gbuitkus said:Homework Statement
y''-4y'-32y={1 when 0<=t<1 and 0 when 1<=t
y(0)=y'(0)=0Homework Equations
The Attempt at a Solution
s2L(y) -4sL(y)-32 L(y)=u1(t)
I am just struggling to figure out if my unit step function is correct.
Solving for L(y) I get:
(e-s) / (s(s2 -4s-32))
but it says I am wrong so I am pretty sure I am doing something wrong with the unit step function
The Laplace Unit Step function, also known as the Heaviside step function, is a mathematical function commonly used in physics and engineering to represent a signal that switches on at a specific time and stays on thereafter.
The Laplace Unit Step function is defined as:
u(t) = 0 for t < 0
u(t) = 1 for t ≥ 0
where t represents time.
The Laplace transform of the Unit Step function is:
L(u(t)) = 1/s
where s is the complex frequency variable.
The Laplace Unit Step function is often used in engineering to model and analyze systems with sudden changes or impulses, such as electrical circuits and control systems.
Yes, the Laplace Unit Step function can be used to solve differential equations through the process of Laplace transform. It helps to simplify the equations and make them easier to solve by transforming them into algebraic equations.