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Calculus- Early Transcendentals, 2021a

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10.7. Second Order Linear Equations -<br />

Variation of Parameters 397<br />

Solve the initial value problem.<br />

Exercise 10.6.17 y ′′ − y = 3t + 5,y(0)=0, y ′ (0)=0<br />

Exercise 10.6.18 y ′′ + 9y = 4t, y(0)=0, y ′ (0)=0<br />

Exercise 10.6.19 y ′′ + 12y ′ + 37y = 10e −4t ,y(0)=4, y ′ (0)=0<br />

Exercise 10.6.20 y ′′ + 6y ′ + 18y = cost − sint, y(0)=0, y ′ (0)=2<br />

Exercise 10.6.21 Find the solution for the mass-spring equation y ′′ + 4y ′ + 29y = 689cos(2t).<br />

Exercise 10.6.22 Find the solution for the mass-spring equation 3y ′′ + 12y ′ + 24y = 2sint.<br />

Exercise 10.6.23 Consider the differential equation my ′′ +by ′ +ky = cos(ωt), with m, b, and k all positive<br />

and b 2 < 2mk; this equation is a model for a damped mass-spring system with external driving force<br />

cos(ωt). Show that the steady state part of the solution has amplitude<br />

1<br />

√<br />

(k − mω 2 ) 2 + ω 2 b 2<br />

√<br />

4mk − 2b 2<br />

Show that this amplitude is largest when ω =<br />

.Thisistheresonant frequency of the system.<br />

2m<br />

10.7 Second Order Linear Equations -<br />

Variation of Parameters<br />

The method of the last section works only when the function f (t) in ay ′′ +by ′ +cy = f (t) has a particularly<br />

nice form, namely, when the derivatives of f look much like f itself. In other cases we can try variation<br />

of parameters as we did in the first order case.<br />

Since as before a ≠ 0, we can always divide by a to make the coefficient of y ′′ equal to 1. Thus, to<br />

simplify the discussion, we assume a = 1. We know that the differential equation y ′′ + by ′ + cy = 0hasa<br />

general solution y = Ay 1 + By 2 . As before, we guess a particular solution to y ′′ + by ′ + cy = f (t);thistime<br />

we use the guess y = u(t)y 1 + v(t)y 2 . Compute the derivatives:<br />

Now substituting:<br />

y ′ = u ′ y 1 + uy ′ 1 + v′ y 2 + vy ′ 2<br />

y ′′ = u ′′ y 1 + u ′ y ′ 1 + u′ y ′ 1 + uy′′ 1 + v′′ y 2 + v ′ y ′ 2 + v′ y ′ 2 + vy′′ 2 .<br />

y ′′ + by ′ + cy = u ′′ y 1 + u ′ y ′ 1 + u′ y ′ 1 + uy′′ 1 + v′′ y 2 + v ′ y ′ 2 + v′ y ′ 2 + vy′′ 2<br />

+bu ′ y 1 + buy ′ 1 + bv′ y 2 + bvy ′ 2 + cuy 1 + cvy 2<br />

= (uy ′′<br />

1 + buy′ 1 + cuy 1)+(vy ′′<br />

2 + bvy′ 2 + cvy 2)<br />

+b(u ′ y 1 + v ′ y 2 )+(u ′′ y 1 + u ′ y ′ 1 + v ′′ y 2 + v ′ y ′ 2)+(u ′ y ′ 1 + v ′ y ′ 2)<br />

= 0 + 0 + b(u ′ y 1 + v ′ y 2 )+(u ′′ y 1 + u ′ y ′ 1 + v′′ y 2 + v ′ y ′ 2 )+(u′ y ′ 1 + v′ y ′ 2 ).

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