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MathSofiya
 3 years ago
Big Round of Applause!!!!
secondorder linear equations
\[y''+4y'+4y=0\]
solve the differential equation
MathSofiya
 3 years ago
Big Round of Applause!!!! secondorder linear equations \[y''+4y'+4y=0\] solve the differential equation

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MathSofiya
 3 years ago
Best ResponseYou've already chosen the best response.1I'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

akitav
 3 years ago
Best ResponseYou've already chosen the best response.1when \[y=e^{3x}\]\[\frac{dy}{dx} = 3e^{3x}\]\[\frac{d^2y}{dx^2}= 9e^{3x}\]This equation works similarly works for the function \[y=e^{3x}\]

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1assume e^rx is a solution; there is an even baser steup that is gone thru that gets you to this part

MathSofiya
 3 years ago
Best ResponseYou've already chosen the best response.1but they started with \[y'9y=0\] where did the e come from?

MathSofiya
 3 years ago
Best ResponseYou've already chosen the best response.1In Paul's example I mean

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1y = 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

akitav
 3 years ago
Best ResponseYou've already chosen the best response.1From 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 \[y = e^{3x} \]

MathSofiya
 3 years ago
Best ResponseYou've already chosen the best response.1@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 \[y'9y=0\]

MathSofiya
 3 years ago
Best ResponseYou've already chosen the best response.1It's a new idea I guess and my brain is being difficult

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1there is a real analysis step that can get us to using e^rx for solutions

MathSofiya
 3 years ago
Best ResponseYou've already chosen the best response.1let's see what it is

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1its been awhile since i had to convince myself of it, so lets see what the google has in store :)

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1this 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/3c32ndOrderLinearEqns_Stu.pdf

MathSofiya
 3 years ago
Best ResponseYou've already chosen the best response.1uhm... 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... \[P(x)\frac{d^2y}{dx^2}+Q(x)\frac{dy}{dx}+R(x)y=c_1y_1(x)+c_2y_2(x)\]

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1Thats just generalizing the setup to include functions as well as constants

MathSofiya
 3 years ago
Best ResponseYou've already chosen the best response.1would this statement be correct? \[P(x)\frac{d^2y}{dx^2}+Q(x)\frac{dy}{dx}+R(x)y=c_1y_1(x)+c_2y_2(x)\]

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1second 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

MathSofiya
 3 years ago
Best ResponseYou've already chosen the best response.1Oh that's what it means

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1im gonna run upstairs to check out some books for the real analysis part. internet explorer is being bad

MathSofiya
 3 years ago
Best ResponseYou've already chosen the best response.1Sure, sounds good. Ok should I just accept that \[y=e^{rx}\] is always a solution?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1in a sense, yes, but x^m can also be used; and there are means of solutions by using power series as well

MathSofiya
 3 years ago
Best ResponseYou've already chosen the best response.1ok so \[y=e^{xr}\] is a solution for \[y''+4y'+4y=0\] I just need to manipulate it?

MathSofiya
 3 years ago
Best ResponseYou've already chosen the best response.1ok 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 ... \[y=c_1e^{2x}+c_2e^{2x}\]

MathSofiya
 3 years ago
Best ResponseYou've already chosen the best response.1I think I got it!!!!!

v4xN0s
 3 years ago
Best ResponseYou've already chosen the best response.0off topic: WTF u guys r on second order diff EQ alredy?? its only been a week

MathSofiya
 3 years ago
Best ResponseYou've already chosen the best response.1No it's my online class that I'm finishing this summer. School starts next week for me @v4xN0s

MathSofiya
 3 years ago
Best ResponseYou've already chosen the best response.1So my final answer is correct right? Or do I need to do more?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1your 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\[y(x)=c_1e^{2x}+c_2~xe^{2x}\]

lgbasallote
 3 years ago
Best ResponseYou've already chosen the best response.0@amistre64 is "almost" correct too... \[y'' + 4y' + 4y = 0\] \[\implies r^2 + 4r + 4 = 0\] \[\implies (r+2)(r+2) = 0\] \[\implies r = 2 \; (2\times)\] \[\large \implies y = y_c = c_1 e^{2x} + c_2 xe^{2x}\]
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