start of lec 8 (sept 22)
last lecture we talked about $ay''+b'y+cy=f(t)$
in the case when $f(t)=0$ :
- $ay''+b'y+cy=0$
then $ar^2+br+c=0$ and solve with quadratic formula
general solutions are:
if $r_{1}\ne r_{2}\Rightarrow y_{h}(t)=c_{1}e^{r_{1}t}+c_{2}e^{r_{2}t}$ overdamped
if $r_{1}=r_{2}\Rightarrow y_{h}(t)=c_{1}e^{rt}+c_{2}te^{rt}$ critically damped
if $r_{1,2}\in \mathbb{C}\Rightarrow y_{h}(t)=e^{\alpha t}(c_{1}\cos(\beta t)+c_{2}\sin(\beta t))$ underdamped
where h means homogenous, (when $f(t)=0$ the equation is homogenous.)
But what about the case when $f(t)\ne 0$ ?
- If $y_{p}(t)$ solves 1) then the general solution to $y(t)$ is $y(t)=y_{h}(t)+y_{p}(t)$
theorem: if $p(t),\ q(t),\ f(t)$ are continuous on $I$ then the following IVP has a unique solution: $y''+p(t)y'+q(t)y=f(t)\quad \text{where}\quad y''(t_{o}),\ y'(t_{o}),\ y(t_{o})\in I$
Method of undetermined coefficients:
ex mouc Find the general solution for:
$$y''-4y'+4y=3t+9$$
The equation is certainly non-homogenous.
First we have to find general solution to the homogenous equation (ie: find $y_{h}(t))$:
- $y''-4y'+4y=0$
characteristic eq: $r^2-4r+4=0$
$r=2$ (repeated root)
$y_{h}(t)=c_{1}e^{2t}+c_{2}te^{2t}$
Good. Now we need $y_{p}(t):$
Look at the equation again: $y''+{-4}y'+4y=3t+9$
We are looking for a particular polynomial where the power is not greater than 1. Because if for example $y_{p}(t)=t^2$ then the LHS would be a degree 2 polynomial and yet the RHS is only a degree one polynomial.
So we guess that the equation will be of the form:
- $y_{p}(t)=At+B$
$y_{p}'=A,\ y_{p}''=0$
$0-4A+4(At+B)=3t+9$
$4A=3,\ -4A+4B=9$
$A=\frac{3}{4},\ B=3$
$y_{p}(t)=\frac{3}{4}t+3$ our guess worked!
general solution: $$y(t)=c_{1}e^{2t}+c_{2}te^{2t}+\frac{3}{4}t+3$$
So the big takeaway from this example is if the RHS of the eq is a polynomial of degree u, we try to find a solution as a polynomial of degree u
ex second_order_nonhomogenous mouc
Find the general solution of the following:
$$y''-4y'+4y=2e^{2t}$$
-
$y_h(t)=c_{1}e^{2t}+c_{2}te^{2t}$ (computed earlier)
-
$y_p(t)=\ ?$
we observe the RHS is some exponential, we need the function + its derivative + its second derivative to equal that, we have no option but suspect that its $Ae^{2t}$
but then the LHS becomes 0! $4Ae^{2t}-4\cdot 2Ae^{2t}+4Ae^{2t}=0$
so $Ae^{2t}$ is a wrong guess.
So what do we do? Let's try $Ate^{2t}$ take $c_{2}=A, c_{1}=0$, this does not work again. LHS becomes 0 again $A(t4e^{2t}+2e^{2t}+2e^{2t})-4A(t2e^{2t}+e^{2t})+4Ate^{2t}=0$
so let's try $At^2e^{2t}$:
$A(\cancel{ t^24e^{2t} }+2e^{2t}2t+e^{2t}2+2t2e^{2t})-4A(\cancel{ t^22e^{2t} }+e^{2t}2t)+\cancel{ 4At^2e^{2t} }$
$=8Ate^{2t}+2Ae^{2t}-8Ate^{2t}$
$=2Ae^{2t}=2e^{2t},\ A=1$ This one works!
we know the homogenous solution and the particular solution. Sum them together to get the general solution:
$$y(t)=c_{1}e^{2t}+c_{2}te^{2t}+t^2e^{2t}$$
Moral of story? if RHS is constant times $e^{2t}$ we guess with an exponent with a constant, if its homogenous we multiply by t, if still not a valid solution then we multiply by t again.
ex IVP second_order_nonhomogenous mouc
$$y''+2y'+2y=2e^{-t}+5\cos t \qquad y(0)=3,\ y'(0)=1$$
We wanna solve this IVP! We know from earlier that it must have a unique solution.
