In class we demonstrated that e^alphat1 sin (beta t) is a solution to the second

Question

In class we demonstrated that e^alphat1 sin (beta t) is a solution to the second order. linear, homogeneous differential equation with constant coefficients whose characteristic equation has complex squareroots. Use the method of 'reduction of order' to find a linearly independent solution to the differential equation described above. Use the Wronksian to confirm that these solutions are indeed linearly independent. (Let it be understood that this method is guaranteed to produce a linearly independent solution and that this exercise is purely academic.)

Explanation / Answer

A solution of the form x(t) = cert to the homogeneous constant coeffi­ cient linear equation a x(n) + (n1) . n an1x + · · · + a1x + a0x = 0 (1) is called a modal solution and cert is called a mode of the system. We saw previously that ert is a solution exactly when r is a root of the characteristic polynomial p a n n 1 (s) = ns + an 1s + · · · + a1s + a0. Warning: This only works for homogeneous constant coefficient linear equations. It does not work for non-constant coefficient or inhomogeneous or nonlinear equations. The roots of polynomials can be real or non-real complex numbers. (We need to be a little careful with our language because a real number is also a complex number with imaginary part 0.) Roots can also be repeated. Studying the second order equation will be enough to help us understand all of these possibilities. So, we study (with a2 = m, a1 = b, a0 = k) .. . mx + bx + kx = 0. (2) which models a spring-mass-dashpot system with no external force. The characteristic equation is ms2 + bs + k = 0. 1. Real Roots We have already done this case earlier in this session. If the characteristic polynomial has real roots r1 and r2 then the modal solutions to (2) are x t 1(t r ) = er1t and x2(t) = e 2 . The general solution is found by superposition x r t r t (t) = c1x 1(t) + c2x 2(t) = c e 1 2 1 + c2e . .. . Example 1. (Real roots) Solve the x + 5x + 4x = 0. Solution. The characteristic equation is s2 + 5s + 4 = 0. This factors as (s +1)(s +4) = 0, so it has roots -1, -4. The modal solutions are x1(t) = et and x2(t 4 ) = e t . Therefore, the general solution is x t 4t (t) = c1e + c2e .

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