Particular Solution for y'' -5y' +6y = te^t with Simple Roots

[flyingpig]

Homework Statement

Find a particular soln

$$y'' -5y' +6y = te^t$$


The characteristic eqtn is $$r^2 - 5r + 6 = (r - 2)(r - 3)$$

r = 2, 3

According to my book

[PLAIN]http://img202.imageshack.us/img202/4539/unledxe.jpg

So I thought

$$y = t(At + A_0)e^{2t} + t(Bt + B_0)e^{3t}$$


But not working...

what exatly is a simple root?

 

[micromass]

Hi flyingpig!

You seemed to use r=2 and r=3, which is incorrect.
What is the correct r? You only need to look at the right-hand side of

$$y^{\prime\prime}-5y^\prime+6y=te^t$$

 

[Mark44]

To elaborate on what micromass said, r = 2 and r = 3 are roots of the characteristic equation, meaning that e^2t and e^3t are solutions of the homogeneous equation y'' - 5y' + 6y = 0. All solutions of this equation have the form y_c = c₁e^2t + c₂e^3t, where y_c is the complementary solution (the solution to the homogeneous problem).

What you're asked for is a particular solution of the equation y'' - 5y' + 6y = te^t. This solution will involve e^t, but won't involve e^2t or e^3t.

 

[hunt_mat]

I would try a particular solution of the form:
$$y_{p}=ate^{t}+be^{t}$$

 

[flyingpig]

But that's what (14) says

 

[hunt_mat]

and did you try it?

 

[flyingpig]

$$y = t(At + A_0)e^{t}$$?

 

[ehild]

No. It is not the same function hunt_mat suggested. Try that. ehild

 

[Mark44]

$$y = t(At + A_0)e^{t}$$?

You're not reading the problem correctly, particularly item (i).
"s = 0 if r is not a root of the associated auxiliary equation."

 

[flyingpig]

You're not reading the problem correctly, particularly item (i).
"s = 0 if r is not a root of the associated auxiliary equation."

Yeah I don't understand what dose that mean? r is a root, i have two??

 

[micromass]

r has nothing to do with the characteristic equation. r is determined by the right-hand side of the equation, that is: $te^t$.
Thus t=1 here.

Now, to determine s, the characteristic equation does come into play. You'll need to check whether r is a root of the characteristic equation.

 

[flyingpig]

r has nothing to do with the characteristic equation. r is determined by the right-hand side of the equation, that is: $te^t$.
Thus t=1 here.

Now, to determine s, the characteristic equation does come into play. You'll need to check whether r is a root of the characteristic equation.

That clears up ALOT.

Just one other thing. Is the table one of those things yo just have to remember?