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MathSofiya
Big Round of Applause!!!! second-order linear equations \[y''+4y'+4y=0\] solve the differential equation
I'm looking at http://tutorial.math.lamar.edu/Classes/DE/SecondOrderConcepts.aspx I don't understand how they came to the e^3t as part of their solution
when \[y=e^{3x}\]\[\frac{dy}{dx} = 3e^{3x}\]\[\frac{d^2y}{dx^2}= 9e^{3x}\]This equation works similarly works for the function \[y=e^{-3x}\]
assume e^rx is a solution; there is an even baser steup that is gone thru that gets you to this part
but they started with \[y'-9y=0\] where did the e come from?
In Paul's example I mean
y = e^rx y' = r e^rx y'' = r^2 e^rx fill it in r^2 e^rx +4r e^rx +4 e^rx =0 factor out the e^rx e^rx (r^2 +4r +4) =0 e^rx is never zero, so that leaves us with the quadratic to solve for zeros, solve for r
From inspection, e^x is a basic function that yields itself as a result of differentiation. So we can directly use this property to assume that it would form a part of the solution to the equation that Paul is discussing. But since we need the function to give 9 times itself after two successive differentiation he used \[y = e^{3x} \]
@amistre64 Is what you have written what Paul did, or is it in response to my question? I'm sorry I'm not trying to be difficult. This is my first DE problem and this is what I'm seeing: \[y'-9y=0\rightarrow y=e^{3x}\] of course with the intermediate steps shown. Lets' just say that the differentiation for y makes sense, but I wouldn't come to think of e^{whatever} when looking at \[y'-9y=0\]
It's a new idea I guess and my brain is being difficult
there is a real analysis step that can get us to using e^rx for solutions
let's see what it is
its been awhile since i had to convince myself of it, so lets see what the google has in store :)
this gives the handwaving that i did, but it might be a better read than what i gave :) still looking for the other stuff http://www.stewartcalculus.com/data/CALCULUS%20Concepts%20and%20Contexts/upfiles/3c3-2ndOrderLinearEqns_Stu.pdf
uhm... Can you tell me what the relationship between \[P(x)\frac{d^2y}{dx^2}+Q(x)\frac{dy}{dx}+R(x)y=0\] and \[y(x)=c_1y_1(x)+c_2y_2(x)\] is? It says that they are the solution for the linear homogeneous equation, so... \[P(x)\frac{d^2y}{dx^2}+Q(x)\frac{dy}{dx}+R(x)y=c_1y_1(x)+c_2y_2(x)\]
Thats just generalizing the setup to include functions as well as constants
would this statement be correct? \[P(x)\frac{d^2y}{dx^2}+Q(x)\frac{dy}{dx}+R(x)y=c_1y_1(x)+c_2y_2(x)\]
second order diffy qs have at most 2 solutions that need to be accounted for; and any linear combination of those solutions is also a solution
Oh that's what it means
im gonna run upstairs to check out some books for the real analysis part. internet explorer is being bad
Sure, sounds good. Ok should I just accept that \[y=e^{rx}\] is always a solution?
in a sense, yes, but x^m can also be used; and there are means of solutions by using power series as well
ok so \[y=e^{xr}\] is a solution for \[y''+4y'+4y=0\] I just need to manipulate it?
ok lets see here, the auxilary equation is \[r^2+4r+4=0\] that factors into \[(r+2)(r+2)=0\] so r=-2 and r=-2 and therefore ... \[y=c_1e^{-2x}+c_2e^{-2x}\]
I think I got it!!!!!
off topic: WTF u guys r on second order diff EQ alredy?? its only been a week
No it's my online class that I'm finishing this summer. School starts next week for me @v4xN0s
So my final answer is correct right? Or do I need to do more?
your solution is "almost" correct notice that the "r" you found for the root of the characteristic equation (also called the auxillary equation) is just double. 2 and 2.\[y_1(x)=c_1e^{2x}\] and all of its constant c1s are a solution; the second y(x) we need to make sure we have ALL the solutions of a "second" degree diffyQ is to simply include an extra "x" in there.\[y_2(x)=c_2~xe^{2x}\] add the two of them together for a complete set of linear combinations of solutions\[y(x)=c_1e^{2x}+c_2~xe^{2x}\]
@amistre64 is "almost" correct too... \[y'' + 4y' + 4y = 0\] \[\implies r^2 + 4r + 4 = 0\] \[\implies (r+2)(r+2) = 0\] \[\implies r = -2 \; (2\times)\] \[\large \implies y = y_c = c_1 e^{-2x} + c_2 xe^{-2x}\]