# Undetermined coefficients why doesn't it work?

1. Apr 12, 2013

### 1MileCrash

1. The problem statement, all variables and given/known data

y''+y=3sin2t+2tcos2t

2. Relevant equations

3. The attempt at a solution

I tried doing this problem by breaking it into 3 others. It doesn't quite work, and I want to know why.

First, I solved for the nonhomog part of:

y''+y=3sin2t

Which is Y1 = -sin2t

Then, I did the same with

y''+y = 2t

which is Y2 = 2t

Lastly, I did

y''+y=cos2t

Which is Y3 = -(1/3)cos2t

Combining these with the homog. parts gives me:

y = c1cost + c2sint - (2/3)tcos2t -sin2t

But, this is not quite right, as the -sin2t should be -(1/9)sin2t.

What is incorrect about breaking the DE up and putting it back together in this way?

Thank you!

2. Apr 12, 2013

### vela

Staff Emeritus
You don't split up products in the forcing function when relying on superposition. Your first solution is fine, but you need to solve
$$y''+y = 2t\cos 2t$$ to get the second solution, which you can then add it to the first to find the particular solution.

3. Apr 12, 2013

### 1MileCrash

I really thought that was the case before starting.

Is this the case because:

d/dt(f(t) + g(t)) = f'(t) + g'(t)

but
d/dt(f(t)g(t)) =/= f'(t)g'(t)

?

4. Apr 12, 2013

### vela

Staff Emeritus
Yeah, pretty much.

5. Apr 12, 2013

### 1MileCrash

This is kind of interesting.

Thanks, now I can solve these problems.

But, I wonder, with these terms I have found before, is there a way to construct the answer? Applying the product rule to the second two terms and adding that to the first doesn't work. I'm trying to think about what's going on and how those terms relate to the correct answer, since it's close. I tend to do that.

6. Apr 13, 2013

### haruspex

Suppose L(f(t)) = u(t) and L(g(t)) = v(t), where L(h) = h"+h.
L(fg) = f"g + 2f'g' + fg" + fg
uv = f"g + fg" + fg + f"g"
L(fg) - uv = 2f'g' - f"g"
In this example, f" = k g' etc., so it almost worked.