Solved on Mar 27, 2024

Find the Laplace transform of 3te5t3te^{5t} using a table of Laplace transforms. Integration by parts may be useful.

STEP 1

1. The function f(t)=3te5tf(t)=3te^{5t} is a piecewise continuous function on [0,)[0, \infty) and of exponential order, which allows us to use the Laplace transform.
2. The Laplace transform of a function f(t)f(t) is given by F(s)=L{f(t)}=0estf(t)dtF(s) = \mathcal{L}\{f(t)\} = \int_0^\infty e^{-st}f(t)dt.
3. A table of Laplace transforms is available to find the transform of basic functions.
4. Integration by parts may be used to find the Laplace transform of products of functions.

STEP 2

1. Recognize the form of f(t)f(t) that matches an entry in the Laplace transform table.
2. If necessary, use integration by parts to rewrite f(t)f(t) in a form that matches an entry in the table.
3. Use the table to find the Laplace transform of f(t)f(t).
4. Simplify the result to get F(s)F(s) in terms of ss.

STEP 3

Identify the form of f(t)f(t) that matches an entry in the Laplace transform table.
The function f(t)=3te5tf(t)=3te^{5t} resembles the form tneatt^n e^{at}, where n=1n=1 and a=5a=5.

STEP 4

Since f(t)f(t) is not exactly in the form of a basic function listed in the table, we will use integration by parts to transform it into a suitable form.
Integration by parts formula is given by udv=uvvdu\int u dv = uv - \int v du.

STEP 5

Choose u=tu = t and dv=3e5tdtdv = 3e^{5t}dt so that we can integrate dvdv easily.
Now we need to find dudu and vv.

STEP 6

Calculate dudu and vv.
Differentiate uu to get du=dtdu = dt and integrate dvdv to get v=35e5tv = \frac{3}{5}e^{5t}.

STEP 7

Apply integration by parts to the Laplace transform of f(t)f(t).
F(s)=L{3te5t}=0est3te5tdt F(s) = \mathcal{L}\{3te^{5t}\} = \int_0^\infty e^{-st}3te^{5t}dt

STEP 8

Substitute uu, dudu, vv into the integration by parts formula.
F(s)=3t35e(5s)t0035e(5s)tdt F(s) = \left.3t \cdot \frac{3}{5}e^{(5-s)t}\right|_0^\infty - \int_0^\infty \frac{3}{5}e^{(5-s)t}dt

STEP 9

Evaluate the first term at the boundaries.
The term 3t35e(5s)t0\left.3t \cdot \frac{3}{5}e^{(5-s)t}\right|_0^\infty goes to zero as tt approaches infinity, assuming s>5s > 5.

STEP 10

Simplify the remaining integral.
F(s)=950e(5s)tdt F(s) = - \frac{9}{5} \int_0^\infty e^{(5-s)t}dt

STEP 11

Use the table of Laplace transforms to find the transform of e(5s)te^{(5-s)t}.
The Laplace transform of eate^{at} is 1sa\frac{1}{s-a}, so the transform of e(5s)te^{(5-s)t} is 1s(5s)\frac{1}{s-(5-s)}.

STEP 12

Substitute the result from the table into the integral.
F(s)=951s(5s) F(s) = - \frac{9}{5} \cdot \frac{1}{s-(5-s)}

STEP 13

Simplify the expression for F(s)F(s).
F(s)=9515=95s25 F(s) = - \frac{9}{5} \cdot \frac{1}{-5} = \frac{9}{5s-25}

STEP 14

Write the final result for F(s)F(s).
F(s)=95s25 F(s) = \frac{9}{5s-25}

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