MATH201/content/Method of undetermined coefficients (lec 8-9).md

#start of lec 8 (sept 22) last lecture we talked about ay''+b'y+cy=f(t)

in the case when f(t)=0 :

1. 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 ? 2) 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)):

1. 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:
2. 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}

1. y_h(t)=c_{1}e^{2t}+c_{2}te^{2t} (computed earlier)
2. 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.

1. 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)) 2) 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? look at it, A must be zero but if A is zero you still get problems. y_{p_{1}}'=-Ae^{-t} y_{p_{1}}''=Ae^{-t} plug in these three and we find that A=2

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

#end of lec 8 #start of lec 9 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)+P_{m}(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: s=0 if \alpha+i\beta is not a root, s=1 if \alpha+i\beta is a root.