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added recommended study material (maybe too verbose now?) on index and changed some tag names

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content/Bernoulli equations (lec 3).md
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@ -1,37 +1,42 @@
# Bernoulli's equation: # Bernoulli's equation:
### $$\frac{ dy }{ dx } +P(x)y=Q(x)y^n \quad\quad n\in\mathbb{R},\quad n\ne0,1$$ ### $$\frac{ dy }{ dx } +P(x)y=Q(x)y^n \quad\quad n\in\mathbb{R},\quad n\ne0,1$$
>I'm calling this #de_b_type1. This is in standard form btw. >For now, I'll tag and refer to these as #de_bernoulli. This is in standard form btw.
It looks almost like a linear equation! In fact if $n=0$ it is by definition. We will see further that if $n=1$ you get a separable equation. So we ignore the cases when $n=0,1$ as these can be solved with prior tools. It looks almost like a linear equation! In fact if $n=0$ it is by definition. We will see further that if $n=1$ you get a separable equation. So we ignore the cases when $n=0,1$ as these can be solved with prior tools.
Bernoulli's equations are important as you will see it in biology and in engineering. Bernoulli's equations are important as you will see it in biology and in engineering.
If $y$ is $+$ then $y(x)=0$ is a solution to the equation: Our goal is to find the general solution to $y$ which is some function of $x$. We should expect one arbitrary constant in our final answer for $y$ since this is a first order differential equation.
$\frac{dy}{dx}+0=0\quad\Rightarrow \quad0=0$ Notice there's an easy solution! $y(x)=0$ is a trivial solution to any Bernoulli equation.
You can verify that by plugging in $y(x)=0$ to the original expression:
$\frac{dy}{dx}+0=0\quad \implies \quad0=0$
Now let's find the general solution.
Let's move the y to the LHS: Let's move the y to the LHS:
$y^{-n}\frac{ dy }{ dx }+P(x)y^{1-n}=Q(x)$ $y^{-n}\frac{ dy }{ dx }+P(x)y^{1-n}=Q(x)$
notice that $y(x)=0$ is no longer a solution! It was lost due to dividing by zero. So from here on out we will have to remember to add it back in our final answers. notice that $y(x)=0$ is no longer a solution! It was lost due to dividing by zero. So from here on out we will have to remember to add it back in our final answers.
let $y^{1-n}=u$ let $y^{1-n}=u$
Differentiating this with respect to x gives us: Differentiating this with respect to $x$ gives us:
$(1-n)y^{-n}\frac{ dy }{ dx }=\frac{du}{dx}$ $(1-n)y^{-n}\frac{ dy }{ dx }=\frac{du}{dx}$
$y^{-n}\frac{ dy }{ dx }=\frac{ du }{ dx }{\frac{1}{1-n}}$ $y^{-n}\frac{ dy }{ dx }=\frac{ du }{ dx }{\frac{1}{1-n}}$
substituting in we get: substituting in we get:
$y^{-n}\frac{ dy }{ dx }+P(x)u=Q(x)=\frac{ du }{ dx }{\frac{1}{1-n}+P(x)u}$ $y^{-n}\frac{ dy }{ dx }+P(x)u=Q(x)=\frac{ du }{ dx }{\frac{1}{1-n}+P(x)u}$
and we get a linear equation again: (Handy formula if you wanna solve Bernoulli equations quick. Just remember that once you find $u(x)$, substitute it back for $y(x)^{1-n}=u(x)$ to get your solution for y.) And that is a linear equation again, which can be solved with prior tools.
Here's a handy formula if you wanna solve Bernoulli equations quick:
$$\frac{1}{1-n}\frac{ du }{ dx }+P(x)u=Q(x)\quad \Box$$ $$\frac{1}{1-n}\frac{ du }{ dx }+P(x)u=Q(x)\quad \Box$$
Just remember that once you find $u(x)$, substitute it back for $y(x)^{1-n}=u(x)$ to get your solution for y, and don't forget to add $y(x)=0$ with your final answer!
