Solve the differential equation by using Laplace transforms. Find the particular solution that satisfies them of:
y''+2y'+y =4e^(-t)
This looks pretty straightforward. How far do you get? Just take L{} of each term and solve for F(s) algebraically.
To solve the given differential equation using Laplace transforms, we first need to take the Laplace transform of both sides of the equation. The Laplace transform of a derivative is given by the formula:
L{f'(t)} = sF(s) - f(0)
where L{f(t)} denotes the Laplace transform of f(t), s is the Laplace variable, and f(0) is the initial value of the function.
Applying the Laplace transform to the given equation, we get:
L{y''} + 2L{y'} + L{y} = 4L{e^(-t)}
Taking the Laplace transform of each term individually, we use the following Laplace transforms:
L{y''} = s^2Y(s) - sy(0) - y'(0)
L{y'} = sY(s) - y(0)
L{y} = Y(s)
L{e^(-t)} = 1/(s+1)
Substituting these Laplace transforms into the equation, we have:
s^2Y(s) - sy(0) - y'(0) + 2(sY(s) - y(0)) + Y(s) = 4/(s+1)
Simplifying this equation, we get:
(s^2 + 2s + 1)Y(s) = 4/(s+1) + sy(0) + y'(0) - 2y(0)
Now, let's find the Laplace transform of the initial conditions y(0) and y'(0):
L{y(0)} = y(0)/s
L{y'(0)} = y'(0)/s
Substituting these transforms into the equation, we have:
(s^2 + 2s + 1)Y(s) = 4/(s+1) + sy(0)/s + y'(0)/s - 2y(0)
Simplifying further, we get:
Y(s) = [4 + sy(0) + y'(0) - 2y(0)] / [(s+1)(s+1)]
Now, we take the inverse Laplace transform of Y(s) to obtain the solution y(t). Since the transform has a repeated factor (s+1), we would need to use partial fraction decomposition to simplify it into simpler terms.
The partial fraction decomposition of Y(s) is given by:
Y(s) = A/(s+1) + B/(s+1)^2
To find the values of A and B, we multiply both sides by the denominator and equate the numerators:
[4 + sy(0) + y'(0) - 2y(0)] = A + B(s+1)
Expanding and equating the coefficients of the like powers of s, we get:
4 + sy(0) + y'(0) - 2y(0) = A + B(s+1)
Comparing the coefficients, we find that A = 3 - 2y(0), and B = 1 + y(0) - y'(0).
Now, we can write the simplified form of Y(s) as:
Y(s) = [3 - 2y(0)]/(s+1) + [1 + y(0) - y'(0)]/(s+1)^2
Taking the inverse Laplace transform of Y(s) using the table of Laplace transform pairs, we can find the solution y(t) in the time domain.