Second Order Non Homogeneous DE

You should try using the annihilator method.

I got Yp=At+B
Yp'=A
Yp"=0 (A and B are constants)

plugging them back into the ODE :

2[A] + [At+B] = t

rearranging:

(2A+B)[1] + A[t] = t

so

A=1

2A+B=0


getting A=1 and B= -2

so then Yp= t-2

add it to Yh and you should get Y(t)

 

oh, and to find Yh just set the ODE = 0 and find the roots

 

That is not the correct thing to try for the homogeneous solution. Try [itex]y = e^{rt}[/itex].

 

No, that is NOT how you solve for Yh.

That's because [itex]B1x^2+ B2x+ B3[/itex] is NOT a solution to the homogeneous equation. Use [itex]y= e^rt[/itex], as LCKurtz suggested, to get the "characteristic equation" to solve for r.

You cannot just let yp= ct. If the right hand side is a power of t, yp can be a polynomial nof degree at most that power- try yp= ct+ d.

 

Actually the annihilator method is correct. In fact I checked with my professor on this. The annihilator method takes out the factor of "guessing" a solution for the particular solution.

I even had my professor solve this and she got the same answer as I did, especially since we just went over this section in class.

Annihilator method is actually quite simple for this problem.

 

As for Yh, what I mean by set the ODE to 0 is:
y''+2y'+y=0

Let y''=r^2, y'=r, y=constant

ending up with:
r^2+2r+1=0

Factoring, you should end up with
(r+1)(r+1)=0

where r=-1,-1

Since this is a repeated root you should use:
Yh= c1*e^(rt) + c2*t*e^(rt)

Once you have Yh, then find Yp (particular solution) and add it to Yh for your general solution Y.

Just remember, since you DON'T have initial conditions

(for example: y'(0)=0, y(0)=1)

Your general solution is going to have constants c1 and c2 since there are no initial conditions given