The Laplace transform is a mathematical tool used to convert a function in the time domain to a function in the complex frequency domain. To find the Laplace transform of \( g(t)=5 t e^{-5 t} \), we can use the formula for the Laplace transform of a function \( f(t) \) given by:
\[ F(s) = \int_0^{\infty} f(t) e^{-s t} dt \]
where \( F(s) \) is the Laplace transform of \( f(t) \) and \( s \) is the complex frequency variable.
Now let's find the Laplace transform of \( g(t) \) step-by-step:
Step 1: Write the function in the standard form for a Laplace transform.
The given function \( g(t) \) is already in a suitable form for the Laplace transform.
Step 2: Apply the Laplace transform.
Using the formula for the Laplace transform, we have:
\[ G(s) = \int_0^{\infty} 5 t e^{-5 t} e^{-s t} dt \]
Simplifying the equation, we can factor out the constant 5:
\[ G(s) = 5 \int_0^{\infty} t e^{-(5+s) t} dt \]
Step 3: Solve the integral.
To solve the integral, we can use integration by parts. Integration by parts states that the integral of the product of two functions, \( u \) and \( v' \), is equal to \( uv \) minus the integral of \( u'v \). Let's choose \( u = t \) and \( v' = e^{-(5+s)t} \):
\[ G(s) = 5 \left[ -\frac{t}{5+s} e^{-(5+s)t} \right]_0^{\infty} + 5 \int_0^{\infty} \frac{1}{5+s} e^{-(5+s)t} dt \]
Evaluating the limits of the first term, we get:
\[ G(s) = 5 \left( 0 - \frac{0}{5+s} \right) + 5 \int_0^{\infty} \frac{1}{5+s} e^{-(5+s)t} dt \]
Since the first term is zero, we are left with:
\[ G(s) = 5 \int_0^{\infty} \frac{1}{5+s} e^{-(5+s)t} dt \]
Step 4: Solve the remaining integral.
To solve the remaining integral, we can use the Laplace transform formula:
\[ \mathcal{L}\{ e^{at} \} = \frac{1}{s-a} \]
In this case, \( a = -(5+s) \), so the Laplace transform of \( e^{-(5+s)t} \) is \( \frac{1}{s + 5 + s} \). Substituting this into the equation, we get:
\[ G(s) = 5 \int_0^{\infty} \frac{1}{5+s} \cdot \frac{1}{s + 5 + s} dt \]
Simplifying the equation, we have:
\[ G(s) = 5 \int_0^{\infty} \frac{1}{(5+s)(10+s)} dt \]
Step 5: Split the fraction into partial fractions.
To evaluate the integral, we can split the fraction into partial fractions. The denominator factors as \( (5+s)(10+s) = 0 \), so we have:
\[ \frac{1}{(5+s)(10+s)} = \frac{A}{5+s} + \frac{B}{10+s} \]
where \( A \) and \( B \) are constants to be determined.
Step 6: Determine the values of \( A \) and \( B \).
To determine the values of \( A \) and \( B \), we can multiply through by the denominator and equate the numerators:
\[ 1 = A(10+s) + B(5+s) \]
Expanding and collecting like terms, we get:
\[ 1 = (10A + 5B) + (A + B)s \]
Comparing the coefficients of \( s \) and the constant term, we have:
\[ 0 = 10A + 5B \quad \text{(equation 1)} \]
\[ 1 = A + B \quad \text{(equation 2)} \]
From equation 2, we can solve for \( A \) in terms of \( B \):
\[ A = 1 - B \]
Substituting this into equation 1, we have:
\[ 0 = 10(1-B) + 5B \]
\[ 0 = 10 - 10B + 5B \]
\[ 0 = 10 - 5B \]
\[ 5B = 10 \]
\[ B = 2 \]
Substituting \( B = 2 \) into equation 2, we have:
\[ A = 1 - 2 \]
\[ A = -1 \]
So, we have \( A = -1 \) and \( B = 2 \).
Step 7: Rewrite the integral with the partial fractions.
Using the values of \( A \) and \( B \), we can rewrite the integral as:
\[ G(s) = 5 \left( \int_0^{\infty} \frac{-1}{5+s} dt + \int_0^{\infty} \frac{2}{10+s} dt \right) \]
Step 8: Evaluate the integral.
Now we can evaluate the integrals:
\[ G(s) = 5 \left( -\ln|5+s| \bigg|_0^{\infty} + 2\ln|10+s| \bigg|_0^{\infty} \right) \]
Applying the limits of integration, we get:
\[ G(s) = 5 \left( -\ln|5+s| - 2\ln|10+s| \right) \]
So, the Laplace transform of \( g(t) = 5 t e^{-5 t} \) is \( G(s) = 5 \left( -\ln|5+s| - 2\ln|10+s| \right) \).
Step-by-step explanation:
To find the Laplace transform of the function \( g(t) = 5 t e^{-5 t} \), we used the formula for the Laplace transform and followed a step-by-step approach. We applied integration by parts to solve the integral, split the fraction into partial fractions, determined the values of the constants, and evaluated the integral. Finally, we obtained the Laplace transform of the function as \( G(s) = 5 \left( -\ln|5+s| - 2\ln|10+s| \right) \).