forked from Sasserisop/MATH201
83 lines
3.6 KiB
Markdown
83 lines
3.6 KiB
Markdown
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#start of lec 8 (sept 22)
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last lecture we talked about $ay''+b'y+cy=f(t)$
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in the case when $f(t)=0$ :
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1) $ay''+b'y+cy=0$
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then $ar^2+br+c=0$ and solve with quadratic formula
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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)
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if $r_{1}=r_{2}$ then $y_{h}(t)=c_{1}e^{r(t)}+c_{2}e^{rt}$
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if imaginary roots:
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$y_{h}(t)=e^{\alpha t}(c_{1}\cos(\beta t)+c_{2}\sin(\beta t))$
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2) If $y_{p}(t)$ solves 1) then its general solution is $y(t)=y_{h}(t)+y_{p}(t)$
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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
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method of undetermined coeffecients:
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#ex
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$y''\pm_{4}y'+4y=3t+9$ lets find general solution, its centainly non homogenous.
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first we have to find general solution to the homogenous equation:
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1) $y''-4y'+4y=0$
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characteristic eq: $r^2-4r+4=0$ what are the roots?
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$r=2$ (repeated solution)
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$y_{h}(t)=c_{1}e^{2t}+c_{2}te^{2t}$
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we are looking for a particular polynomial where the power is not greater than 1 (?)
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2) $y_{p}(t)=At+B$
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$y_{p}'=A,\ y_{p}''=0$
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$-4A+4(At+B)=3t+9$
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$4A=3,\ -4A+4B=9$
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$A=\frac{3}{4},\ B=3$
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$y(t)=\frac{3}{4}t+3$
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general solution: $$y(t)=c_{1}e^{2t}+c_{2}te^{2t}+\frac{3}{4}t+3$$
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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
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#ex
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$$y''-4y'+4y=e^{2t}$$
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find general solution.
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1) $y_h(t)=c_{1}e^{2t}+c_{2}te^{2t}$ (computed earlier)
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2) $y_p(t)$
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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 $Ae^{2t}$
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but then the LHS becomes 0!
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so $Ae^{2t}$ is a wrong guess.
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so what do we do? try $Ate^{2t}$ take $c_{2}=A, c_{1}=0$, this does not work again. LHS becomes 0 again
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so try $At^2e^{2t}$
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$2Ae^{2t}=2e^{2t},\ A=1$ This one works!
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we know the homogeenous solution.
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$y(t)=c_{1}e^{2t}+c_{2}te^{2t}+t^2e^{2t}$ is the general solution
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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.
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Ex:
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$y''+2y'+2y=2e^{-t}+\cos t,\ y(0)=3,\ y'(0)=1$ I wanna solve this IVP! it must have a unique solution.
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1) set RHS to 0: $r^2+2r+2=0$
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$r_{1,2}=-1\pm i$ sqrt(i) is interesting, but not the topic for today.
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$y_{h}(t)=e^{-t}(c_{1}\cos(t)+c_{2}\sin(t))$
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2) $y_{p}(t)=$
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RHS is much more complicated, sum of 2 functions. Lets use principle of super position
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$y_{p}(t)=y_{p_{1}}+y_{p_{2}}$
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where $y_{p_{1}}$ solves $y''+2y'+2y=2e^{-t}$
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$y_{p_{2}}$ solves $y''+2y'+2y=5\cos (t)$
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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.
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$y_{p_{1}}'=-Ae^{-t}$
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$y_{p_{1}}''=Ae^-t$ plug in these three and we find that A=2
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second equation, not so easy:
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solution of cos t doenst quite work
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$y_{p_{2}}=A\cos(t)+B\sin(t)$
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$y_{p_{2}}'=-A\sin(t)+b\cos (t)$
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$y_{p_{2}}''=-A\cos t-B\sin t$
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$(A+2B)\cos(t)+(-2A+B)\sin(t)=5\cos(t)$
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$A+2B=0$
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$-2A+B=0$ -> A=1, B=2
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but $y_{p_{1}}\ne y_{p_{2}}$ because of the $e^{-t}$ term
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$y(t)=c_{1}e^{-t}\cos(t)+c_{2}e^{-t}\sin t+2e^{-t}+\cos t+2\sin t$
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$y(0)=3=c_{1}+3=3\implies c_{1}=0$
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$y'(0)=1=c_{2}$
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final solution $y(t)=e^{-t}(\sin t+2)+\cos t+2\sin t$
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If we have an equation of the from:
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1) $ay''+by'+cy=P_{m}(t)e^{rt}$
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where $p_{m}(t)=a_{m}t^m+a_{m-1}t^{m-1}+ \dots +a_{0}$
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then the guess is: $y_{p}(t)=t^s(b_{mt}t^m+b_{m-1}t^{m-1}+\dots+b_{0})e^{rt}$
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(i) s=0 if r is not a characteristic polynomial
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(ii)) s=1 if r is a single root
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(iii) s=2 if r is a double root
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we will talk about this more in the coming lecture.
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#end of lec 8 |