>Remember when I said that when n=1 the equation becomes a separable equation?: >Remember when I said that when n=1 the equation becomes a separable equation?:
>$y^{-n}\frac{ dy }{ dx }+P(x)y^{1-n}=Q(x)$ >$y^{-n}\frac{ dy }{ dx }+P(x)y^{1-n}=Q(x)$
>let $n=1$ >let $n=1$
>$y^{-1}\frac{ dy }{ dx }+P(x)=Q(x)$ >$y^{-1}\frac{ dy }{ dx }+P(x)=Q(x)$
>$y^{-1}dy=dx(Q(x)-P(x))$ <-This is indeed a separable equation #de_s_type1 >$y^{-1}dy=dx(Q(x)-P(x))$ <-This is indeed a separable equation #de_separable
--- ---
# Examples of Bernoulli's equation: # Examples of Bernoulli's equation:
#ex #de_b_type1 Find the general solution to: #ex #de_bernoulli Find the general solution to:
$y'+y=(xy)^2$ $y'+y=(xy)^2$
Looks like a Bernoulli equation because when we distribute the $^2$ we get $x^2y^2$ on the RHS. This also tells us that n=2 Looks like a Bernoulli equation because when we distribute the $^2$ we get $x^2y^2$ on the RHS. This also tells us that n=2
$y'+y=x^2y^2$ $y'+y=x^2y^2$
$y'y^{-2}+y^{-1}=x^2$ $y'y^{-2}+y^{-1}=x^2$
>Note that we lost the y(x)=0 solution here, we will have to add it back in the end. >Note that we lost the $y(x)=0$ solution here, we will have to add it back in the end.
let $u=y^{1-n}=y^{-1}$ let $u=y^{1-n}=y^{-1}$
Differentiating wrt. $x$ we get: $\frac{du}{dx}=-y^{-2}{\frac{dy}{dx}}$ Differentiating wrt. $x$ we get: $\frac{du}{dx}=-y^{-2}{\frac{dy}{dx}}$

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content/Exact equations (lec 4-5).md
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@ -6,7 +6,7 @@ $dF=\frac{ \partial F }{ \partial x }dx+\frac{ \partial F }{ \partial y }dy=0$ s
so $F(x,y)=C$ so $F(x,y)=C$
the solution to these exact equations is given by $F()$ but how do we recover $F$ from it's partial derivatives? the solution to these exact equations is given by $F()$ but how do we recover $F$ from it's partial derivatives?
Equation of the form: $$M(x,y)dx+N(x,y)dy=0$$ Equation of the form: $$M(x,y)dx+N(x,y)dy=0$$
>I'm calling this #de_e_type1 >I'm calling this #de_exact
is called exact if $M(x,y)=\frac{ \partial F }{ \partial x }$ and $N(x,y)=\frac{ \partial F }{ \partial y }$ for some function $F(x,y)$ is called exact if $M(x,y)=\frac{ \partial F }{ \partial x }$ and $N(x,y)=\frac{ \partial F }{ \partial y }$ for some function $F(x,y)$
then differentiating we get: then differentiating we get:
@ -31,7 +31,7 @@ $F(x,y)=\int M(x,y) \, dx+g(y)$ where g is any function of y. The constant of in
now 2nd condition: $N=\frac{ \partial F }{ \partial y }=\frac{ \partial }{ \partial y }\int M(x,y) \, dx+g'(y)=N(x,y)$ now 2nd condition: $N=\frac{ \partial F }{ \partial y }=\frac{ \partial }{ \partial y }\int M(x,y) \, dx+g'(y)=N(x,y)$
to reiterate, first test if equation is exact, then take m or n and integrate with x or y respectively then differentiate with respect to y or x respectively. to reiterate, first test if equation is exact, then take m or n and integrate with x or y respectively then differentiate with respect to y or x respectively.
