Math / Calculus

QuestionPrevious Problem Problem List Next Problem
Standard 16: Problem 6 (1 point)
Find the Laplace transform of $5 \cos (a t)-2 e^{m t}$ : $\square$
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Studdy Solution

STEP 1

What is this asking? We need to find the Laplace transform of the expression $5 \cos(at) - 2e^{mt}$ . Watch out! Remember the specific formulas for the Laplace transforms of cosine and exponential functions!
Don't mix them up.

STEP 2

1. Laplace Transform of Cosine
2. Laplace Transform of Exponential
3. Combine the Results

STEP 3

Let's start with the Laplace transform of $5 \cos(at)$ .
Remember, the Laplace transform of $\cos(at)$ is given by $\frac{s}{s^2 + a^2}$ .
So, we're going to use that!

STEP 4

Now, we multiply that by 5 to get the Laplace transform of $5 \cos(at)$ , which is $5 \cdot \frac{s}{s^2 + a^2} = \frac{5s}{s^2 + a^2}$ .
See? Not so bad!

STEP 5

Next up is the Laplace transform of $-2e^{mt}$ .
The Laplace transform of $e^{mt}$ is $\frac{1}{s - m}$ .
Keep this formula in mind, it's a good one!

STEP 6

We multiply by $-2$ to find the Laplace transform of $-2e^{mt}$ , which gives us $-2 \cdot \frac{1}{s - m} = \frac{-2}{s - m}$ .
Awesome!

STEP 7

Finally, we put it all together.
Since the Laplace transform is a linear operator, we can just add the individual transforms we found.

STEP 8

The Laplace transform of $5 \cos(at) - 2e^{mt}$ is the sum of the transforms we calculated: $\frac{5s}{s^2 + a^2} + \frac{-2}{s - m}$ .
We can rewrite this as $\frac{5s}{s^2 + a^2} - \frac{2}{s - m}$ .

STEP 9

The Laplace transform of $5 \cos(at) - 2e^{mt}$ is $\frac{5s}{s^2 + a^2} - \frac{2}{s - m}$ .

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QuestionPrevious Problem Problem List Next Problem
Standard 16: Problem 6 (1 point)
Find the Laplace transform of $5 \cos (a t)-2 e^{m t}$ : $\square$
Preview My Answers Submit Answers You have attempted this problem 0 times. You have unlimited attempts remaining.

Studdy Solution

STEP 1

What is this asking? We need to find the Laplace transform of the expression $5 \cos(at) - 2e^{mt}$ . Watch out! Remember the specific formulas for the Laplace transforms of cosine and exponential functions!
Don't mix them up.

STEP 2

1. Laplace Transform of Cosine
2. Laplace Transform of Exponential
3. Combine the Results

STEP 3

Let's start with the Laplace transform of $5 \cos(at)$ .
Remember, the Laplace transform of $\cos(at)$ is given by $\frac{s}{s^2 + a^2}$ .
So, we're going to use that!

STEP 4

Now, we multiply that by 5 to get the Laplace transform of $5 \cos(at)$ , which is $5 \cdot \frac{s}{s^2 + a^2} = \frac{5s}{s^2 + a^2}$ .
See? Not so bad!

STEP 5

Next up is the Laplace transform of $-2e^{mt}$ .
The Laplace transform of $e^{mt}$ is $\frac{1}{s - m}$ .
Keep this formula in mind, it's a good one!

STEP 6

We multiply by $-2$ to find the Laplace transform of $-2e^{mt}$ , which gives us $-2 \cdot \frac{1}{s - m} = \frac{-2}{s - m}$ .
Awesome!

STEP 7

Finally, we put it all together.
Since the Laplace transform is a linear operator, we can just add the individual transforms we found.

STEP 8

The Laplace transform of $5 \cos(at) - 2e^{mt}$ is the sum of the transforms we calculated: $\frac{5s}{s^2 + a^2} + \frac{-2}{s - m}$ .
We can rewrite this as $\frac{5s}{s^2 + a^2} - \frac{2}{s - m}$ .

STEP 9

The Laplace transform of $5 \cos(at) - 2e^{mt}$ is $\frac{5s}{s^2 + a^2} - \frac{2}{s - m}$ .

