4.2 nth Order Linear Equations with Constant Coefficients PDF

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4.2 nth Order Linear Equations with Constant Coefficients...

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Section 4.2: nth Order Linear Equations with Constant Coefficients A third order homogeneous equation with constant coefficients takes the form: a y ′′′ (t) + b y ′′(t) + c y ′ (t) + d y (t) = 0, and has characteristic equation: a r 3 + b r 2 + c r + d = 0. This pattern continues for higher order equations. When solving second order homogeneous equations with constant coefficients, a real root could only be a solution to the characteristic equation twice, and complex roots couldn’t be repeated. For a general n th order equation, a real value of r can solve the characteristic equation n times, and complex roots can also be repeated.

Repeated Roots for Higher Order Characteristic Equations If r = r0 is a real root of the characteristic equation m times, then: e r0 t , te r0 t , . . . , t m−1 e r0 t

solve the differential equation.

If r = λ ± µi is a root of the characteristic equation m times, then: e λt cos(µt), e λt sin(µt ), . . . , t m−1 e λt cos(µt), t m−1 e λt sin(µt) solve the differential equation. If r = ±µi is a root of the characteristic equation m times, then: cos(µt), sin(µt), . . . , t m−1 cos(µt), t m−1 sin(µt) solve the differential equation.

Section 4.2: nth Order Linear Equations with Constant Coefficients Finding Solutions to nth Order Characteristic Equations A characteristic equation of order n will always have n roots (counting repeated roots). Useful Fact: If r = λ + µi is a root of a polynomial with real-valued coefficients, then r = λ − µi is also a root.

Example Consider the equation y (4) + 81y = 0. Find all four roots to the characteristic equation, and use those roots to find the general solution.

Solution The characteristic equation is r 4 + 81 = 0 =⇒ r 4 = −81. This doesn’t have any real solutions because a real number r raised to the fourth power cannot equal a negative number. We start by rewriting r 4 = −81 as r 4 = 34 (−1) = 34 e iθ . Next, use Euler’s formula to find a value of θ that solves e iθ = −1 e iθ = cos(θ) + i sin(θ)

=⇒

e iπ = −1

=⇒

θ=π

Taking θ = π will give the following solution to the characteristic equation:      r 4 = 34 e iπ =⇒ r = 3e (π/4)i = 3 cos 4π + i sin π4 =

√ 3 2 (1 2

+ i)

We still need to find three more solutions to the characteristic equation. Because cos(θ) and sin(θ) are periodic, the solution θ = π is not unique. In fact, θ can be π + 2πk for any integer value of k . Taking k = 0 gives us π, so we’ll look at k = 1, k = 2, and k = 3 for the rest of our solutions: k = 1: k = 2: k = 3:

θ = π + 2π(1) = 3π θ = π + 2π(2) = 5π θ = π + 2π(3) = 7π

r 4 = 34 e 3πi =⇒ r = 3e (3π/4)i      = = 3 cos 3π + i sin 3π 4 4

r 4 = 34 e 5πi =⇒ r = 3e (5π/4)i     5π  = = 3 cos 5π + i sin 4 4

r 4 = 34 e 7πi =⇒ r = 3e (7π/4)i     7π  = 3 cos 7π + i sin = 4 4

The four solutions, in conjugate pairs are:

r=

√ 3 2 (1 2

± i)

and

r=

√ 3 2 (−1 2

√ 3 2 (−1 2

+ i)

√ 3 2 (−1 2

− i)

√ 3 2 (1 2

− i)

± i)

So our general solution is: √  √ i  √ i √ h  √  h  √  3 2 3 2 y (t) = e 2 t c1 cos 3 2 2 t + c2 sin 3 2 2 t + e − 2 t c3 cos 3 2 2 t + c4 sin 3 2 2 t √

Note: We could have also just used k = 0 and k = 1 to find r = 32 2(1 + i) and r = then used the fact that having r = λ + µi as a root means r = λ − µi is also a root.

√ 3 2 (−1 2

+ i), and...