Finding solution of first order non linear equation

Question

I have $$m\frac{dv}{dt}=mg-kv^2$$

and I want to find v(t). I tried to separate the derivative over both sides but I am getting no where. At the moment I have

$$v+\frac{v^2}{gt}=\frac{kt}{m}$$

Can someone please detail how to solve this

Answer 1

well, you can write your equation as follows

$$ \frac{ \mathrm{d}v}{\mathrm{d}t} = g - \frac{k}{m} v^2 \iff \frac{\mathrm{d}v}{g - \frac{k}{m} v^2} = \mathrm{d}t \iff \int \frac{\mathrm{d}v}{g - \frac{k}{m} v^2} = t + C $$

To sove the integral, write $$ \frac{k}{m} v^2 = \left( \sqrt{ \frac{k}{m}} v \right)^2 $$

and put $ z = \sqrt{ \frac{k}{m}} v \implies \mathrm{d}z = \sqrt{ \frac{k}{m}} \mathrm{d}v $ Hence your integral become

$$ \sqrt{\frac{m}{k}} \int \frac{\mathrm{d}z}{g - z^2}$$

I will solve the integral for the general case, and then it should easy for you to solve the required integral:

$$ \int \frac{ \mathrm{d}x}{a^2 - x^2 } = \int \frac{ \mathrm{d} x}{(a-x)(a+x)} = \int \frac{\mathrm{d}x}{2a(a-x)} - \int \frac{ \mathrm{d}x}{2a(a+x)} = \frac{1}{2a} \ln \left( \frac{a-x}{a+x} \right) + K$$

Answer 2

$$ \begin{align} m\frac{dv}{dt}&=mg-kv^2\\ m\frac{dv}{mg-kv^2}&=dt\\ \frac1g\int\frac{dv}{1-\dfrac k{mg} v^2}&=\int\ dt\\ \frac1g\int\frac{dv}{1-a^2v^2}&=\int\ dt\quad\Rightarrow\quad a^2=\dfrac k{mg}\\ \frac1{2g}\left(\int\frac{dv}{1+av}+\int\frac{dv}{1-av}\right)&=\int\ dt\\ \frac1{2g}\left(\ln(1+av)-\ln(1-av)\right)&=t+C_1\\ \ln\left(\frac{1+av}{1-av}\right)&=2gt+C_2\quad\Rightarrow\quad C_2=2gC_1\\ \frac{1+av}{1-av}&=e^{\large 2gt+C_2}\\ \frac{1+av}{1-av}&=Ke^{\large 2gt}\quad\Rightarrow\quad K=e^{\large C_2}\\ 1+av&=Ke^{\large 2gt}-aKe^{\large 2gt}v\\ av+aKe^{\large 2gt}v&=Ke^{\large 2gt}-1\\ \left(Ke^{\large 2gt}+1\right)av&=Ke^{\large 2gt}-1\\ v&=\frac{Ke^{\large 2gt}-1}{a\left(Ke^{\large 2gt}+1\right)}\\ v(t)&=\sqrt{\frac{mg}{k}}\left(\frac{Ke^{\large 2gt}-1}{Ke^{\large 2gt}+1}\right). \end{align} $$