# Find the general solution?

1. Jul 25, 2013

### Success

Find the general solution of y"'-y"-y'+y=2e-t+3.

Here's the work:
r3-r2-r+1=r2(r-1)-(r-1)=(r-1)(r2-1)=(r2-1)2(r+1)
r=1, -1
y=c1et+c2tet+c3e-t

The answer in the textbook is y=c1et+c2tet+c3e-t+(1/2)te-t+3 but I don't know how to get the last 2 terms. Help me...

2. Jul 25, 2013

### Zondrina

I believe what they have there is the particular solution $y_p$.

Last edited: Jul 25, 2013
3. Jul 25, 2013

### Success

Let me try to find the particular solution and see if it makes sense then.

4. Jul 25, 2013

### Boorglar

Yes, what you've found is the solution to the homogeneous equation (where the right hand side is 0). You need to add the particular solution.

5. Jul 26, 2013

### HallsofIvy

Staff Emeritus
There are two standard ways of finding a "specific solution" to a non-homogeneous linear equation to be added to the general solution to the associated homogeneous equation:
1) "Variation of Parameters". Given independent solutions y1 and y2 to the associated homogeneous equation, look for a specific solution to the entire equation of the form uy1+ vy2, solving for u and v.

2) "Undetermined Coefficients". Guess the correct form for the specific solution up to coefficients that have to be determined.

The second only works if you can guess the correct form and that will be only when the "right hand side" (or "non-homogeous part") of the equation is itself one of the various kinds of functions one expects to get as solutions to a linear homogeneous equation with constant coefficients. The first works for any "right hand side" but is much more difficult computationally.

The " kinds of functions one expects to get as solutions to a linear homogeneous equation with constant coefficients" are (1) exponentials, (2) polynomials, (3) sines and cosines, and (4) combinations of those.

Here, the "right hand side" is $2e^{-t}+ 3$ which is of that kind so we ca use "undetermined coefficients". Yes, the general solution to the "associated homogeneous equation" is $C_1e^t+ C_2te^{t}+ C_3e^{-t}$. For a "right hand side" involving $e^{-t}$ we would normally try $$Ae^{-t}$$ but that is already a solution of the associated homogeneous equation so we try, instead, $$Ate^{-t}$$ (just as the fact that "1" is a double root of the characteristic equation gives $e^t$ and $te^t$ as solutions to the associated homogeneous equation).

That is, set $y(t)= Ate^{-t}+ B$ in the equation and solve for A and B.

6. Jul 26, 2013

### Success

HallsofIvy, how do I do that? Can you show complete work?

7. Jul 26, 2013

### Zondrina

He's already done most of the work for you by guessing the correct form of the particular solution. All you would have to do at this point is plug it in and solve for the coefficients.

Undetermined coefficients relies on the fact that you can guess the form of the particular solution.

8. Jul 26, 2013

### Redbelly98

Staff Emeritus
No, he can't. Or he shouldn't -- since showing the complete work would be against our forum guidelines. To proceed, you should:

1. Substitute the expression HallsofIvy wrote into the differential equation you have.
2. Take derivatives, as necessary, of HallsofIvy's expression.
3. Combine like terms together, and examine them to figure out what A and B are.

If you get stuck, post all the work you were able to do up until the point where you got stuck.

9. Jul 26, 2013

### Staff: Mentor

You have a mistake here that I don't think anyone noticed. The final factorization should be (r - 1)2(r + 1), NOT (r2 - 1)2(r + 1).

10. Jul 26, 2013

### Success

Mark44, yeah, you're right.

11. Jul 26, 2013

### Success

Okay, I've took the derivatives for all of them and I got y'=Ce^(-t)(-t+1), y"=Ce^(-t)(t-2) and y'''=Ce^(-t)(-t+3). I combined like terms and got Ce^(-t)(-t+3-t+2+t-1+t)=2e^-t, C=1/2, so yeah, the answer makes sense. But can anyone tell me how to get t in (-t+3-t+2+t-1+t)? The last term.