General solution to 2nd order ODE

[forty]

Find the general solution of the ordinary differential equation.

y'' - 7y'+ 6y = 2e^(3t) + te^(t)

First i found GS(H) by lettings y = e^(cx)

and got GS(H) = Ae^(6t) + Be^(t)

i then found PS(IH) y'' - 7y'+ 6y = 2e^(3t) by letting y = ae^(3t)

and got PS(IH) = -(1/3)e^(3t)

Now my problem
I am trying to find the other PS(IH) y'' - 7y'+ 6y = te^(t)
do i do this by letting y = (a+bt)e^(t) ??

 

[Defennder]

This is what I have from my notes:

http://img164.imageshack.us/img164/6935/notesom1.jpg

 

[HallsofIvy]

Find the general solution of the ordinary differential equation.

y'' - 7y'+ 6y = 2e^(3t) + te^(t)

First i found GS(H) by lettings y = e^(cx)

and got GS(H) = Ae^(6t) + Be^(t)

i then found PS(IH) y'' - 7y'+ 6y = 2e^(3t) by letting y = ae^(3t)

and got PS(IH) = -(1/3)e^(3t)

Now my problem
I am trying to find the other PS(IH) y'' - 7y'+ 6y = te^(t)
do i do this by letting y = (a+bt)e^(t) ??

Normally, you would try that but since e^t is already a solution to the homogeneous equation, this will not work. (Surely you have already tried and discovered that?) The standard step for this situation is to multiply by t: try y= (at+ bt²e^t.

 

[forty]

you say try y= (at+ bt2et.

is that bracket suppose to be there? I am not sure if that's (at+bt^2)e^t or at+b(t^2).e^t

 

[forty]

I used y = (at+bt^2)e^t
y' = (at+bt^2)e^t + (a+2bt)e^t
y'' = (at+bt^2)e^t + (a+2bt)e^t + (a+2bt)e^t + 2be^t

i substituted those in and equated coefficients and got the result a = o and b = 1/10

have i done this right ?

 

[Defennder]

I didn't get that. For me, neither a nor b is zero. Why don't you check this by substituting it into the modified DE to see if it satifies the equation?

 

[HallsofIvy]

I used y = (at+bt^2)e^t
y' = (at+bt^2)e^t + (a+2bt)e^t
y'' = (at+bt^2)e^t + (a+2bt)e^t + (a+2bt)e^t + 2be^t

= (at+ bt^2)e^t+ 2(a+ 2bt)e^t+ 2be^t

i substituted those in and equated coefficients and got the result a = o and b = 1/10

have i done this right ?

You want y'' - 7y'+ 6y = 2e^(3t) + te^(t) and you've already done the "2e^(3t)" part.

y'' = (at+ bt^2)e^t+ 2(a+ 2bt)e^t+ 2be^t
-7y'= -7(at+bt^2)e^t - 7(a+ 2bt)e^t
+6y= 6(at+ bt^2)e^t

so y"- 7y'+ 6y= -5(a+ 2bt)e^t+ 2be^t= (-5a+ 2b)e^t+ -10bte^t= te^t. Equating "like coefficients, -10b= 1 so b= -1/10, not 1/10.