integration help

Question

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How do you solve the Initial value probelm $dp/dt = 10p(1-p), p(0)=0.1$?

i need to integrate the following: $$ \frac{dr}{dt}=r(u-r) $$ u is a constant and $$t_0=0, r(0)=r_0$$ this is part of the work required to answer an example in my textbook. it being an example, the solution for r is given, but the working out is not shown. the textbook is on solving nonlinear ordinary differential equations, so the focus is not so much on showing how to integrate. iv been trying to get it, but i cant. i tried to use a method in zill for integrating bernoulli equations, that got me close. i tried integration by separation, but that left me very confused. please help. the answer given is: $$ r=\frac{ur_0}{r_0+(u-r_0)e^{-ut}} $$

Answer 1

rewrite the equation as $\frac{dr}{r(u-r)}=dt$ and observe that $\frac{1}{r(u-r)}=\frac{1}{ur}+\frac{1}{u(u-r)}$. So you now have $\int\frac{dr}{r(u-r)}dr=dt$ becomes; $\int\frac{1}{r(u-r)}dr=\int(\frac{1}{ur}+\frac{1}{u(u-r)})dr=dt$. So we have $\frac{1}{u}\ln(r)-\frac{1}{u}\ln(u-r)=t+c$ where $c$ is a constant of integration. Thus equation reduces to $\ln(\frac{r}{u-r})=u(t+c)$. Using the given conditions you find that $c=\frac{1}{u}\ln(\frac{r_0}{u-r_0})$. Substituting back into the equation you get $\ln(\frac{r}{u-r})-\ln(\frac{r_0}{u-r_0})=ut$ and taking expential you get expression ; $\frac{r}{u-r}-\frac{r_0}{u-r_0}=e^{ut}$ and simplifying gives you the required expression.

Answer 2

Using separation of variables, we get $$\frac{dr}{r(u-r)}=dt.$$ Note that $\frac{1}{r(u-r)}=\frac{1}{u}(\frac{1}{r}+\frac{1}{u-r})$. Therefore, if we integrate the above equality from 0 to $t$ and if $r(0)=r_0$, we get $$\frac{1}{u}\big((\ln r-\ln(u-r)-(\ln r_0-\ln(u-r_0))\big)=t,$$ which implies that $$\ln(\frac{r}{u-r})-\ln(\frac{r_0}{u-r_0})=ut.$$ Taking exponential, we get $$\frac{r}{u-r}=\frac{r_0}{u-r_0}e^{ut}.$$ This implies that $$r=\frac{ur_0}{r_0+(u-r_0)e^{-ut}}.$$