Question

Take the Laplace transform of the following initial value and solve for Y(s)=L{y(t)}: sin(π t), 0 ≤ t;1

y''+9y= { y(0)=0, y′(0)= 0, 1 ≤ t
a) Y(s)= ? (Hint: write the right-handed side in terms of the Heaviside function)
b) Now find the inverse transform to find y. (Use step(t-c) for uc(t) .)
Note= π/ (s²+π²)(s²+9) = π/ π² -9 (1/(s²+9) - 1/(s²+ π²)

Answer 1

The task involves solving for Y(s) using the Laplace transform of the piecewise function given by sin(πt), modified by the Heaviside function, and then determining y(t) by taking the inverse Laplace transform.

Step-by-step explanation:

The question asks us to solve a differential equation using Laplace transforms with given initial conditions.

a) By applying the Heaviside function, we can represent the piecewise function on the right-hand side as sin(πt)u(t-1), where u(t-c) is the step function. The Laplace transform of sin(πt) multiplied by the Heaviside function shifted by 1 is L{sin(πt)u(t-1)} = πe-s / (s2 + π2). Since we have initial conditions y(0) = 0 and y'(0) = 0, the transformed equation is s2Y(s) - sy(0) - y'(0) + 9Y(s) = πe-s / (s2 + π2). Simplified, this gives Y(s) = πe-s / (s2 + 9)(s2 + π2).

b) The inverse Laplace transform will then be used to find y(t). The given hint suggests the partial fraction decomposition of Y(s), which leads to y(t) = L-1{π/(π2 - 9)(1/(s2 + 9) - 1/(s2 + π2))}. After calculating the inverse Laplace, we find the solution for the initial value problem in terms of t.