Math  /  Calculus

QuestionStandard 16: Problem 4 (1 point) a. Find the Laplace transform F(s)=L{f(t)}F(s)=\mathcal{L}\{f(t)\} of the function f(t)=2e9t+3t+6e10tf(t)=2 e^{-9 t}+3 t+6 e^{10 t}, defined on the interval t0t \geq 0. F(s)=L{2e9t+3t+6e10t}=F(s)=\mathcal{L}\left\{2 e^{-9 t}+3 t+6 e^{10 t}\right\}= \square help (formulas) b. For what values of ss does the Laplace transform exist? \square help (inequalities)
Note: You can earn partial credit on this problem.

Studdy Solution

STEP 1

What is this asking? We need to find the Laplace transform of a function that's a mix of exponential and linear terms, and then figure out for which values of ss this transform actually works! Watch out! Don't mix up the Laplace transform formulas for different function types, and be careful with those signs in the exponents!
Also, remember to consider the conditions under which the Laplace transform converges.

STEP 2

1. Transform Each Term
2. Combine and Simplify
3. Determine Convergence

STEP 3

Alright, let's **break down** this function piece by piece!
We've got f(t)=2e9t+3t+6e10tf(t) = 2e^{-9t} + 3t + 6e^{10t}.
We'll use the *linearity* of the Laplace transform, meaning we can transform each part separately and then just add them up!

STEP 4

First, the Laplace transform of 2e9t2e^{-9t}.
Remember the formula L{eat}=1sa\mathcal{L}\{e^{at}\} = \frac{1}{s - a}.
Here, our aa is 9-9, so we get: L{2e9t}=2L{e9t}=21s(9)=2s+9 \mathcal{L}\{2e^{-9t}\} = 2 \cdot \mathcal{L}\{e^{-9t}\} = 2 \cdot \frac{1}{s - (-9)} = \frac{2}{s + 9} See how we pulled the **constant 2** out front?
That's the linearity property in action!

STEP 5

Next up: 3t3t.
The Laplace transform of tnt^n is n!sn+1\frac{n!}{s^{n+1}}.
In our case, n=1n = 1, so we have: L{3t}=3L{t}=31!s1+1=3s2 \mathcal{L}\{3t\} = 3 \cdot \mathcal{L}\{t\} = 3 \cdot \frac{1!}{s^{1+1}} = \frac{3}{s^2}

STEP 6

Finally, let's tackle 6e10t6e^{10t}.
This one's similar to the first term.
Our aa is now **10**, so: L{6e10t}=6L{e10t}=61s10=6s10 \mathcal{L}\{6e^{10t}\} = 6 \cdot \mathcal{L}\{e^{10t}\} = 6 \cdot \frac{1}{s - 10} = \frac{6}{s - 10}

STEP 7

Now, let's **put it all together**!
Adding the results from the previous calculations, we get: F(s)=L{f(t)}=2s+9+3s2+6s10 F(s) = \mathcal{L}\{f(t)\} = \frac{2}{s + 9} + \frac{3}{s^2} + \frac{6}{s - 10} That's our Laplace transform!

STEP 8

For the Laplace transform to exist, all individual transforms must exist.
Looking back at our individual transforms, we see that 2s+9\frac{2}{s+9} exists for s>9s > -9, 3s2\frac{3}{s^2} exists for s>0s > 0, and 6s10\frac{6}{s-10} exists for s>10s > 10.

STEP 9

We need all three conditions to be true *simultaneously*.
So, we need ss to be greater than 9-9, greater than 00, *and* greater than 1010.
The strictest of these is s>10s > 10, so that's our **final condition** for convergence.

STEP 10

a. F(s)=2s+9+3s2+6s10F(s) = \frac{2}{s + 9} + \frac{3}{s^2} + \frac{6}{s - 10} b. s>10s > 10

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