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QuestionPrevious Problem Problem List Next ProblemStandard 16: Problem 6 (1 point)Find the Laplace transform of 5cos⁡(at)−2emt5 \cos (a t)-2 e^{m t}5cos(at)−2emt : □\square□Preview My Answers Submit Answers You have attempted this problem 0 times. You have unlimited attempts remaining.

Studdy Solution

STEP 1

What is this asking? We need to find the Laplace transform of the expression 5cos⁡(at)−2emt5 \cos(at) - 2e^{mt}5cos(at)−2emt. Watch out! Remember the specific formulas for the Laplace transforms of cosine and exponential functions!Don't mix them up.

STEP 2

1. Laplace Transform of Cosine2. Laplace Transform of Exponential3. Combine the Results

STEP 3

Let's **start** with the Laplace transform of 5cos⁡(at)5 \cos(at)5cos(at).Remember, the Laplace transform of cos⁡(at)\cos(at)cos(at) is given by ss2+a2\frac{s}{s^2 + a^2}s2+a2s​.So, we're going to use that!

STEP 4

Now, we **multiply** that by **5** to get the Laplace transform of 5cos⁡(at)5 \cos(at)5cos(at), which is 5⋅ss2+a2=5ss2+a25 \cdot \frac{s}{s^2 + a^2} = \frac{5s}{s^2 + a^2}5⋅s2+a2s​=s2+a25s​.See? Not so bad!

STEP 5

Next up is the Laplace transform of −2emt-2e^{mt}−2emt.The Laplace transform of emte^{mt}emt is 1s−m\frac{1}{s - m}s−m1​.Keep this formula in mind, it's a good one!

STEP 6

We **multiply** by −2-2−2 to find the Laplace transform of −2emt-2e^{mt}−2emt, which gives us −2⋅1s−m=−2s−m-2 \cdot \frac{1}{s - m} = \frac{-2}{s - m}−2⋅s−m1​=s−m−2​.Awesome!

STEP 7

Finally, we **put it all together**.Since the Laplace transform is a linear operator, we can just add the individual transforms we found.

STEP 8

STEP 9

The Laplace transform of 5cos⁡(at)−2emt5 \cos(at) - 2e^{mt}5cos(at)−2emt is 5ss2+a2−2s−m\frac{5s}{s^2 + a^2} - \frac{2}{s - m}s2+a25s​−s−m2​.

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Start learning now

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QuestionPrevious Problem Problem List Next Problem
Standard 16: Problem 6 (1 point)
Find the Laplace transform of $5 \cos (a t)-2 e^{m t}$ : $\square$
Preview My Answers Submit Answers You have attempted this problem 0 times. You have unlimited attempts remaining.

What is this asking? We need to find the Laplace transform of the expression $5 \cos(at) - 2e^{mt}$ . Watch out! Remember the specific formulas for the Laplace transforms of cosine and exponential functions!
Don't mix them up.

1. Laplace Transform of Cosine
2. Laplace Transform of Exponential
3. Combine the Results

Let's start with the Laplace transform of $5 \cos(at)$ .
Remember, the Laplace transform of $\cos(at)$ is given by $\frac{s}{s^2 + a^2}$ .
So, we're going to use that!

Now, we multiply that by 5 to get the Laplace transform of $5 \cos(at)$ , which is $5 \cdot \frac{s}{s^2 + a^2} = \frac{5s}{s^2 + a^2}$ .
See? Not so bad!

Next up is the Laplace transform of $-2e^{mt}$ .
The Laplace transform of $e^{mt}$ is $\frac{1}{s - m}$ .
Keep this formula in mind, it's a good one!

We multiply by $-2$ to find the Laplace transform of $-2e^{mt}$ , which gives us $-2 \cdot \frac{1}{s - m} = \frac{-2}{s - m}$ .
Awesome!

Finally, we put it all together.
Since the Laplace transform is a linear operator, we can just add the individual transforms we found.

The Laplace transform of $5 \cos(at) - 2e^{mt}$ is $\frac{5s}{s^2 + a^2} - \frac{2}{s - m}$ .