Hi, could someone please explain how to find the complementary solution?
For example with this equation y'' + 8y' +16y= e^(-4t) + ((t^2)+5)e^(-4t)

Question

Hi, could someone please explain how to find the complementary solution? 
For example with this equation y'' + 8y' +16y= e^(-4t) + ((t^2)+5)e^(-4t)

solomonzelman · Answer

first the homogeneous, can you find it?

efsthathia · Answer

I'm not really sure if I know.

amorfide · Answer

@SolomonZelman  please help the guy, I can give him step by step for the answer but I can't explain why, so I would prefer to leave it to you

solomonzelman · Answer

You are trying to find a solution to a non-homogeneous equation, without knowing the solution to the homogeneous one?

solomonzelman · Answer

Can you solve  $\color{black}{\displaystyle y''+8y'+16y=0}$  ?

solomonzelman · Answer

Set the characteristic equation (or the auxiliary equation).
Can you do this?

solomonzelman · Answer

Look up Lamar Tutorial (or maybe other material) on this topic.
As of now you seem to have insufficient background for this question.
(Sorry to disappoint you.)

When you solve this (whether you need to look anything up or not), note that you will have a repeated root $r$, and having $e^{rt}$, you will have a general solution of the form c_1e^rt+tc_2e^rt+t^2c_3e^rt

amorfide · Answer

solomon, if i may, I can solve this question but I have no idea what or where the homogenous is, care to explain?

efsthathia · Answer

I was following the notes for how to solve a nonhomogeneous equation using undetermined coefficients, but I guess I'll just have to review. Is the complementary solution the same as the homogeneous?

loser66 · Answer

You have to solve the homogeneous part to make sure that you don't have repeated solution.

loser66 · Answer

For the homogeneous part, that is y"-8y +16 y =0

The characteristic equation is $r^2-8r =16=0$, gives you the solution is??

loser66 · Answer

sorry $r^2-8r+16=0$

loser66 · Answer

Oh, Solomon is here. Let him explain. 
@SolomonZelman , please

solomonzelman · Answer

I'm on another problem ... just viewing in case...

solomonzelman · Answer

Did you find the homogeneous solution though?

amorfide · Answer

use the characteristic equation 
and you will end up with something similar to

y=1, y'=m, y''=m^2

substitute into the ode and make it equal to 0, that is
y''+8y'+16y=g(t), let g(t)=0

so you get

m^2 + 8m +16=0
solve for m

then you have a complimentary solution based on the values of m

if m is two distinct roots, of the quadratic then
|dw:1478746539681:dw|

if m is one repeated root
|dw:1478746573810:dw|

if m is imaginary
|dw:1478746748794:dw|
where m=a+bi