I have a problem with understanding solution some differential equation. This equation is: $$z'(t) = \frac{\partial f}{\partial x_0} (t, \phi(t,x_0)) \cdot z(t)$$ with initial condition $z(0) = 1$.

It comes from the book Differential equations, dynamical systems, and an introduction to chaos, see page 13.

The authors report that the solution of this differential equation is function: $$z(t) = \exp\left(\int_0^t \frac{\partial f}{\partial x_0} (s, \phi(s,x_0)\,{\rm d}s\right)$$

I'm trying slove that. In first step via separation of variablos I can compute that:

$$\frac{{\rm d}z}{z} = \frac{\partial f}{\partial x_0} (t, \phi(t,x_0))\,{\rm d}t$$

In next step I'm trying integrate both sides od the equation in limits from $0$ to $t$: $$\left.\ln(z)\right|_0^t = \int_0^t \frac{\partial f}{\partial x_0} (s, \phi(s,x_0)\,{\rm d}s$$

Then from definition of logarithm I obtain the following result:

$$z|_0^t = \exp\left(\int_0^t \frac{\partial f}{\partial x_0} (s, \phi(s,x_0)\,{\rm d}s\right)$$

Here is a problem, because on the left side of the equation I have in final form: $z(t) -1$. Could someone exlain me why my solution is other than solution from book? I will be grateful for your help Best regards ;)

  • 2
    $\begingroup$ Note that $\left.\ln(z)\right|_0^t = \ln(z(t)) - \ln(z(0)) = \ln[z(t) / z(0)]$ $\endgroup$ – Winther Sep 25 '16 at 19:56
  • 2
    $\begingroup$ Inserting bounds should be done in the LHS in the 2nd last equation: $\ln(z(t))-\ln(z(0))=\ln(z(t))-\ln(1) = \ln(z(t))$ $\endgroup$ – H. H. Rugh Sep 25 '16 at 19:56
  • $\begingroup$ Thanks so much for answers ;) $\endgroup$ – Krzysztof Michalski Sep 25 '16 at 20:01

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