start of lecture 1
Intro (Newton example):
The prof decided to open with a real-world problem where we find the equations that describe a falling object using differential equations (DE's):
We know $F=ma$
$F=m\frac{dv}{dt}=mg-kv$ we account for air resistance here. We can approximate that the force of air resistance is proportional to the speed times a constant k.
We can rearrange and solve it as it is a separable DE:
$\frac{dv}{mg-kv}=\frac{dt}{m}$
integrating both sides:
$\int \frac{dv}{mg-kv}=\frac{t}{m}+C$
let $u=mg-kv \quad du=-kdv$
$\int \frac{dv}{mg-kv}=\int \frac{du}{-k*u}=\frac{1}{-k}\ln\mid mg-kv\mid=\frac{t}{m}+C$
Very cool, but I want the velocity as a function of time, isolate v
$\ln\mid mg-kv\mid=-\frac{kt}{m}+C$
$\mid mg-kv\mid=e^{\frac{-kt}{m}+C}=e^{\frac{-kt}{m}}e^C$
$e^C$ is a + constant, the absolute value will multiply the inside expression by -1 when the inside is negative, so we can replace the $e^C$ constant with an arbitrary constant A that can be + or -
$mg-kv=Ae^{\frac{-kt}{m}}$
so, the general solution is $$v(t)=\frac{1}{k}(mg-Ae^{\frac{-kt}{m}})$$
Separable DE:
$$\frac{dy}{dx}=f(y)g(x) \rightarrow \frac{dy}{f(y)}=g(x)dx\quad where\quad f(y)\ne0$$
Since these are so similar, I'm calling these two de_separable Note that $\frac{1}{f(y)}$ is still an arbitrary function of y. So you could also say: $k(y)dy=g(x)dx$ is a separable equation.
$$\frac{dy}{dt}=\frac{1-t^2}{y^2}$$
$y^2dy=dt(1-t^2)$
integrating both sides yields:
$\frac{y^3}{3}=t-\frac{t^3}{3}+C$
finally we get:
$$y=(3t-t^3+C)^\frac{1}{3}$$
Initial value problem (IVP):
A Differential equation with provided initial conditions.
$$\frac{dy}{dx}=2x\cos^2(y), \quad y(0)=\frac{\pi}{4}$$
$\frac{dy}{\cos^2(y)}=2xdx$
integrate both sides yields:
$\int \frac{dy}{\cos^2(y)}=\tan(y)+C=x^2$
plug in $y(0)=\frac{\pi}{4}$
$\tan\left( \frac{\pi}{4} \right)+C=0$
$1+C=0$
$C=-1$
So, the answer is:
$$y=\arctan(x^2+1)$$
end of Lecture 1