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Consider Thiele's differential equation for $t\in[0,\infty)$ (all the other functions are continuous on $[0,\infty)$, too.)

$$ \begin{align} V'(t)&=\Big(\phi(t)+\lambda(t)\Big)V(t)+\pi(t)-\lambda(t)A(t)\\ V(0)&=0 \end{align} $$

I am reading a proof about the unique solution being

$$V(t)=\int_0^t \big(\pi(s)-\lambda(s)A(s)\big)\exp\Big(\int_s^t \big(\phi(u)+\lambda(u)\big)du\Big)ds$$

So the first thing happening in the proof is that the author solves the equation

$$V'(t)=\big(\phi(t)+\lambda(t)\big)V(t)$$

and finding the solution by variation of the constant afterwards. The last equation is equivalent to

$$\frac{V'(t)}{V(t)}=\big(\phi(t)+\lambda(t)\big)$$

and therefore

$$\int_0^t\frac{V'(s)}{V(s)}ds=\int_0^t \big(\phi(s)+\lambda(s)\big)ds$$

Now he states something I do not understand:

$$\log V(t)=\int_0^t \big(\phi(s)+\lambda(s)\big)ds + c$$

In my opinion, using that $\log' V(t)= \frac{V'(t)}{V(t)}$, it should be

$$\log V(t) -\log V(0)=\int_0^t \big(\phi(s)+\lambda(s)\big)ds,$$

which seems to be not well defined, since $V(0)=0$. Is this some sort of method to solve this equation or is this just wrong? What is the procedure here?

Thanks in advance for any help!

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  • $\begingroup$ The method is not wrong in itself, but dividing by $V(t)$ requires additional explanation. It is much better to multiply both sides of $V'(t)-(\phi(t)+\lambda(t))V(t)=\pi(t)-\lambda(t)A(t)$ by the integrating factor $\exp(-\int\limits_s^t(\phi(u)+\lambda(u))\,\mathrm{d}u)$ and notice that the LHS is just the derivative of $V(t)\exp(-\int\limits_s^t(\phi(u)+\lambda(u))\,\mathrm{d}u)$. Perhaps you could take another textbook? $\endgroup$ – user539887 May 1 at 6:40
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You can't apply $V(0)=0$ because this boundary condition applies to the full equation, not the equation without the forcing term.

Instead solve this auxiliary equation with a random constant, use this solution to obtain a solution to the full equation and only then apply the boundary condition.

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