I'm currently teaching myself how to do ODE's and I have come across some questions which I'm a bit lost with:
The First:
$ x \frac {dy} {dx} + 2y = x \sqrt {y^3} $ (for x > 0).
The second I have seen very often on exam papers and am unsure how to go about it:
The population of fish in a pond is described by the Verhulst equation $$ \frac {dP} {dt} = rP(1 − \frac {P}{K}). $$ Initially, the population of fish in the pond is 800 and the carrying capacity of the fish population is K = 2000. After 2 months, the fish population has grown to 1000.
(a) How many fish will be in this pond after 5 months?
(b) After how many months will the fish population be 1300?
Any help for either topic would be greatly appreciated. Thanks
The first is a Bernoulli equation. Write it as $$\dfrac{dy}{dx}+\frac2xy=y^{3/2}\tag{1}$$ Let $v=y^{1-3/2}=y^{-1/2}$, by Chain Rule we have $$\frac{dv}{dx}=-\frac12y^{-3/2}\frac{dy}{dx}$$ So, multiplying $(1)$ by $-\frac12y^{-3/2}$ we get \begin{align*} -\frac12y^{-3/2}-\frac1xy^{-1/2}&=-\frac12\\[3pt] \frac{dv}{dx}-\frac1xv&=-\frac12\tag{2} \end{align*} Equation $(2)$ is linear. An integrating factor for this is $\mu(x)=e^{\int -\frac1xdx}=e^{-\ln x}=x^{-1}$. After multiply eq $(2)$ by $\mu(x)=x^{-1}$ becomes \begin{align*} x^{-1}\frac{dv}{dx}-x^{-2}v&=-\frac12x^{-1}\\[3pt] \dfrac{d}{dx}\left\{x^{-1}v\right\}&=-\frac12x^{-1} \end{align*} Integrating both sides give us $$x^{-1}v=-\frac12\ln x+c\quad\implies\quad v=-\frac12x\ln x+cx$$ Then $$\boxed{\color{blue}{y=\left(-\tfrac12x\ln x+cx\right)^{-2}}}$$