added lec 18 and fixed broken link

content/Heaviside Unit step function (lec 18).md

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+
+# (Heaviside) Unit step function
+We got a joke! Heaviside (British electrician) was told during a restaurant that mathematicians are finally using his formula, he replied: I don't need the chef to tell me the food was good. (a lil bit cynical!)
+Anyways, the Heaviside step function is defined as:
+
+$u(t-a)=\begin{cases}0,\quad t<a \\1,\quad a\leq t\end{cases}$
+graph it, it just "switches on" when $t=a$
+if we use laplace transforms we will see that solving equations with this Heaviside function is very natural. Using #mouc wont work, #voparam might work but it's messy.
+
+#ex
+compute:
+$\mathcal{L}\{u(t-a)f(t-a)\}$
+$\int _{0} ^\infty e^{-st}u(t-a)f(t-a)\, dt$
+$=\int _{a}^\infty e^{-st}\cdot 1f(t-a)\, dt$
+let $x=t-a$
+$=\int _{0} ^\infty e^{-s(x+a)}f(x)\, dx$
+$=e^{-as}\int _{0} ^\infty e^{-sx}f(x)\, dx$
+the right side is nothing but the LT of f. How nice!
+$$\mathcal{L}\{u(t-a)f(t-a)\}=e^{-as}F(s)$$
+Very useful formula! ^
+$\mathcal{L}\{u(t-a)f(t)\}=\mathcal{L}\{u(t-a)f(t+a-a)\}=e^{-as}\mathcal{L}\{f(t+a)\}$
+$\mathcal{L}^{-1}\{e^{-as}F(s)\}=u(t-a)f(t-a)$
+#ex 
+multiple functions being switched:
+$f(t)=\begin{cases} 1, \quad t<1 \\ t, \quad 1\leq t\leq \pi \\ \sin(t), \quad \pi\leq t\end{cases}$
+then we express it using heaviside functions:
+$f(t)=1-u(t-1)+tu(t-1)-tu(t-\pi)+\sin(t)u(t-\pi)$
+think it about switching on certain parts of the equation at certain times t.
+
+#ex 
+current in electric circuit I(t) is defined by:
+$I''+I=g(t),\quad I(0)=I'(0)=0$
+$g(t)=\begin{cases}1,\quad 0\leq t<\pi \\ -1,\quad \pi\leq t\end{cases}$
+find $I(t)$
+the capacitance here is huge, 1F! (I believe it comes from the coefficient of $y$ term)
+and we are switching it instantaneously, lets imagine we don't blow anything up.
+$I''+I=1-2u(t-\pi)$
+$\mathcal{L}\{I(t)\}=J(s)$
+$s\cdot I(0)=s\cdot I'(0)=0$
+using formula we derived in earlier ex:
+$s^2J(s)+J=\frac{1}{s}-\frac{2e^{-\pi s}}{s}$
+$J=\underbrace{ \frac{1}{s(s^2+1)} }_{ =F(s) }-2 \frac{1}{s(s^2+1)}e^{-\pi s}$
+$\mathcal{L}^{-1}\left\{ \frac{1}{s(s^2+1)} \right\}=\mathcal{L}^{-1}\{\frac{1}{s}-\frac{s}{s^2+1}\}=1-\cos t$
+$=1-\cos(t)-2u(t-\pi)(1-\cos(t+\pi))$
+$$=1-\cos(t)-2u(t-\pi)(1+\cos(t))$$
+#end of lec 18

content/Reduction of order (lec 11).md

@@ -15,7 +15,7 @@ $v''+\left( \frac{2y_{1}'}{y_{1}}+p(x) \right)v'=\frac{f(x)}{y_{1}}$
 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 
 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.
 Isn't this nice? some kind of magic. We made some guesses and we arrived somewhere.
 ## What you need to remember:
@@ -24,10 +24,10 @@ I know memorizing formulas robs the richness of mathematics, but that is just th
 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.
 3) $\mu(x)=y_{1}^2\cdot e^{\int p(x) \, dx}$ <- where $y_{1}$ is one of your homogenous solutions.
-4) $u(x)=\frac{1}{\mu(x)}\int \mu(x) \frac{f(x)}{y_{1}} \, dx$ <-don't confuse $q(x)$ with $\frac{f(x)}{y_{1}}$
+4) $u(x)=\frac{1}{\mu(x)}\int \mu(x) \frac{f(x)}{y_{1}} dx$ <-don't confuse $q(x)$ with $\frac{f(x)}{y_{1}}$
-5) $y(x)=y_{1}(x)\int u(x) \, dx$ is your general solution.
+5) $y(x)=y_{1}(x)\int u(x) dx$ is your general solution.
 
