MATH201/content/More second order stuff (lec 8).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 solution is: y_{h}(t)=c_{1}e^{r_{1}(t)}+c_{2}e^{r_{2}t} where h means homogenous, ( because when =0 its homogenous)

if r_{1}=r_{2} then y_{h}(t)=c_{1}e^{r(t)}+c_{2}e^{rt} if imaginary roots: y_{h}(t)=e^{\alpha t}(c_{1}\cos(\beta t)+c_{2}\sin(\beta t)) 2) If y_{p}(t) solves 1) then its general solution is y(t)=y_{h}(t)+y_{p}(t) theorem: if p(t),\ g(t),\ f(t) are continuous on I then the IVP y''+p(t)y'+q(t)y=f(t), y(t_{o}),\ y'(t_{o})=y_{1} t_{o}\in I has a unique solution method of undetermined coeffecients: #ex y''\pm_{4}y'+4y=3t+9 lets find general solution, its centainly non homogenous. first we have to find general solution to the homogenous equation:

1. y''-4y'+4y=0 characteristic eq: r^2-4r+4=0 what are the roots? r=2 (repeated solution) y_{h}(t)=c_{1}e^{2t}+c_{2}te^{2t} we are looking for a particular polynomial where the power is not greater than 1 (?)
2. y_{p}(t)=At+B y_{p}'=A,\ y_{p}''=0 -4A+4(At+B)=3t+9 4A=3,\ -4A+4B=9 A=\frac{3}{4},\ B=3 y(t)=\frac{3}{4}t+3 general solution: $y(t)=c_{1}e^{2t}+c_{2}te^{2t}+\frac{3}{4}t+3$ so big takeaway is if the RHS of eq is a polynomial of degree u, we try to find a solution as a polynomial of degree u

#ex

y''-4y'+4y=e^{2t}

find general solution.

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 derivative + its second derivative to equal that, we have no option but suspect that its Ae^{2t} but then the LHS becomes 0! so Ae^{2t} is a wrong guess. so what do we do? try Ate^{2t} take c_{2}=A, c_{1}=0, this does not work again. LHS becomes 0 again so try At^2e^{2t} 2Ae^{2t}=2e^{2t},\ A=1 This one works! we know the homogeenous solution. y(t)=c_{1}e^{2t}+c_{2}te^{2t}+t^2e^{2t} is the general solution moral of sotry? 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: y''+2y'+2y=2e^{-t}+\cos t,\ y(0)=3,\ y'(0)=1 I wanna solve this IVP! it must have a unique solution.
3. set RHS to 0: r^2+2r+2=0 r_{1,2}=-1\pm i sqrt(i) is interesting, but not the topic for today. y_{h}(t)=e^{-t}(c_{1}\cos(t)+c_{2}\sin(t))
4. y_{p}(t)= RHS is much more complicated, sum of 2 functions. Lets use principle of super position y_{p}(t)=y_{p_{1}}+y_{p_{2}} 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 doenst quite work 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=0 -2A+B=0 -> A=1, B=2 but y_{p_{1}}\ne y_{p_{2}} because of the e^{-t} term y(t)=c_{1}e^{-t}\cos(t)+c_{2}e^{-t}\sin t+2e^{-t}+\cos t+2\sin t y(0)=3=c_{1}+3=3\implies c_{1}=0 y'(0)=1=c_{2} final solution y(t)=e^{-t}(\sin t+2)+\cos t+2\sin t

If we have an equation of the from:

1. 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} then the guess is: y_{p}(t)=t^s(b_{mt}t^m+b_{m-1}t^{m-1}+\dots+b_{0})e^{rt} (i) s=0 if r is not a characteristic polynomial (ii)) s=1 if r is a single root (iii) s=2 if r is a double root we will talk about this more in the coming lecture. #end of lec 8