# Laplace Transform, what am I doing wrong?

1. Apr 20, 2010

### raytrace

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

Solve using the Laplace Transforms (can not use partial fractions)

f '(t) + $$\int2f(u) du$$ = 2 + 3f(t)

2. Relevant equations

Using Laplace

f '(t) gets replaced with sF(s) -f(0)

$$\int2f(u) du$$ gets replaced with $$\frac{2F(s)}{s}$$

Please correct me if I'm wrong on the replacements here.

3. The attempt at a solution

After using Laplace on both sides I get

$$sF(s)-f(0)+\frac{2F(s)}{s} = \frac{2}{s} + 3F(s)[\tex] [tex]sF(s)-3F(s)+\frac{2F(s)}{s} = \frac{2}{s} + f(0)[\tex] [tex]F(S)(s-3+\frac{2}{s}) = \frac{2}{s} + f(0)[\tex] Divide through and manipulate a little to get: [tex]F(S) = \frac{2}{(s-2)(s-1)} + f(0)\frac{s}{(s-2)(s-1)}[\tex] OK, here is where I get stuck. The first half I can figure out, it's the s/((s-2)(s-1)) that I can't figure out. I did find a transform in the Laplace tables in the back of the book but this particular transform was not on the list of approved transforms we could use freely (without proving). So, I've either screwed up in my math here somewhere's or I have to prove the Inverse Laplace Transform of s/((s-2)(s-1)). Now someone mentioned using the L'Hopitals rule on it but I don't see how. I'm completely at a loss. Please help. 2. Apr 20, 2010 ### Mark44 ### Staff: Mentor Fixed your tex tags. You have the closing tex tag wrong -- [ \tex] instead of [ /tex] Your work looks fine. I don't get why you can't use partial fractions, and it sounds like maybe you can use this technique. The restriction was to not use "nonapproved" transforms unless you can prove them. For this problem you would be evaluating L-1 {2/((s - 2)(s - 1))} which is exactly the same as evaluating L-1{2/(s -2) + -2/(s -1)}. If you have to prove that 2/((s-2)(s -1)) is identically equal to 2/(s -2) + -2/(s - 1), that's very easy and is a very small price to pay. 3. Apr 21, 2010 ### raytrace Sorry for the confusion there. It's the [tex]L^{-1}\frac{s}{(s-2)(s-1)}$$ that I can't figure out.

From the transform table at the back of the book I know it turns into $$\frac{ae^{at}-be^{bt}}{a-b}$$

However, I am forbidden to use partial fractions for this part and I can only use the transform in the back if, and only if, I can prove it... which means I have to know how to do the proof from the first to the second... and this is where I'm completely lost.

4. Apr 21, 2010

### Count Iblis

Well, unreasonable demands call for unreasonable solutions. You can just compute the Laplace transform of the answer and then show that it yields the correct formula.

5. Apr 21, 2010