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

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MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
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
 2 years ago

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

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

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

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
In Paul's example I mean
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
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
 2 years ago

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

MathSofiya Group TitleBest 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 \[y'9y=0\]
 2 years ago

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

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

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
let's see what it is
 2 years ago

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

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
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/3c32ndOrderLinearEqns_Stu.pdf
 2 years ago

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

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

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

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
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
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
Oh that's what it means
 2 years ago

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

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

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

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

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

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
I think I got it!!!!!
 2 years ago

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

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

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

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

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