Find the La Place Transform of
a.) f(t)=t
b.) f(x)=y"+2y'+Y
To find the Laplace transform of a function, we will first need to recall the definition of the Laplace transform and then apply the basic transforms we have at hand.
a.) f(t) = t:
The Laplace transform of f(t) can be found using the formula:
F(s) = L[f(t)] = ∫[0,∞] of e^(-st) * f(t) dt
where F(s) is the Laplace transform of f(t), s is the complex variable, t is the time variable, and ∫[0,∞] denotes the integral from 0 to infinity.
In this case, we need to determine F(s) for f(t) = t. To do this, we plug t into the formula and evaluate the integral:
F(s) = L[t] = ∫[0,∞] of e^(-st) * t dt
To evaluate this integral, we can use integration by parts. Let's denote u = t and dv = e^(-st) dt. Then we can find du and v by taking the derivative and integral of our choices for u and dv.
du = dt (derivative of u)
v = (-1/s) * e^(-st) (integral of dv)
Now we can use the formula for integration by parts:
∫ u dv = uv - ∫ v du
Applying this formula, we have:
F(s) = [-t/s * e^(-st)] evaluated from t = 0 to t = ∞ - ∫[0,∞] (-1/s) e^(-st) dt
Evaluating the limits of the first term:
F(s) = [(-∞)/s * e^(-∞s) - (0)/s * e^(-0s)] - ∫[0,∞] (-1/s) e^(-st) dt
Since e^(-∞s) approaches zero as s goes to infinity, and e^(-0s) is equal to 1, the first term simplifies to:
F(s) = [0 - 0] - ∫[0,∞] (-1/s) e^(-st) dt
This becomes:
F(s) = ∫[0,∞] (1/s) e^(-st) dt
Now we need to evaluate this integral. Notice that this is the Laplace transform of 1, which is 1/s.
Therefore, we have:
F(s) = 1/s
So, the Laplace transform of f(t) = t is F(s) = 1/s.
b.) f(x) = y" + 2y' + y:
To find the Laplace transform of f(x), we need to use the property that the Laplace transform of a derivative is equal to the product of the Laplace transform of the function and s. Specifically, for a derivative of order n, we have:
L[f^(n)(t)] = s^n * F(s) - s^(n-1) * f(0) - s^(n-2) * f'(0) - ... - f^(n-1)(0)
where F(s) is the Laplace transform of f(t), f'(0) is the initial value of the first derivative of f(t), f''(0) is the initial value of the second derivative of f(t), and so on.
In this case, we have f(x) = y'' + 2y' + y. Let's assume y(0) = a and y'(0) = b for simplicity.
Taking the Laplace transform of both sides, we have:
L[f(x)] = L[y'' + 2y' + y]
F(s) = L[y''] + 2L[y'] + L[y]
Using the property mentioned earlier, and remembering that L[y] = Y, we can express this as:
F(s) = s^2 * Y - sy(0) - y'(0) + 2(s * Y - y(0)) + Y
Substituting y(0) = a and y'(0) = b, we have:
F(s) = s^2 * Y - sa - b + 2s * Y - 2a + Y
Grouping the terms with Y, we obtain:
F(s) = (s^2 + 2s + 1)Y - (sa + 2a + b)
Now let's analyze the term in parentheses: s^2 + 2s + 1. This is actually a perfect square trinomial that can be factored as (s + 1)^2.
So we have:
F(s) = (s + 1)^2 * Y - (sa + 2a + b)
Therefore, the Laplace transform of f(x) = y'' + 2y' + y, denoted as F(s), is:
F(s) = (s + 1)^2 * Y - (sa + 2a + b).