across a weird way of calculating a initial value problem without explicity having to calculate the constant that made that initial value problem true. And I'm not sure why you can do this or why does it work, but apparently it does. I will start with an example straight from Wikipedia:

Imagine that you were given this first order differential equation:
$$ y'(t) = 0,85. y(t) $$ with a initial value of $ y(0) = 19 $

The way my teacher did this was, instead of simply integrate both sides, he used a definite integral to cover the initial value, ending up with something like this: $$ log(|y(s)|) \Big|_{t_0}^t = \int_{t_0}^t 0,85\;\mathrm{d}s $$ Where $t_0$ is the $t$ of the initial value.
Solving to this: $$ |y(t)|=e^{0,85.t}.e^{-0,85.0}.e^{log(|19|)} \implies |y(t)|=19.e^{0.85t} $$ So we got to the solution without ever needing to explicity calculate the constant. You would depend on the initial value to solve the absolute value, but still no $C$'s were written.

Conclusion, apparently if you have something like this: $$ y'(t) = a(t).y(t) $$ with initial value $y(t_0)=y_0$
This would solve to: $$ |y(t)|=|y_0|.e^{\int_{t_0}^t a(t) \;\mathrm{d}t} $$ My question is, why is this valid? Why are we evaluating the integral? And why does that find our constant without having to explicity search for her?

Thank you for your time, and I hope I can clear this out.


Your teacher is just applying the fundamental theorem of calculus (FTC): if $F'(x) = f(x)$ for all $x \in [a, b]$, then $$\int_a^b f(x)\ dx = F(b) - F(a)$$

It is common when solving first-order ODEs to state things in terms of antiderivatives instead of integrals, but it is not necessary. They are essentially the same thing. And note that the FTC only requires one anti-derivative of $f$, not the entire family. So as long as you know a function $F$ such that $F' = f$, it does not matter that there are others.

In this case, your teacher expressed $$\frac{y'(t)}{y(t)} = 0,85$$ So, $$\int_0^t \frac{y'(t)}{y(t)}\ dt = \int_0^t 0,85\ dt = 0,85t$$ Now by the chain rule for integration, $$\int_0^t \frac{y'(t)}{y(t)}\ dt = \int_{y(0)}^{y(t)} \frac{dy}{y}= \int_{19}^{y(t)} \frac{dy}{y}$$ and by the FTC, since $\frac{d}{dx}(\log |x|) = \frac{1}{x}$, $$\int_{19}^{y(t)} \frac{dy}{y} = \log |y|\left|_{19}^{y(t)}\right. = \log|y(t)| - \log 19$$ $$ \log|y(t)| - \log 19 = 0,85t$$$$\log|y(t)| = 0,85t + \log 19$$ In this case, it is not really true that we have not had to find $C$. We have found it. It is $\log 19$. There is no need to estimate a decimal expression for it, as we aren't planning to use it in this form. We want an equation solved for $y$, so: $$e^{\log |y(t)|} = e^{0,85t + \log 19} = e^{0,85t}e^{\log 19}$$ $$|y(t)| = 19e^{0,85t}$$ We can go one step further: since $y$ is differentiable, it is also continuous, and as $|y|$ is never zero, $y$ is either always positive, or always negative. Since $y(0) = 19 > 0$, $y$ is always positive. Therefore, $$y(t) = 19e^{0,85t}$$

  • $\begingroup$ Thank you very much! Makes sence now! $\endgroup$ – PatriqDesigns Dec 17 '15 at 8:20

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