Some questions for non/homogeneous equations

  • Thread starter brandy
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what do you do when m=0, do you just have 1 term + a constant?

what do you do when there is only 1 value for m. do you have Ae^mt + Ate^mt or what?

im a little confused.

EDIT
-for differential equations i mean, when youre trying to solve for y and the equation has differentials of y in varying degrees.
 
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tiny-tim

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hi brandy! :smile:

(try using the X2 icon just above the Reply box :wink:)

what's m ? :confused:

can you write out the equation you're referring to?
 
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im just trying to teach myself how to do these equations. they are called either homogeneous differential equations or nonhomogeneous differential equations (which have a homogeneous solution and a particular solution which is combined)
questions like these:
y'' + 2y'+ 3y=0
3y'' +y'= 5 cos (t/4)
for the homogeneous stuff i'm assuming y=e^(mt) t is the variable. and going from there.
sorry i didnt define m :S


btw do i have to use latex for powers like that? does it bother people? i was going to use latex for more complicated things but if it bothers people....
 

tiny-tim

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you mean m is the root of the characteristic polynomial, and you're asking what if m = 0, or if m is a multiple root?

if m = 0, the solution is a constant (isn't that obvious? :wink:)

if m is a root, n times, then the solution is emt times a polynomial of degree n-1 in t (works for n = 1 also! :biggrin:)
btw do i have to use latex for powers like that? does it bother people? i was going to use latex for more complicated things but if it bothers people....
just use the SUP tags :smile:
 
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i dont quite understand.
y'' + 2y'+ 1y=0 solving for y
let y=emt
therefore
m2+2m+3=0 (differentiating y and multiplying out emt
which is
(m+1)2=0
m=-1
so i thought you write it as this, but im not sure. this is part of my question.
y=Ae-t+Bte-t

for
y'' + 3y'+ 2y=0 solving for y
(m+1)(m+2)=0
m=-1 and m=-2
i think you write it as this
y=Ae-t + Be-2t



and yea i had a constant for m=0 but the answer in the textbook didnt include that term.
 
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for
y'' + 2y'+ 1y=0 solving for y
and let y=emt
therefore
m2+2m+3=0 (differentiating y and multiplying out e^mt
which is
(m+1)2=0
m=-1
NOTE: We didnt take e^mt because there is no value of m such can bring an answer equal to zero
y=Ae-t+Bte-t

for
y'' + 3y'+ 2y=0 solving for y
(m+1)(m+2)=0
m=-1 and m=-2
i think you write it as this
y=Ae-t + Be-2t

so for your last statement that you have a value for constant for m=0, this also doesn't exist..
 

tiny-tim

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hi brandy! :wink:
y'' + 2y'+ 1y=0

y=Ae-t+Bte-t

y'' + 3y'+ 2y=0

y=Ae-t + Be-2t
yes, that's all correct :smile:

isn't that what i said?​
and yea i had a constant for m=0 but the answer in the textbook didnt include that term.
sorry, i'm not following you …

what differential equation are you referring to here? :confused:
 

HallsofIvy

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If, for example, you had the equation
[tex]\frac{d^2y}{dx^2}+ 2\frac{dy}{dx} + y= e^{-t}[/tex]

Then the associated homogeous differential equation is
[tex]\frac{d^2y}{dx^2}+ 2\frac{dy}{dx} + y= 0[/tex]
which, as you say, has characteristic equation [itex]m^2+ 2+ 1= (m+1)^2= 0[/itex] which has m= -1 as a double root. Yes, the general solution to that associated homogenous equation is [itex]Ce^{-x}+ Dxe^{-x}[/itex].

Now, to use "undetermined coefficients" to seek a solution to the entire equation, normally, we would try a solution of the form [itex]Ae^{-t}[/itex] but that already satisfies the homogeneous equation. In fact, [itex]Ate^{-t}[/itex] so we would try [itex]At^2e^{-t}[/itex].

Try it yourself: if [itex]y= At^2e^{-t}[/itex], what are y' and y''? What do you get when you put those into the equation? What must A be in order that the equation be satisfied?
 

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