# Thread: Undetermined Coefficients

1. ## Undetermined Coefficients

Find a particular solution to y+25y=10sin(5t)

I chose y_p to be Acos(5t)+Bsin(5t).
After simplifying, my equation becomes 0=0, which leads to nothing.

Would I multiply my particular solution by t or a higher power of t? In general, it is still unclear to me when to modify particular solutions.

2. You have to modify the particular solution when the homogeneous solution contains what you would normally have tried for the particular solution. For this equation, what is the homogeneous solution?

3. Originally Posted by krtica
Find a particular solution to y+25y=10sin(5t)

I chose y_p to be Acos(5t)+Bsin(5t).
After simplifying, my equation becomes 0=0, which leads to nothing.

Would I multiply my particular solution by t or a higher power of t? In general, it is still unclear to me when to modify particular solutions.
Well...

I assume you meant $y'+25y=10\sin(5t)$

Ok then,

If $y_p = A\cos(5t) + B\sin(5t)$ then

$y_p' = -5A\sin(5t) + 5B\cos(5t)$

So we have...

$y'+25y= -5A\sin(5t) + 5B\cos(5t) + 25A\cos(5t) + 25B\sin(5t)$

$\sin(5t)(-5A + 25B) + \cos(5t)(5B + 25A)$

So you get, $25B - 5A = 10$ and $5B + 25A = 0$

=> $B = -5A$

=> $-125A - 5A = 10$

=> $A = -\frac{1}{13}$

$B = \frac{5}{13}$

4. I apologize, I meant to type y"+25y=10sin(5t).

5. You need to find the homogeneous solution. That is, solve the following DE: $\ddot{y}(t)+25 y(t)=0$.

I think you'll find, if you do so, that the solution includes the particular solution you were thinking of using as an ansatz. Therefore, you need to modify your ansatz. (Ansatz, in case you haven't heard the term, is a German word that essentially means "your guess" or "your working hypothesis".)