Big Round of Applause!!!!
second-order linear equations
\[y''+4y'+4y=0\]
solve the differential equation

- anonymous

- jamiebookeater

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- 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

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

assume e^rx is a solution; there is an even baser steup that is gone thru that gets you to this part

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## More answers

- anonymous

but they started with
\[y'-9y=0\] where did the e come from?

- anonymous

In Paul's example I mean

- 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

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

@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

It's a new idea I guess and my brain is being difficult

- amistre64

there is a real analysis step that can get us to using e^rx for solutions

- anonymous

let's see what it is

- amistre64

its been awhile since i had to convince myself of it, so lets see what the google has in store :)

- anonymous

sure :)

- 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

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

Thats just generalizing the setup to include functions as well as constants

- 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

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

Oh that's what it means

- amistre64

im gonna run upstairs to check out some books for the real analysis part. internet explorer is being bad

- anonymous

Sure, sounds good.
Ok should I just accept that
\[y=e^{rx}\] is always a solution?

- 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

ok so
\[y=e^{xr}\] is a solution for
\[y''+4y'+4y=0\]
I just need to manipulate it?

- 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

I think I got it!!!!!

- anonymous

off topic: WTF u guys r on second order diff EQ alredy?? its only been a week

- anonymous

No it's my online class that I'm finishing this summer. School starts next week for me @v4xN0s

- anonymous

So my final answer is correct right? Or do I need to do more?

- 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

@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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