- set RHS to 0: $r^2+2r+2=0$
$r_{1,2}=-1\pm i$
someone mentions sqrt(i). sqrt(i) is interesting, but not the topic for today.
$y_{h}(t)=e^{-t}(c_{1}\cos(t)+c_{2}\sin(t))$
- $y_{p}(t)=\ ?$
RHS is much more complicated, sum of 2 functions. Lets use principle of super position:
$y_{p}(t)=y_{p_{1}}(t)+y_{p_{2}}(t)$
where $y_{p_{1}}$ solves $y''+2y'+2y=2e^{-t}$
$y_{p_{2}}$ solves $y''+2y'+2y=5\cos (t)$
lets try $y_{p_{1}}=Ae^{-t}$ Does this work?
$y_{p_{1}}'=-Ae^{-t}$
$y_{p_{1}}''=Ae^{-t}$ plug in these three and we find that A=2. Yes it works!
second equation, not so easy:
solution of $\cos(t)$ doesn't quite work because the LHS will obtain a $\sin$ term. Lets try this instead:
$y_{p_{2}}=A\cos(t)+B\sin(t)$
$y_{p_{2}}'=-A\sin(t)+B\cos (t)$
$y_{p_{2}}''=-A\cos t-B\sin t$
$(A+2B)\cos(t)+(-2A+B)\sin(t)=5\cos(t)$
$A+2B=5$
$-2A+B=0$ solving the system of linear equations yields: A=1, B=2
but $y_{p_{1}}\ne y_{p_{2}}$ because of the $e^{-t}$ term.
The general solution is:
$y(t)=c_{1}e^{-t}\cos(t)+c_{2}e^{-t}\sin t+2e^{-t}+\cos t+2\sin t$
now we solve the IVP:
$y(0)=3=c_{1}+3=3\implies c_{1}=0$
$y'(t)=c_{2}e^{-t}\cos(t)+\sin(t)(-1)e^{-t}-2e^t-\sin(t)+2\cos(t)$
$y'(0)=c_{2}+0-2+0+2$
$y'(0)=1=c_{2}$
final solution to IVP:
$$y(t)=e^{-t}(\sin t+2)+\cos t+2\sin t$$
remember in a previous example when we had to guess that $y_{p}=At^2e^{2t}$? Here is a generalized algorithm that can find $y_{p}$ when the RHS falls under the following form. Reducing the guess work to zero:
Generalized guesses for undetermined coefficients:
case i) $ay''+by'+cy=P_m(t)e^{rt}$
where $P_{m}(t)=a_{m}t^m+a_{m-1}t^{m-1}+ \dots +a_{0}$ i.e. P is a polynomial degree m.
Then we guess the particular solution is of the form: $y_{p}(t)=t^s(b_{m}t^m+b_{m-1}t^{m-1}+\dots+b_{0})e^{rt}$
where:
s=0, if r is not a root,
s=1 if r is a single root,
s=2 if r is a double root.
case ii) $ay''+by'+cy=P_{m}(t)e^{\alpha t}\cos(\beta t)+Q_{n}(t)e^{\alpha t}\sin(\beta t)$
Then we guess the particular solution is of the form: $y_{p}(t)=t^s[(A_{k}t^k+A_{k-1}t^{k-1}+\dots+A_{0})e^{\alpha t}\cos(\beta t)+(B_{k}t^k+B_{k-1}t^{k-1}+\dots+B_{0})e^{\alpha t}\sin(\beta t)]$
where:
$k=\max(m,n)$
$s=0$ if $\alpha+i\beta$ is not a root,
$s=1$ if $\alpha+i\beta$ is a root.