Use the Laplace transform to solve the given initial value p

Use the Laplace transform to solve the given initial value problem. y\" + 3y\' + 2y = 0; y(0) = 1. y\'(0) = 0 y\" + 2y\' + 5y = 0; y(0) = 2. y\'(0) = -1 Express the following function in terms of the Heaviside function u_c(t). f(t) = {sint, 0 lessthanorequalto t, pi t^3, pi lessthanorequalto t

Solution

5 )

a) y\"+ 3y\'+2y=0 , y(0)=0, y\'(0)=1

r^2 + 3r + 2 = 0

(r + 2)(r + 1) = 0

r + 2 = 0, r + 1 = 0
r = - 2, r = -1

general solution

y = ce^(x) + ce^(2x)

when y(0) = 0,

0 = ce^(0) + ce^[2(0)]
0 = c + c-------------- eq1

when y\'(0) = 1

y = ce^(x) + ce^(2x)
y\' = ce^(x) + 2ce^(2x)
1 = ce^(0) + 2ce^[2(0)]
1 = c + 2c ------------------------ eq2

solving for c and c bu elimination

0 = c + c
-[1 = c + 2c]
--------------------
- 1 = - c
c = 1

solving for c from eq1

0 = c + c
0 = c + 1
c = - 1

so

y = ce^(x) + ce^(2x)
y = - e^(x) + e^(2x)
y = e^(x)[e^(x) - 1] ---------------- Answer

b) for equation y\'\' + 2y\' + 5y = 0

Applying L to both sides:

[s^2 Y(s) - s * 2 - (-1)] + 2 [sY(s) - 2] + 5 Y(s) = 0

Solve for Y(s):

(s^2 + 2s + 5) Y(s) = 2s + 3
Y(s) = (2s + 3)/(s^2 + 2s + 5).

Complete the square:

Y(s) = (2s + 3) / [(s + 1)^2 + 4]
= (2(s + 1) + 1) / [(s + 1)^2 + 4]
= 2(s + 1)/[(s + 1)^2 + 4] + 1/[(s + 1)^2 + 4]

Inverting term by term yields
y(t) = e^(-t) [2 cos(2t) + (1/2) sin(2t)].