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6.9: Discontinuous Forcing

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In the last section we looked at the Heaviside function its Laplace transform. Now we will use this tool to solve differential equations.

Example 6.9.1

A 1 kg bar is attached to a spring with spring constant 5 and damping constant 2. It is pulled down 3 inches from it equilibrium position and the released. After one second a constant force of 10 Newtons is exerted on the bar. The force remains turned on indefinitely. Determine the equation of motion of the bar.

Solution

We have the differential equation

y+2y+5y=10u1(t)

with

y(0)=3

y(0)=0.

We solve this by the method of Laplace transforms. We have

L{y}+2L{y}+5L{y}=10L{u1(t)}s2L{y}3s0+2(sL{y}3)+5L{y}=10es/s(s2+2s+5)L{y}3s6=10es/sL{y}=3s+6s2+2s+5+10s(s2+2s+5).

We use partial fractions on the second term

10s(s2+2s+4)=As+Bs+Cs2+2s+5

A(s2+2s+5)+(Bs+C)s=10

(A+B)s2+(2A+C)s+5A=10

The constant term gives

A=2

Thus

B=2 C=4

We can complete the square

s2+2s+5=(s+1)2+4

Putting all the algebra together we get

L{y}=3s+6(s+1)2+4+es2s+es2s4(s+1)2+4=3s+1+1(s+1)2+4+es2s2es(s+1)+1(s+1)2+4=3s+1(s+1)2+4+3/22(s+1)2+4+es2s2es(s+1)(s+1)2+4es2(s+1)2+4

Now use the table to take the inverse Laplace transform to get

y=3etcos(2t)+3/2et1sin(2t)+2u12u1(t)et+1cos(2t2)u1(t)et+1sin(2t2).

Below is the graph

discon2.gif

Contributors and Attributions


This page titled 6.9: Discontinuous Forcing is shared under a not declared license and was authored, remixed, and/or curated by Larry Green.

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