Final answer:
To solve the differential equation y''-8y'+20y = te^t using the Laplace Transform, we first take the transform of each term and apply initial conditions, then we solve the resulting algebraic equation for Y(s). The solution y(t) is obtained by taking the inverse Laplace Transform of Y(s).
Step-by-step explanation:
Solving a Differential Equation using Laplace Transform
To solve the differential equation y''-8y'+20y = te^t with initial conditions y(0) = 0 and y'(0) = 0 using the Laplace Transform, we must follow a series of steps. First, we take the Laplace Transform of each term in the equation and apply the initial conditions. Using properties of Laplace Transforms, we can express this in terms of Y(s), the Laplace transform of y(t). We then have an algebraic equation which we can solve for Y(s). Once we have Y(s), we find y(t) by taking the inverse Laplace Transform.
We start by taking the Laplace Transform of each term:
- Laplace Transform of y'': L{y''(t)} = s^2Y(s) - sy(0) - y'(0)
- Laplace Transform of y': L{y'(t)} = sY(s) - y(0)
- Laplace Transform of y: L{y(t)} = Y(s)
- Laplace Transform of te^t: L{te^t} = \frac{1}{(s-1)^2}
Substitute these into the transformed equation:
s^2Y(s) - 8sY(s) + 20Y(s) = \frac{1}{(s-1)^2}
Next, we solve for Y(s):
Y(s) = \frac{1}{(s^2 - 8s + 20)} \frac{1}{(s-1)^2}
Finally, use partial fraction decomposition and inverse Laplace transform to find y(t). The particular solution can be found using convolution or applying the properties of the Laplace Transform for products of functions.
Given the initial conditions y(0) = 0 and y'(0) = 0, the resulting transforms simplify, and we are left with an equation where we can easily use the inverse Laplace Transform to get the solution y(t).