revised reduction of order

content/Free vibrations (lec 11-12).md

@@ -31,7 +31,7 @@ and $\tan \phi=\frac{c_{1}}{c_{2}}$
 so we can get A and phi from c_1 and c_2.
 this under damped case also reaches 0 as t->$\infty$
 
-this system in the drawing is in free vibrattion (RHS=0 means no external force=free vibration.)
+this system in the drawing is in free vibration (RHS=0 means no external force=free vibration.)
 #ex 
 $y''+by'+25y=0 \qquad y(0)=1\quad y'(0)=0$
 1) b=0 -> no friction in the system (undamped)

content/Laplace transform (lec 14-16).md

@@ -24,11 +24,12 @@ $\mathcal{L}\{\alpha f(t)+\beta g(t)\}=\alpha \mathcal{L}\{f\}+\beta y\mathcal{L
 the LT is a linear operator. as shown above. Proof:
 $\mathcal{L}\{\alpha f(t)+\beta g(t)\}=\alpha \int_{0}^{\infty}f(t)e^{-st}  \, dt+\beta \int_{0}^{\infty}g(t)e^{-st}  \, dt \quad \Box$
 
-Def: $f(t)$ is piecewise continuous on an interval $I$ if $f(t)$ is continuous on $I$, except possibly at a /finite/ number of points of /jump/ discontinuity
+Def: $f(t)$ is piecewise continuous on an interval $I$ if $f(t)$ is continuous on $I$, except possibly at a <u>finite</u> number of points of <u>jump</u> discontinuity
 what is continuouity? the limit exists and equals the value at that point.
+So what is jump discontinuity look at the picture:
 ![[Drawing 2023-10-11 13.17.32.excalidraw]]
 Def: f(t) is of exponential order $\alpha$ if there $\exists\ T,\ M, T\geq0$ such that $\forall\ t>T : f(t)\leq Me^{\alpha t}$
-this is important so that f(t)e^{-st} doesn't go into infinity,
+this is important so that $f(t)e^{-st}$ doesn't go into infinity,
 this can be proven, but not necessary
 Theorem: If $f(t)$ is piecewise continuous and of an exponential order $\alpha$ on $[0, \infty)$, then $\mathcal{L}\{f\}$ exists $s>\alpha$
 and we are guarenteed that $s>\alpha$
@@ -133,11 +134,11 @@ $\mathcal{L}^{-1}\left\{ \frac{A}{s}+\frac{B}{s^2}+\frac{C}{s^3}+\frac{D}{s+1} \
 $=\mathcal{L}^{-1}\left\{ \frac{{As^2(s+1)+Bs(s+1)+C(s+1)+Ds^3}}{(s+1)s^3} \right\}$
 we get a linear system:
 $\begin{matrix}A+D=0  \\A+B=3 \\B+C=5 \\C=3\end{matrix}$
-after some computation for solving linear system:
+solving the linear system yeilds:
 $A=1,\ B=2,\ C=3,\ D=-1$
 so:
 $\mathcal{L}^{-1}\left\{ \frac{1}{s}+\frac{2}{s^2}+\frac{3}{s^3}+\frac{-1}{s+1} \right\}$
-$=1+2t+\frac{3}{2}t^2-e^{-t}$
+$$=1+2t+\frac{3}{2}t^2-e^{-t}$$
 is the final answer.
 
 #ex #inv_LT

content/Reduction of order (lec 11).md

@@ -18,7 +18,18 @@ This can be solved with prior tools now, We compute the integrating factor $\mu$
 $\mu=e^{\int(2y_{1}'/y_{1}+p)dx}=e^{\ln(y_{1}^2)}e^{\int P(x) \, dx}=y_{1}^2\cdot e^{\int p(x) \, dx}$
 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:
+#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:
+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 $\frac{f(x)}{y_{1}}$
+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 
+
+# Reduction of order example
 #ex #second_order_nonhomogenous  #reduction_of_order 
 Find the general solution to the equation:
 $$y''+4xy'+(4x^2+2)y=8e^{-x(x+2)}$$
@@ -32,9 +43,10 @@ $u'+\left( \frac{2y_{1}'}{y_{1}}+4x \right)u=\frac{8{e^{-x^2}e^{-2x}}}{e^{-x^2}}
 $u'+\underbrace{ \left( \frac{2{e^{-x^2}(-2x)}}{e^{-x^2}}+4x \right) }_{ =0 }u=8e^{-2x}$
 $u'=8e^{-2x}$
 > Lucky us! This is just a separable equation. No need to treat it like a linear equation.
-
+integrating both sides:
 $u=-4e^{-2x}+c_{1}$
 $v'=u=-4e^{-2x}+c_{1}$
+integrating both sides again:
 $v(x)=2e^{-2x}+c_{1}x+c_{2}$
 general solution:
 $$y(x)=v(x)y_{1}(x)=(2e^{-2x}+c_{1}x+c_{2})e^{-x^2}$$

content/Variation of parameters (lec 9-10).md

@@ -4,7 +4,6 @@
 ## Variation of parameters is a method to solve:
 ## $$ay''+by'+cy=f(t)$$
 #voparam
-Variation of parameters is generally known to be a more powerful replacement for method of undetermined coefficients.
 First, find the homogenous solution:
 $y_{h}=c_{1}y_{1}(t)+c_{2}y_{2}(t)$
 Now we need the particular solution, let $y_{p}$ be in the following form: