When solving a simple first-order differential equation, say $$\frac {z^{\prime}(w)} {z(w)}=-\frac a w \text ,$$ my lecture note shows first $$\int^{w}\frac{z^{\prime}(r)} {z(r)} \mathrm d r=-a \int^{w}\frac 1 r \mathrm d r+C$$ and then $$\ln z(w)=-a \ln w+C \text .$$ I want to know how is the integral in last step done? And why do we write things like $\int^{w}$? If for example it is $\int^{w}_0$, isn't the integral undefined?
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- $\begingroup$ The point is that you can choose any starting point for your integration (for exampe 0 as you suggest), since this will only change the already arbitrary costant $C$. $\endgroup$Hugo– Hugo2020-07-26 19:19:13 +00:00Commented Jul 26, 2020 at 19:19
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You have $$ (\ln z(r))'=z'(r)/z(r)=-a/r. $$ Integrate on both sides $$ \ln(w)=-a\ln(r)+C. $$ If you use lower bound, for example v, then we do not need to add the constant $C$ and we have $$ \int_{v}^w(\ln z(r))'dr=\ln z(w)-\ln z(v) $$ equals $$ -a\int_{v}^w\frac{1}{r}dr=-a\ln w+a\ln v. $$ i.e $$ \ln z(w)=-a\ln w+\ln z(v)+a\ln v. $$ where $C=\ln z(v)+a\ln v$.
