Math / Calculus

Questionii. $\mathcal{L}\left\{\cosh 2 t e^{4 t}\right\} \quad$ ; First Shift Theorem ; Teorem Anjakan Pertama

Studdy Solution

STEP 1

What is this asking? Find the Laplace transform of the function $\cosh(2t)e^{4t}$ . Watch out! Don't forget the definition of hyperbolic cosine and the First Shift Theorem!

STEP 2

1. Define the function
2. Apply the First Shift Theorem

STEP 3

Alright, let's define our function!
We've got $f(t) = \cosh(2t)e^{4t}$ .
Remember that hyperbolic cosine is defined as $\cosh(x) = \frac{e^x + e^{-x}}{2}$ .
So, let's rewrite our function using this definition:
$f(t) = \frac{e^{2t} + e^{-2t}}{2} \cdot e^{4t}.$

STEP 4

Now, let's distribute that $e^{4t}$ to both terms in the numerator:
$f(t) = \frac{e^{2t} \cdot e^{4t} + e^{-2t} \cdot e^{4t}}{2}.$

STEP 5

Using the exponent rule $e^a \cdot e^b = e^{a+b}$ , we can simplify this to:
$f(t) = \frac{e^{2t+4t} + e^{-2t+4t}}{2} = \frac{e^{6t} + e^{2t}}{2}.$

STEP 6

Now for the main event: the First Shift Theorem!
This theorem states that if $\mathcal{L}\{f(t)\} = F(s)$ , then $\mathcal{L}\{e^{at}f(t)\} = F(s-a)$ .
This is super useful because it lets us deal with those pesky exponential terms.

STEP 7

Let's break down our function into pieces we can work with.
We have $f(t) = \frac{1}{2}(e^{6t} + e^{2t})$ .
We know that $\mathcal{L}\{e^{at}\} = \frac{1}{s-a}$ for $s > a$ .

STEP 8

Applying the Laplace transform to our function, we get:
$\mathcal{L}\{f(t)\} = \frac{1}{2} \left( \mathcal{L}\{e^{6t}\} + \mathcal{L}\{e^{2t}\} \right).$

STEP 9

Using our knowledge of the Laplace transform of $e^{at}$ , we have:
$\mathcal{L}\{f(t)\} = \frac{1}{2} \left( \frac{1}{s-6} + \frac{1}{s-2} \right).$

STEP 10

And there we have it!
We've successfully transformed our function!

STEP 11

The Laplace transform of $\cosh(2t)e^{4t}$ is $\frac{1}{2} \left( \frac{1}{s-6} + \frac{1}{s-2} \right)$ .

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Questionii. $\mathcal{L}\left\{\cosh 2 t e^{4 t}\right\} \quad$ ; First Shift Theorem ; Teorem Anjakan Pertama

Studdy Solution

STEP 1

What is this asking? Find the Laplace transform of the function $\cosh(2t)e^{4t}$ . Watch out! Don't forget the definition of hyperbolic cosine and the First Shift Theorem!

STEP 2

1. Define the function
2. Apply the First Shift Theorem

STEP 3

Alright, let's define our function!
We've got $f(t) = \cosh(2t)e^{4t}$ .
Remember that hyperbolic cosine is defined as $\cosh(x) = \frac{e^x + e^{-x}}{2}$ .
So, let's rewrite our function using this definition:
$f(t) = \frac{e^{2t} + e^{-2t}}{2} \cdot e^{4t}.$

STEP 4

Now, let's distribute that $e^{4t}$ to both terms in the numerator:
$f(t) = \frac{e^{2t} \cdot e^{4t} + e^{-2t} \cdot e^{4t}}{2}.$

STEP 5

Using the exponent rule $e^a \cdot e^b = e^{a+b}$ , we can simplify this to:
$f(t) = \frac{e^{2t+4t} + e^{-2t+4t}}{2} = \frac{e^{6t} + e^{2t}}{2}.$

STEP 6

Now for the main event: the First Shift Theorem!
This theorem states that if $\mathcal{L}\{f(t)\} = F(s)$ , then $\mathcal{L}\{e^{at}f(t)\} = F(s-a)$ .
This is super useful because it lets us deal with those pesky exponential terms.

STEP 7

Let's break down our function into pieces we can work with.
We have $f(t) = \frac{1}{2}(e^{6t} + e^{2t})$ .
We know that $\mathcal{L}\{e^{at}\} = \frac{1}{s-a}$ for $s > a$ .

STEP 8

Applying the Laplace transform to our function, we get:
$\mathcal{L}\{f(t)\} = \frac{1}{2} \left( \mathcal{L}\{e^{6t}\} + \mathcal{L}\{e^{2t}\} \right).$

STEP 9

Using our knowledge of the Laplace transform of $e^{at}$ , we have:
$\mathcal{L}\{f(t)\} = \frac{1}{2} \left( \frac{1}{s-6} + \frac{1}{s-2} \right).$

STEP 10

And there we have it!
We've successfully transformed our function!

STEP 11

The Laplace transform of $\cosh(2t)e^{4t}$ is $\frac{1}{2} \left( \frac{1}{s-6} + \frac{1}{s-2} \right)$ .

