Differential EquationsSolving - Guide

Solving ay 00 + by 0 + cy = 0 To solve ay 00 + by 0 + cy = 0, we use the ansatz y = eλt . Substitution of the ansatz into the ODE,   eλt aλ2 + bλ + c = 0 The equation aλ2 + bλ + c = 0 is the characteristic equation associated with the DE. Its solutions are from the quadratic formula, √ −b ± b2 − 4ac λ1,2 = 2a A Summary of the Cases (details afterward): 1. If the roots are real, distinct: y(t) = C1 eλ1 t + C2 eλ2 t . 2. If we get a double root (in that case, b2 − 4ac = 0, and λ = −b/2a, then we need a second linearly independent solution. We will show below that the general solution is: y = C1 eλt + C2 teλt = eλt (C1 + tC2 ) 3. If we get complex roots, b2 − 4ac < 0, let λ1,2 = α ± βi, then: y = eαt (C1 cos(βt) + C2 sin(βt)) Details: Case 2 In this case, b2 − 4ac = 0, and we had only one real root, λ = −b/2a. We can obtain the second solution using variation of parameters (Section 4.2), and assume that the second solution is of the form y2 = u(t)eλt . Substituting y2 into the ODE gives:     u00 aeλt + u0 (b + 2aλ)eλt = 0 Divide by aeλt . Substitute λ = −b/2a to simplify: ! 00 u +u 0 b + 2λ = 0 a 00 ⇒u +u 0 b −b +2· a 2a ! =0 ⇒ u00 = 0 Now take u0 = 1 (we don’t need a generic antiderivative- Choose one). And u = t. Therefore, y2 (t) = u(t)eλt = teλt Details: Case 2 (Also see the Practice Sheet) If the roots are complex, b2 − 4ac < 0, we can write √ √ −b ± b2 − 4ac −b 4ac − b2 λ1,2 = = ± i = α ± βi 2a 2a 2a From the Practice Sheet, e(α±βi)t = eαt e±βt i = eαt (cos(±βt) + i sin(±βt)) = eαt (cos(βt) ± i sin(βt)) 1 This last equality is because the cosine is an even function, cos(−w) = cos(w), and sine is an odd function, sin(−w) = − sin(w). If we let y1 = eαt (cos(βt) + i sin(βt)) , y2 = eαt (cos(βt) − i sin(βt)) We know that we can find a fundamental set of solutions by taking the first function as the solution to: ay 00 + by 0 + cy = 0, y(0) = 1, y 0 (0) = α and the second as the solution to: ay 00 + by 0 + cy = 0, y(0) = 0, y 0 (0) = β (Note that the Wronskian at t = 0 is β 6= 0, so that y1 , y2 will form a fundamental set). Let y = C1 y1 + C2 y2 . In the practice problems, we showed that y10 (0) = α + iβ, y20 (0) = α − iβ For the case that y(0) = 1 and y 0 (0) = 0, we get the two equations: C1 + C2 = 1, (α + iβ)C1 + (α − iβ)C2 = α and the solutions are: C1 = C2 = 12 , and y = eαt cos(βt). For the case that y(0) = 0 and y 0 (0) = 1, we get the two equations: C1 + C2 = 0, (α + iβ)C1 + (α − iβ)C2 = β and the solutions are: C1 = − 2i , C2 = 2i , and C1 y1 + c2 y2 simplifies to eαt sin(βt). which is a multiple of just eαt sin(βt). Some Comments on Roots of Polynomials The problem: an y n + . . . + a1 y 0 + a0 y = 0 will always reduce to solving the nth degree polynomial: an λn + an−1 λn−1 + . . . + a1 λ + a0 = 0 Notes: • If pn (x) has a root at x = a, then (x − a) is a factor. We could perform long division (or synthetic division) to factor this term out. • If the coefficients of pn (x) are real, then the polynomial can be factored into a product of linear and irreducible quadratic terms. The roots of the quadratics will always come in complex conjugate pairs. • If a real root a is repeated k times, the independent solutions will be: eat , teat , . . . , tk−1 ekt This will also work in the complex case- Multiply one fundamental set by t to get the second. For example, if the characteristic equation is: (λ2 + 1)2 = 0, a fundamental set would be: sin(t), cos(t), t sin(t), t cos(t) • In general, it is difficult to find the roots of an nth degree polynomial. There are formulas when n = 2, and n = 3. In Abstract Algebra, it is shown that there is no (closed form) formula if n ≥ 5. In Maple, use the roots command. 2