Answer:
y(t) = c₁ e^(-1/2 t) cos(√3/2 t) + c₂ e^(-1/2 t) sin(√3/2 t) + 1
Step-by-step explanation:
y" + y' + y = 1
This is a second order nonhomogenous differential equation with constant coefficients.
First, find the roots of the complementary solution.
y" + y' + y = 0
r² + r + 1 = 0
r = [ -1 ± √(1² − 4(1)(1)) ] / 2(1)
r = [ -1 ± √(1 − 4) ] / 2
r = -1/2 ± i√3/2
These roots are complex, so the complementary solution is:
y = c₁ e^(-1/2 t) cos(√3/2 t) + c₂ e^(-1/2 t) sin(√3/2 t)
Next, assume the particular solution has the form of the right hand side of the differential equation. In this case, a constant.
y = c
Plug this into the differential equation and use undetermined coefficients to solve:
y" + y' + y = 1
0 + 0 + c = 1
c = 1
So the total solution is:
y(t) = c₁ e^(-1/2 t) cos(√3/2 t) + c₂ e^(-1/2 t) sin(√3/2 t) + 1
To solve for c₁ and c₂, you need to be given initial conditions.
Less than 45° F is potentially hazardous food.
HI= 15 i think i hope that is right Your welcome