-Notice the gotcha, we have three nested integrals. If you notice any of the integrals is difficult to compute, you might have to give up and use another method like #mouc or #voparam. No method is better than the other. For example, solving $y''+y'+y=xe^x$ is trivial using #mouc but practically impossible using #reduction_of_order 
+Notice the gotcha, we have three nested integrals. If you notice any of the integrals is difficult to compute, you might have to give up and use another method like #mouc or #voparam . No method is better than the other. For example, solving $y''+y'+y=xe^x$ is trivial using #mouc but practically impossible using #reduction_of_order .
 
 # Reduction of order example
 #ex #second_order_nonhomogenous  #reduction_of_order 

content/Solving IVP's using Laplace transform (lec 17-18).md

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+#start of lec 17
+summary of all the equations (separable linear exact, ...) available on eclass
+(homogenous (2 forms), Bernoulli equation, linear coefficients,)
+
+second order equations:
+put it in standard form
+solve the homogenous eq, 
+with constant coefficients
+3 cases:  overdamped critically damped under damped
+now find particular solution to get general solution $y=y_{h}+y_{p}$
+either use #mouc  or #voparam 
+
+we also have Cauchy-Euler equations (two methods, either guess y=x^r or use the general formula)
+
+reduction of order is a technique when you're given a solution and you have to find the other
+
+mechanical systems $my''+by'+ky=F_{o}\cos(\gamma t)$ (he recommends solving with mouc) that gives you the transient solution.
+the particular solution is the steady part.
+recall $\mu$=amplitude $\gamma_{r}$=resonance freq $\mu(\gamma_{_{r}})$=amplitude at resonance
+
+Laplace transforms: definition and 4 important properties
+That's all that will be covered on the midterm. 25 minutes spent covering what will be on the midterm!
+
+So, why did we learn all this stuff about laplace transforms? We will now see how its useful:
+
+#ex #LT #second_order_nonhomogenous 
+$$y''+y=t^2+2 \qquad y(0)=1 \qquad y'(0)=-1$$
+LHS=RHS, so the Laplace transforms of each side must also be equal:
+$\mathcal{L}\{y''+y\}=\mathcal{L}\{t^2+2\}$
+applying linearity:
+$\mathcal{L}\{y''\}+\mathcal{L}\{y\}=\mathcal{L}\{t^2\}+\mathcal{L}\{2\}$
+using properties:
+$s^2Y(s)-s+1+Y(s)=\frac{2!}{s^3}+\frac{2}{s}$
+factor out Y(s)
+$Y(s)(s^2+1)=\frac{2(s^2+1)}{s^3}+s-1$
+$Y(s)=\frac{2}{s^3}+\frac{s}{s^2+1}-\frac{1}{s^2+1}$
+now we take the inverse LT to obtain y(s)
+$$y(t)=t^2+\cos(t)-\sin(t)$$
+Done!
+
+second example:
+#ex #LT #second_order_nonhomogenous 
+$$y''-y=(t-2)e^{t-2} \qquad y(2)=0 \qquad y'(2)=0$$
+we need y at 0 but its as 2.
+so we make a substitution: $x=t-2$
+$\frac{d^2y}{dx^2}-y=xe^x$ <- notice the y here is not the same as the y above, lousy notation again :) 
+where $y(0)=0 \qquad \frac{dy}{dx}(0)=0$
+hit it with the LT!
+notice 1/s^2 is LT of x so we have to shift it
+$s^2Y(s)-Y(s) =\frac{1}{(s-1)^2}$
+$Y(s)=\frac{1}{(s-1)^3(s+1)}=\frac{A}{s-1}+\frac{B}{(s-1)^2}+\frac{C}{(s-1)^3}+\frac{D}{s+1}$
+$\frac{{A(s-1)^2(s+1)+B(s-1)(s+1)+C(s+1)+D(s-1)^3}}{(s-1)^3(s+1)}$
+$\begin{matrix}A+D=0  \\-A-B-3D=0 \\ -A+c+3D=0 \\ A-B+C-D=1\end{matrix}$
+$A=\frac{1}{8}$
+$B=-\frac{1}{4}$
+$C=\frac{1}{2}$
+$D=-\frac{1}{8}$
+plug into expression then take inv LT to obtain y(t):
+final solution: $y(x)=\frac{1}{8}e^x-\frac{1}{4}xe^x+\frac{1}{4}x^2e^x-\frac{1}{8}e^-x$
+where x=t-2
+all done!
+#end of lec 17 #start of lec 18
+
+#ex 
+$$y''+ty'-2y=2 \qquad y(0)=y'(0)=0$$
+hit it with the LT!
+$\mathcal{L}\{y''\}+\mathcal{L}\{ty'\}-2\mathcal{L}\{y\}=\frac{2}{s}$
+
+$s^2Y(s)-\frac{d}{ds}\mathcal{L}\{y'\}-2Y(s)=\frac{2}{s}$
+$s^2Y-\frac{d}{ds}(sY(s))-2Y=\frac{2}{s}$
+apply product rule:
+$s^2Y-Y-s\frac{dY}{ds}=\frac{2}{s}$
+^ Boooo! another differential equation! :(
+$\frac{dY}{ds}$ lies in the s "phase space"
+$-s\frac{dY}{ds}+s\left( s-\frac{3}{5} \right)=\frac{2}{s}$
+This is a linear equation!
+divide by -s to get it in standard form
+$\frac{dY}{ds}-\left( s-\frac{3}{5} \right)y=-\frac{2}{s^2}$
+compute integrating factor:
+$\mu(s)=e^{-\int (s-3/s) \, ds}=e^{-s^2/2}e^{\ln{s^3}}=s^3e^{-s^2/2}$
+$\frac{d}{ds}(s^3e^{-s^2/2}Y)=-2se^{-s^2/2}$
+
+$s^3e^{-s^2/2}Y=-2\int se^{-s^2/2} \, ds$
+use u sub.
+$u=\frac{s^2}{2}$
+$-2\int e^{-u}  \, du$
+$=2e^{-s^2/2}+C$
+$Y(s)=2s^3+C \frac{e^{s^2/2}}{s^3}$
+is this even a legitimate thing to take an inverse of?
+the lim of the expression approaches inf as s approaches inf
+So what do we do? Well we have that C term. We have to set $C=0$
+then:
+$Y(s)=\frac{2}{s^3}$
+$$y(t)=t^2$$
+we just solved a new equation that hasn't fit into our previous equation types using LT. How cool is that!