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Questionii. L{cosh⁡2te4t}\mathcal{L}\left\{\cosh 2 t e^{4 t}\right\} \quadL{cosh2te4t}; First Shift Theorem ; Teorem Anjakan Pertama

Studdy Solution

STEP 1

What is this asking? Find the Laplace transform of the function cosh⁡(2t)e4t\cosh(2t)e^{4t}cosh(2t)e4t. Watch out! Don't forget the definition of hyperbolic cosine and the First Shift Theorem!

STEP 2

1. Define the function2. Apply the First Shift Theorem

STEP 3

STEP 4

Now, let's **distribute** that e4te^{4t}e4t to both terms in the numerator:f(t)=e2t⋅e4t+e−2t⋅e4t2.f(t) = \frac{e^{2t} \cdot e^{4t} + e^{-2t} \cdot e^{4t}}{2}.f(t)=2e2t⋅e4t+e−2t⋅e4t​.

STEP 5

Using the **exponent rule** ea⋅eb=ea+be^a \cdot e^b = e^{a+b}ea⋅eb=ea+b, we can simplify this to:f(t)=e2t+4t+e−2t+4t2=e6t+e2t2.f(t) = \frac{e^{2t+4t} + e^{-2t+4t}}{2} = \frac{e^{6t} + e^{2t}}{2}.f(t)=2e2t+4t+e−2t+4t​=2e6t+e2t​.

STEP 6

Now for the **main event**: the First Shift Theorem!This theorem states that if L{f(t)}=F(s)\mathcal{L}\{f(t)\} = F(s)L{f(t)}=F(s), then L{eatf(t)}=F(s−a)\mathcal{L}\{e^{at}f(t)\} = F(s-a)L{eatf(t)}=F(s−a).This is super useful because it lets us deal with those pesky exponential terms.

STEP 7

Let's **break down** our function into pieces we can work with.We have f(t)=12(e6t+e2t)f(t) = \frac{1}{2}(e^{6t} + e^{2t})f(t)=21​(e6t+e2t).We know that L{eat}=1s−a\mathcal{L}\{e^{at}\} = \frac{1}{s-a}L{eat}=s−a1​ for s>as > as>a.

STEP 8

Applying the **Laplace transform** to our function, we get:L{f(t)}=12(L{e6t}+L{e2t}).\mathcal{L}\{f(t)\} = \frac{1}{2} \left( \mathcal{L}\{e^{6t}\} + \mathcal{L}\{e^{2t}\} \right).L{f(t)}=21​(L{e6t}+L{e2t}).

STEP 9

Using our **knowledge** of the Laplace transform of eate^{at}eat, we have:L{f(t)}=12(1s−6+1s−2).\mathcal{L}\{f(t)\} = \frac{1}{2} \left( \frac{1}{s-6} + \frac{1}{s-2} \right).L{f(t)}=21​(s−61​+s−21​).

STEP 10

And there we have it!We've successfully **transformed** our function!

STEP 11

The Laplace transform of cosh⁡(2t)e4t\cosh(2t)e^{4t}cosh(2t)e4t is 12(1s−6+1s−2)\frac{1}{2} \left( \frac{1}{s-6} + \frac{1}{s-2} \right)21​(s−61​+s−21​).

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Questionii. $\mathcal{L}\left\{\cosh 2 t e^{4 t}\right\} \quad$ ; First Shift Theorem ; Teorem Anjakan Pertama

What is this asking? Find the Laplace transform of the function $\cosh(2t)e^{4t}$ . Watch out! Don't forget the definition of hyperbolic cosine and the First Shift Theorem!

1. Define the function
2. Apply the First Shift Theorem

Now, let's distribute that $e^{4t}$ to both terms in the numerator:
$f(t) = \frac{e^{2t} \cdot e^{4t} + e^{-2t} \cdot e^{4t}}{2}.$

Using the exponent rule $e^a \cdot e^b = e^{a+b}$ , we can simplify this to:
$f(t) = \frac{e^{2t+4t} + e^{-2t+4t}}{2} = \frac{e^{6t} + e^{2t}}{2}.$

Now for the main event: the First Shift Theorem!
This theorem states that if $\mathcal{L}\{f(t)\} = F(s)$ , then $\mathcal{L}\{e^{at}f(t)\} = F(s-a)$ .
This is super useful because it lets us deal with those pesky exponential terms.

Let's break down our function into pieces we can work with.
We have $f(t) = \frac{1}{2}(e^{6t} + e^{2t})$ .
We know that $\mathcal{L}\{e^{at}\} = \frac{1}{s-a}$ for $s > a$ .

Applying the Laplace transform to our function, we get:
$\mathcal{L}\{f(t)\} = \frac{1}{2} \left( \mathcal{L}\{e^{6t}\} + \mathcal{L}\{e^{2t}\} \right).$

Using our knowledge of the Laplace transform of $e^{at}$ , we have:
$\mathcal{L}\{f(t)\} = \frac{1}{2} \left( \frac{1}{s-6} + \frac{1}{s-2} \right).$

And there we have it!
We've successfully transformed our function!

The Laplace transform of $\cosh(2t)e^{4t}$ is $\frac{1}{2} \left( \frac{1}{s-6} + \frac{1}{s-2} \right)$ .