#ex #de_e_type1 #ex #de_exact
$$\underbrace{( 2xy+3 )}_{ M }dx+\underbrace{ (x^2-1) }_{N}dy=0$$ $$\underbrace{( 2xy+3 )}_{ M }dx+\underbrace{ (x^2-1) }_{N}dy=0$$
$\frac{ \partial M }{ \partial y }=2x=\frac{ \partial N }{ \partial x }=2x$ so its exact! $\frac{ \partial M }{ \partial y }=2x=\frac{ \partial N }{ \partial x }=2x$ so its exact!
$\frac{ \partial F }{ \partial y }=N(x,y)=x^2-1$ $\frac{ \partial F }{ \partial y }=N(x,y)=x^2-1$

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content/Linear coefficient equations (lec 4).md
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@ -2,7 +2,7 @@
#start of lecture 4 #start of lecture 4
## Linear coefficients equations ## Linear coefficients equations
$$(a_{1}x+b_{1}y+c_{1})dx+(a_{2}x+b_{2}y+c_{2})dy=0 \qquad a_{1},b_{1},c_{1},a_{2},b_{2},c_{2}\in \mathbb{R}$$ $$(a_{1}x+b_{1}y+c_{1})dx+(a_{2}x+b_{2}y+c_{2})dy=0 \qquad a_{1},b_{1},c_{1},a_{2},b_{2},c_{2}\in \mathbb{R}$$
> I'm calling this #de_LC_type1 > I'm calling this #de_LC
imagine $c_{1},c_{2}=0$ It becomes a homogenous equation! #de_h_type2 imagine $c_{1},c_{2}=0$ It becomes a homogenous equation! #de_h_type2
@ -19,7 +19,7 @@ if $\det\begin{pmatrix}a_{1} & b_{1} \\a_{2} & b_{2}\end{pmatrix}\ne0$ the syste
if $\det\begin{pmatrix}a_{1} & b_{1} \\a_{2} & b_{2}\end{pmatrix}=0 \Rightarrow$ the system is unsolvable but we get an equation of type $\frac{ dy }{ dx }=G(ax+by)$ (also homogenous) if $\det\begin{pmatrix}a_{1} & b_{1} \\a_{2} & b_{2}\end{pmatrix}=0 \Rightarrow$ the system is unsolvable but we get an equation of type $\frac{ dy }{ dx }=G(ax+by)$ (also homogenous)
### Example ### Example
#ex #de_LC_type1 #ex #de_LC
$$(-3x+y+6)dx+(x+y+2)dy=0$$ $$(-3x+y+6)dx+(x+y+2)dy=0$$
let $x=u+k$ let $x=u+k$
$y=v+l$ $y=v+l$

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content/Linear equations (lec 2-3).md
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@ -3,7 +3,7 @@
# Linear equation: # Linear equation:
$$a(x)\frac{ dy }{ dx }+b(x)y=f(x)$$ $$a(x)\frac{ dy }{ dx }+b(x)y=f(x)$$
>I'm calling this #de_L_type1 >I'm calling this #de_linear_intro
if we assume $b(x)=a'(x)$ it kinda starts to look like a product rule if we assume $b(x)=a'(x)$ it kinda starts to look like a product rule
$a(x)y'+a'(x)y=f(x)=(ay)'$ $a(x)y'+a'(x)y=f(x)=(ay)'$
@ -11,7 +11,7 @@ $ay=\int f(x) \, dx$ <-yay! We can find the solutions to y.
