Math Help - help with this IVT problem..

1. help with this IVT problem..

Find the general solution for the problem y'' + 2y' + y= e^-t.. solve with initial conditions y(0)=0, y'(0)=1

i get the eigen values are both -1, which is e^-t.. that is the term on the right side of the original equation. im completely lost at this step.. really appreciate it if someone could help me out with this (sorry for the vague title, wasn't really sure what to call it)

2. Originally Posted by twostep08
Find the general solution for the problem y'' + 2y' + y= e^-t.. solve with initial conditions y(0)=0, y'(0)=1

i get the eigen values are both -1, which is e^-t.. that is the term on the right side of the original equation. im completely lost at this step.. really appreciate it if someone could help me out with this (sorry for the vague title, wasn't really sure what to call it)
What method are you using? You mention eigenvalues are you decomposing the equation is to a first order system? I can think of at least there different distinct ways to solve this ODE. Please show some work so we know where you are stuck and what method you are trying to use.

3. i not very good with the DiffEQ jargon, so bear with me.. i broke the left side into the characteristic polynomial and found the roots (which ive, been told is lamba, which is the notation i see for eigen) to both be -1. since this is the same exponent of t on the right, i believe this is some special case, but i cant remember how to do it next.. i believe "guess" ate^-t? but even if that true, i dont know where to go from there... if its not too hard of a problem, do you think you could walk me through the rest, cuz i got an exam on it tomorrow..

4. Originally Posted by twostep08
Find the general solution for the problem y'' + 2y' + y= e^-t.. solve with initial conditions y(0)=0, y'(0)=1

i get the eigen values are both -1, which is e^-t.. that is the term on the right side of the original equation. im completely lost at this step.. really appreciate it if someone could help me out with this (sorry for the vague title, wasn't really sure what to call it)
The 'characteristic equation' is...

(1)

... and it has the only solution r=-1 but with multeplicity two, so that the general solution of the 'incomplete' DE is...

(2)

The particular solution of the 'complete' DE independent from (2) is found with little effort to be...

(3)

Kind regards

$\chi$ $\sigma$

5. that wouldn't be the final answer to the question tho? or would it? because isnt there usually something along the lines of k_1 e^(first root) + k_2 e^(second root) + (what you just found in your above post)? like when we "guess" something and plug it (and its first and 2nd derivatives) into the given equation?

i keep getting different types of problems confused with each other, so hopefully you know what im thinking of if im wrong

6. Originally Posted by twostep08
... because isnt there usually something along the lines of k_1 e^(first root) + k_2 e^(second root) + (what you just found in your above post)?...
That is true only if the second order 'characteristic equation' has two distinct roots... in your case there one root r=-1 with multeplicity two so that the solution is...

(1)

Kind regards

$\chi$ $\sigma$