# Thread: How to make a proper guess at a particular solution

1. ## How to make a proper guess at a particular solution

]My question is how do you from a proper guess at the particuliar solution of a differential equation for example.

Say this is the right hand of the equation

$\displaystyle sin(2t)+te^t+4$

I know it helps to break it up into seperate parts, but I am working with a book solution here and my guess are just off it seems from the books proper guess.

My guess for
$\displaystyle sin(2t)$

$\displaystyle Acos(2t)+Bsin(2t)$

Book's guess
$\displaystyle t(Acos(2t)+Bsin(2t)$

and the for the term of 4 the book's guess was
$\displaystyle Dt^2$

How do you properly determine a guess like this?

2. Originally Posted by sk8erboyla2004
]My question is how do you from a proper guess at the particuliar solution of a differential equation for example.

Say this is the right hand of the equation

$\displaystyle sin(2t)+te^t+4$

I know it helps to break it up into seperate parts, but I am working with a book solution here and my guess are just off it seems from the books proper guess.

My guess for
$\displaystyle sin(2t)$

$\displaystyle Acos(2t)+Bsin(2t)$

Book's guess
$\displaystyle t(Acos(2t)+Bsin(2t)$
Look at the roots of the characteristic equation, or the general solution to the homogeneous equation. In the first case I bet (but cant guarantee since you have not posted the left hand side of the problem) that 2i is a root, and in the second case the general solution includes sin and cos functions of 2t.

CB

3. Why , Yes, it is (+,-) 2i and 0,0

4. Say, could you take the time to post the left hand side? I'm a newbie in a class, and I would be very interested in seeing the whole problem.
Thanks

5. Originally Posted by sk8erboyla2004
Why , Yes, it is (+,-) 2i and 0,0
Then $\displaystyle A\cos(2t)+B\sin(2t)$ cannot be a particular solution since it is a solution of the homogeneous equation.

CB

6. Sorry for a late reply, very busy.
Thanks for the informative reply Captain Black.
So it would seem so to take a look at the complimentary/homogenous solution to perform your guess on the particuliar.

My guess would not be correct because, let me try to see why. But you can help explain.

1. The obvious you stated, a term in the complimentary solution cannot be a term in the particuliar solution
hence why the addition of the variable, t, in the particuliar solution ?

Now another question as well. For the term of (4) in the right-hand side of the differiential equation's solution.
Why not an undetermined coeff. as such would not be suitable?

$\displaystyle Dt$

Is that because it would conflict at our guess in the particuliar solution of the term

$\displaystyle t(Asin(2t)+Bcos(2t))$