we can rewrite the linear equation in what's called standard form: we can rewrite the linear equation in what's called standard form:
$$\frac{ dy }{ dx }+P(x)y=Q(x)$$ $$\frac{ dy }{ dx }+P(x)y=Q(x)$$
>I'm calling this #de_L_type2 ) >I'm calling this #de_linear )
we will define a function $\mu(x)$ called the integration factor, also expressed as $I(x)$ we will define a function $\mu(x)$ called the integration factor, also expressed as $I(x)$
Multiply both sides by $\mu(x)$ Multiply both sides by $\mu(x)$
@ -29,10 +29,10 @@ finally we get that $\mu(x)=I(x)=e^{\int P(x) \, dx}\quad \Box$ #remember
--- ---
#end of lecture 2 #start of lecture 3 #end of lecture 2 #start of lecture 3
# Examples of linear equations: # Examples of linear equations:
#ex #de_L_type2 Find the general solution to the equation: #ex #de_linear Find the general solution to the equation:
## $$(1+\sin(x))y'+2\cos(x)y=\tan(x)$$ ## $$(1+\sin(x))y'+2\cos(x)y=\tan(x)$$
let $a(x)=1+sin(x)\qquad b(x)=2\cos(x)$ let $a(x)=1+sin(x)\qquad b(x)=2\cos(x)$
we can see that $b(x)\ne a'(x)$ :( so we cant use #de_L_type1 we can see that $b(x)\ne a'(x)$ :( so we cant use #de_linear_intro
let's rearrange it into standard form: let's rearrange it into standard form:
$y'+\frac{{2\cos(x)}}{1+\sin(x)}=\frac{\tan(x)}{1+\sin(x)}$ $y'+\frac{{2\cos(x)}}{1+\sin(x)}=\frac{\tan(x)}{1+\sin(x)}$
$P(x):=\frac{2\cos(x)}{1+\sin(x)} \qquad Q(x)=\frac{\tan(x)}{1+\sin(x)}$ $P(x):=\frac{2\cos(x)}{1+\sin(x)} \qquad Q(x)=\frac{\tan(x)}{1+\sin(x)}$
@ -58,7 +58,7 @@ Albeit a bit ugly, we have found the general solution to the DE:
$$y=\frac{1}{(1+\sin(x))^2}(\ln\mid sec(x)\mid+\ln\mid sec(x)+\tan(x)\mid-\sin(x)+C)$$ $$y=\frac{1}{(1+\sin(x))^2}(\ln\mid sec(x)\mid+\ln\mid sec(x)+\tan(x)\mid-\sin(x)+C)$$
--- ---
#ex #IVP #de_L_type2 #ex #IVP #de_linear
## $$y'+\tan(x)y=\cos^2(x) \qquad y\left( \frac{\pi}{4} \right)=\frac{1}{2}$$ ## $$y'+\tan(x)y=\cos^2(x) \qquad y\left( \frac{\pi}{4} \right)=\frac{1}{2}$$
Looks like a linear equation with an initial value. Looks like a linear equation with an initial value.
$P(x)=\tan(x) \qquad Q(x)=\cos^2(x) \qquad I(x)=e^{\int \tan(x) \, dx}$ $P(x)=\tan(x) \qquad Q(x)=\cos^2(x) \qquad I(x)=e^{\int \tan(x) \, dx}$

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content/Reduction of order (lec 11).md
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@ -13,14 +13,14 @@ $v\cancelto{ 0 }{ (y_{1}''+p(x)y_{1}'+q(x)y_{1}) }+v''y_{1}+(2y_{1}'+p(x)y_{1})v
$y_{1}v''+(2y_{1}'+p(x)y_{1})=f(x)$ $y_{1}v''+(2y_{1}'+p(x)y_{1})=f(x)$
$v''+\left( \frac{2y_{1}'}{y_{1}}+p(x) \right)v'=\frac{f(x)}{y_{1}}$ $v''+\left( \frac{2y_{1}'}{y_{1}}+p(x) \right)v'=\frac{f(x)}{y_{1}}$
substitute $v'=u$ substitute $v'=u$
$u'+\left( \frac{2y_{1}'}{y_{1}}+p(x) \right)u=\frac{f(x)}{y_{1}}$<- This is now a linear first order equation #de_L_type2 $u'+\left( \frac{2y_{1}'}{y_{1}}+p(x) \right)u=\frac{f(x)}{y_{1}}$<- This is now a linear first order equation #de_linear
This can be solved with prior tools now, We compute the integrating factor $\mu$ This can be solved with prior tools now, We compute the integrating factor $\mu$
$\mu=e^{\int(2y_{1}'/y_{1}+p)dx}=e^{\ln(y_{1}^2)}e^{\int p(x) \, dx}=y_{1}^2\cdot e^{\int p(x) \, dx}$ $\mu=e^{\int(2y_{1}'/y_{1}+p)dx}=e^{\ln(y_{1}^2)}e^{\int p(x) \, dx}=y_{1}^2\cdot e^{\int p(x) \, dx}$
From there, continue on as you would with any linear first order equation. From there, continue on as you would with any linear first order equation.
