Some questions for non/homogeneous equations

[brandy]

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.

 

[tiny-tim]

hi brandy!

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

what's m ?

can you write out the equation you're referring to?

 

[brandy]

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]

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? )

if m is a root, n times, then the solution is e^mt times a polynomial of degree n-1 in t (works for n = 1 also! )

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

 

[brandy]

i don't quite understand.
y'' + 2y'+ 1y=0 solving for y
let y=e^mt
therefore
m²+2m+3=0 (differentiating y and multiplying out e^mt
which is
(m+1)²=0
m=-1
so i thought you write it as this, but I am 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.

 

[alube1]

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]

hi brandy!

y'' + 2y'+ 1y=0
…
y=Ae^-t+Bte^-t

y'' + 3y'+ 2y=0
…
y=Ae^-t + Be^-2t

yes, that's all correct …

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? ​

 

[HallsofIvy]

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

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

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

Try it yourself: if $y= At^2e^{-t}$, 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?