52 lines
2.0 KiB
Markdown
52 lines
2.0 KiB
Markdown
|
|
#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_s_type1 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.
|
|
|
|
#ex #de_s_type1
|
|
$$\frac{dy}{dt}=\frac{1-t^2}{y^2}$$
|
|
$y^2dy=dt(1-t^2)$
|
|
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.
|
|
|
|
#ex #IVP #de_s_type1
|
|
$$\frac{dy}{dx}=2x\cos^2(y), \quad y(0)=\frac{\pi}{4}$$
|
|
$\frac{dy}{\cos^2(y)}=2xdx$
|
|
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
|
|
|
|
$\int \frac{1}{x^5} \, dx$
|
|
$a\in\mathbb{C}$
|
|
$\Rightarrow$
|