Isn't this nice? some kind of magic. We made some guesses and we arrived somewhere. Isn't this nice? some kind of magic. We made some guesses and we arrived somewhere.
## What you need to remember: ## What you need to remember:
#remember #remember
I know memorizing formulas robs the richness of mathematics, but that is just the nature of test taking imo. If you want to minimize the amount of work to the lowest possible level, this would be the fastest algorithm, it's a little heavy on memorization: I know memorizing formulas robs the richness of mathematics, but that is just the nature of test taking imo. If you want to minimize the amount of work to the lowest possible level, this would be the fastest algorithm (that I know of), it's a little heavy on memorization:
1) $y''+p(x)y'+q(x)y=f(x)$ 1) $y''+p(x)y'+q(x)y=f(x)$
2) $u'+\left( \frac{2y_{1}'}{y_{1}}+p(x) \right)u=\frac{f(x)}{y_{1}}$ <- Notice, if the coefficient for the $u$ term is $0$, you can treat the equation as a separable equation to minimize computation (integrate both sides to get u, then move on to step 5). Otherwise, move on to step 3. 2) $u'+\left( \frac{2y_{1}'}{y_{1}}+p(x) \right)u=\frac{f(x)}{y_{1}}$ <- Notice, if the coefficient for the $u$ term is $0$, you can treat the equation as a separable equation to minimize computation (integrate both sides to get u, then move on to step 5). Otherwise, move on to step 3.
3) $\mu(x)=y_{1}^2\cdot e^{\int p(x) \, dx}$ <- where $y_{1}$ is one of your homogenous solutions. 3) $\mu(x)=y_{1}^2\cdot e^{\int p(x) \, dx}$ <- where $y_{1}$ is one of your homogenous solutions.
@ -43,6 +43,7 @@ $u'+\left( \frac{2y_{1}'}{y_{1}}+4x \right)u=\frac{8{e^{-x^2}e^{-2x}}}{e^{-x^2}}
$u'+\underbrace{ \left( \frac{2{e^{-x^2}(-2x)}}{e^{-x^2}}+4x \right) }_{ =0 }u=8e^{-2x}$ $u'+\underbrace{ \left( \frac{2{e^{-x^2}(-2x)}}{e^{-x^2}}+4x \right) }_{ =0 }u=8e^{-2x}$
$u'=8e^{-2x}$ $u'=8e^{-2x}$
> Lucky us! This is just a separable equation. No need to treat it like a linear equation. > Lucky us! This is just a separable equation. No need to treat it like a linear equation.
integrating both sides: integrating both sides:
$u=-4e^{-2x}+c_{1}$ $u=-4e^{-2x}+c_{1}$
$v'=u=-4e^{-2x}+c_{1}$ $v'=u=-4e^{-2x}+c_{1}$

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content/Separable equations (lec 1).md
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@ -19,9 +19,9 @@ so, the general solution is $$v(t)=\frac{1}{k}(mg-Ae^{\frac{-kt}{m}})$$
## Separable DE: ## Separable DE:
$$\frac{dy}{dx}=f(y)g(x) \rightarrow \frac{dy}{f(y)}=g(x)dx\quad where\quad f(y)\ne0$$ $$\frac{dy}{dx}=f(y)g(x) \rightarrow \frac{dy}{f(y)}=g(x)dx\quad where\quad f(y)\ne0$$
>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. >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.
