anonymous
  • anonymous
Big Round of Applause!!!! second-order linear equations \[y''+4y'+4y=0\] solve the differential equation
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
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
anonymous
  • anonymous
when \[y=e^{3x}\]\[\frac{dy}{dx} = 3e^{3x}\]\[\frac{d^2y}{dx^2}= 9e^{3x}\]This equation works similarly works for the function \[y=e^{-3x}\]
amistre64
  • amistre64
assume e^rx is a solution; there is an even baser steup that is gone thru that gets you to this part

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
but they started with \[y'-9y=0\] where did the e come from?
anonymous
  • anonymous
In Paul's example I mean
amistre64
  • amistre64
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
anonymous
  • anonymous
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 \[y = e^{3x} \]
anonymous
  • anonymous
@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 \[y'-9y=0\]
anonymous
  • anonymous
It's a new idea I guess and my brain is being difficult
amistre64
  • amistre64
there is a real analysis step that can get us to using e^rx for solutions
anonymous
  • anonymous
let's see what it is
amistre64
  • amistre64
its been awhile since i had to convince myself of it, so lets see what the google has in store :)
anonymous
  • anonymous
sure :)
amistre64
  • amistre64
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
anonymous
  • anonymous
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... \[P(x)\frac{d^2y}{dx^2}+Q(x)\frac{dy}{dx}+R(x)y=c_1y_1(x)+c_2y_2(x)\]
amistre64
  • amistre64
Thats just generalizing the setup to include functions as well as constants
anonymous
  • anonymous
would this statement be correct? \[P(x)\frac{d^2y}{dx^2}+Q(x)\frac{dy}{dx}+R(x)y=c_1y_1(x)+c_2y_2(x)\]
amistre64
  • amistre64
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
anonymous
  • anonymous
Oh that's what it means
amistre64
  • amistre64
im gonna run upstairs to check out some books for the real analysis part. internet explorer is being bad
anonymous
  • anonymous
Sure, sounds good. Ok should I just accept that \[y=e^{rx}\] is always a solution?
amistre64
  • amistre64
in a sense, yes, but x^m can also be used; and there are means of solutions by using power series as well
anonymous
  • anonymous
ok so \[y=e^{xr}\] is a solution for \[y''+4y'+4y=0\] I just need to manipulate it?
anonymous
  • anonymous
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 ... \[y=c_1e^{-2x}+c_2e^{-2x}\]
anonymous
  • anonymous
I think I got it!!!!!
anonymous
  • anonymous
off topic: WTF u guys r on second order diff EQ alredy?? its only been a week
anonymous
  • anonymous
No it's my online class that I'm finishing this summer. School starts next week for me @v4xN0s
anonymous
  • anonymous
So my final answer is correct right? Or do I need to do more?
amistre64
  • amistre64
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\[y(x)=c_1e^{2x}+c_2~xe^{2x}\]
lgbasallote
  • lgbasallote
@amistre64 is "almost" correct too... \[y'' + 4y' + 4y = 0\] \[\implies r^2 + 4r + 4 = 0\] \[\implies (r+2)(r+2) = 0\] \[\implies r = -2 \; (2\times)\] \[\large \implies y = y_c = c_1 e^{-2x} + c_2 xe^{-2x}\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.