Problem # 1 In class, we considered differential equations of the form 9(t) + a1

Question

Problem # 1 In class, we considered differential equations of the form 9(t) + a19(t) + a29(t) = bo(t) If al > 0 and a2 0, and al 2 a2, then we chose to write the solution in terms of the (wn, 5) parameters, which are derived from a1 and a2. If the forcing function u is a constant, v(t) v, we derived that all particular solutions are of the form Cl2 where A and B are free parameters Suppose the initial conditions are y(0) o and (0) o. Find the correct values for A and B so that the initial conditions are satisfied. Your answer should be in terms of the given initial conditions, and system parameters (wn, 5,b)

Explanation / Answer

Given differential equation is in the form of inhomogeneous equation

y"(t)+a1y'(t)+a2y(t)=bv(t).

First of all,take homogenous equation as second order equation take constant zeta as £

Now equation is

y"(t)+2£wny'(t)+wn2y(t)=0.

So,a1=2£wn and a2=wn2.

Take the condition

a1<2a2---->2£wn<2wn2

After calculation £<1. If £<1 it will satisfy above condition

Now,roots of above quadratic equation are

y(t)=(-2£wn+-(4£2wn2-4wn2)/2.After cancellation

y(t)=-£wn+-wn(1-£2. Because £<1.

If we consider inhomogeneous equation

Given that v(t) is constant that is v(t)=v-..derivative of constant is zero.

So,from differential equation

0+0+a2y(t)=bv-

y(t)=(b/a2)v-.

Finally solution is

y(t)=e-£wnt(Acoswn((1-£2)t+Bsinwn((1-£2)t)+(b/a2)v-.

A and B are constant coefficients.

Given initial conditions are

y(0)=y0   y'(0)=y0'

Substitute these conditions

y(0)=(b/a2)v'+A=y0

A=y0-(b/a2)v-.

y'(0)=(b/a2)v-+B=y0'.

B=y0'-(b/a2)v-. We know that a2=wn2.

Finally solution in terms of initial conditions and system parameters (£,wn,b) is

y(t)=e-£wnt((y0-(b/wn2)v-)coswn((1-£2)t+(y0'-(b/wn2)v-)sinwn((1-£2)t)+(b/wn2)v-.

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.