#ex #de_s_type1 #ex #de_separable
$$\frac{dy}{dt}=\frac{1-t^2}{y^2}$$ $$\frac{dy}{dt}=\frac{1-t^2}{y^2}$$
$y^2dy=dt(1-t^2)$ $y^2dy=dt(1-t^2)$
integrating both sides yields: integrating both sides yields:
@ -32,7 +32,7 @@ $$y=(3t-t^3+C)^\frac{1}{3}$$
## Initial value problem (IVP): ## Initial value problem (IVP):
A Differential equation with provided initial conditions. A Differential equation with provided initial conditions.
#ex #IVP #de_s_type1 #ex #IVP #de_separable
$$\frac{dy}{dx}=2x\cos^2(y), \quad y(0)=\frac{\pi}{4}$$ $$\frac{dy}{dx}=2x\cos^2(y), \quad y(0)=\frac{\pi}{4}$$
$\frac{dy}{\cos^2(y)}=2xdx$ $\frac{dy}{\cos^2(y)}=2xdx$
integrate both sides yields: integrate both sides yields:

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content/_index.md
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@ -32,9 +32,14 @@ I have written these notes for myself, I thought it would be cool to share them.
[Big LT table (.png)](drawings/bigLTtable.png) [Big LT table (.png)](drawings/bigLTtable.png)
[Small LT table (.png)](drawings/smallLTtable.png) [Small LT table (.png)](drawings/smallLTtable.png)
</br> </br>
# Recommended material: # Recommended study material
For the midterm exam, I highly recommend watching this video by The Math Sorcerer: [youtube.com/watch?v=kIZpbeE_yTc](https://youtube.com/watch?v=kIZpbeE_yTc) For the midterm exam, I highly recommend watching this video by The Math Sorcerer: [youtube.com/watch?v=kIZpbeE_yTc](https://youtube.com/watch?v=kIZpbeE_yTc)
From my experience, studying off this video was by far the best use of my time. Try each question yourself and follow his solution to check. From my experience, studying off this video was by far the best use of my time. Try each question yourself and follow his solution to check.
</br> </br>
For the final exam, I unfortunately couldn't find good study videos. I recommend studying PDE's hard, solidify your understanding of heat eq, driven heat eq, heat eq with non-zero end points, wave eq, and driven wave eq. Afterwards, I recommend studying power series since it's the next biggest scary monster. Finally, go over the rest of the past topics to fill your understanding and memory if you have the time. I'm thinking I should record a final exam review guide, hmmm. I'll update this text if I ever make one. For the final exam, I unfortunately couldn't find good study videos. I recommend studying PDE's hard, solidify your understanding of heat eq, driven heat eq, heat eq with non-zero end points, wave eq, and driven wave eq. Afterwards, I recommend studying power series since it's the next biggest scary monster. Finally, go over the rest of the past topics to fill your understanding and memory if you have the time.
</br>
The recommended course textbook when I took the class was: <i>Fundamentals of Differential Equations, R. Kent Nagle, Edward B. Saff and Arthur D. Snider, 9th Edition</i> Which is a good textbook imo, although I didn't use it often.
</br>
I mostly studied the material by attending the lectures and then reviewing/revising these typed notes on the bus or at home, often relying on my prof's notes on eclass in case I copied off the whiteboard wrong/couldn't keep up. (eclass is the name of my university's online class management system.)
Of course there may still be mistakes riddled throughout so as of Jan 5th 2024, <b>I'm offering 1$ CAD in bounties for every mistake reported to my email/git repo, at least until supplies last.</b> General editing and formatting changes are also gladly welcomed through the git repository below or by email. Seeing people use these notes and benefitting from it makes me happy, so thanks for sticking around :) and remember to use what you learn for good! And to lead life with honor and integrity and be ethical engineers. Dr. Minev used to never forget to stress the importance of this in his lectures, and I wholeheartedly agree.
</br> </br>