#start of lecture 1

Intro (Newton example):

Newton example where we find the equations to 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 the force of air resistance is proportional to the speed times a constant. 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:

$\frac{dy}{dx}=f(y)g(x) \rightarrow \frac{dy}{f(y)}=g(x)dx\quad where\quad f(y)\ne0$
(I'm calling this <a class="hashtag" onclick="focusTag(this)">de_s_type1)</a>

ex: $\frac{dy}{dt}=\frac{1-t^2}{y^2}$ $y^2dy=dt(1-t^2)$ integrating both sides yields: $\frac{y^3}{3}=t-\frac{t^3}{3}+C$ $y=(3t-t^3+C)^\frac{1}{3}$

Initial value problem (IVP):

A Differential equation with provided initial conditions.

ex IVP de_s_type1 ex: $\frac{dy}{dx}=2x\cos^2(y), \quad y(0)=\frac{\pi}{4}$ $\frac{dy}{\cos^2(y)}=2xdx$ 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

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