MATH201/content/Separable equations (lec 1).md

#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