# Varation of Parameters fun diff EQ question, where do i go next?

1. Feb 17, 2006

### mr_coffee

Varation of Parameters fun!! diff EQ question, where do i go next?

This is my first attempt at doing Variation of parameters, didn't go to bad, things cancled out pretty well but now i'm almost done but i'm stuck!
The problem says: Find the solution of y''+15y'+56y = 54*e^(-5t), with y(0) = 8, and y'(0) = 2, y = ?

Here is my work:

If you don't follow me or see an error right off the bat, please let me know! Thanks!

2. Feb 18, 2006

### HallsofIvy

Staff Emeritus
Now integrate u1= -54e3t and u2'= 54e2t to find u1 and u2 of course! Once you know those put them into "u1e-8t+ u2e-7t to get a "specific solution" and add that to the general solution of the homogenous equation.

3. Feb 18, 2006

### mr_coffee

ahh thanks Ivey, i'm alittle confused on what u mean when u said add it to the homogenous equation, the orginal equation wasn't homogenous was set equal to 54e^(-5t), are you saying add it to the r^2+15r+56 = 0? Thanks here is what i got now:

4. Feb 18, 2006

### HallsofIvy

Staff Emeritus
Well, that's pretty much the whole idea! If $y(t)= C_1e^{-8t}+ C_2e^{-7t}$ is the general solution to y"+ 15y'+ 56y= 0 and $y(t)= 9e^{-5t}$ satisfies y"+ 15y'+ 56y= 54e-5t, then
The general solution to y"+ 15y'+56y= 54e-5t is
$y(t)= C_1e^{-8t}+ C_2e^{-7t}+ 9e^{-5t}$.

5. Feb 18, 2006

### mr_coffee

Thanks again Ivey!
I don't know if its right because webhw is still down, yay!
but this is what I got:

6. Feb 18, 2006

### assyrian_77

I personally think it's easier to solve it like this: First, as usual, you solve the homogeneous equation which is basically setting the right hand side (RHS) to zero, i.e. $y"+ 15y'+56y=0$. This will give you $y_{h}(t)=C_1e^{-8t}+ C_2e^{-7t}$. Next you solve the "special" or "particular" equation by making an ansatz (educated guess) depending on what expression you have on the RHS. In this case I would choose the ansatz $y_{p}(t)=Ae^{-5t}$. Differentiate this and insert into the original equation gives you a value on A. The total solution is simply $y=y_{h}+y_{p}$. And then use your initial conditions to get the values of $C_1$ and $C_2$.

Sound complicated, but that's how I used to solve these things.

And yes, I too got $C_1=-40$ and $C_2=39$.

7. Feb 19, 2006

### mr_coffee

i like ur method alot better! But how did you make an educational guess of $y_{p}(t)=Ae^{-5t}$ ? I understand why u would guess an e^(t) but how did u konw it should be e^(-5t)? Thanks!

8. Feb 19, 2006

### Benny

You can find out more about that method if you look up 'method of undetermined coefficients.' I find that reduction of order is generally faster because you don't need to memorise specific forms for certain situations.

9. Feb 19, 2006

### assyrian_77

I made the ansatz $y_{p}(t)=Ae^{-5t}$ because the the RHS of the original equation had the exponential term $e^{-5t}$. More generally, I could have chosen the ansatz $y_{p}(t)=Ae^{-5t}+Be^{+5t}$ but the second term is redundant.