content/_index.md

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 # This is the main index
 I have written these notes for myself, I thought it would be cool to share them. These notes may be inaccurate, incomplete, or incoherent. No warranty is expressed or implied. Reader assumes all risk and liabilities.
 </br>
+Good luck on midterms! <3 -Oct 18 2023
+</br>
 [Separable equations (lec 1)](separable-equations-lec-1.html)
 [Homogenous equations (lec 2)](homogenous-equations-lec-2.html)
 [Linear equations (lec 2-3)](linear-equations-lec-2-3.html)
@@ -15,6 +17,9 @@ I have written these notes for myself, I thought it would be cool to share them.
 [Free vibrations (lec 11-12)](free-vibrations-lec-11-12.html) (raw notes, not reviewed or revised yet.)
 [Resonance & AM (lec 13-14)](resonance-am-lec-13-14.html)
 [Laplace transform (lec 14-16)](laplace-transform-lec-14-16.html) (raw notes, not reviewed or revised yet.)
+[Solving IVP's using Laplace transform (lec 17-18)](solving-ivps-using-laplace-transform-lec-17-18.html) (raw notes, not reviewed or revised yet.)
+[(Heaviside) Unit step function (lec 18)](unit-step-function) (raw notes, not reviewed or revised yet.)
+
 
 </br>
 [How to solve any DE, a flow chart](Solve-any-DE.png)