SAM Mathematics for Engineers Lecture 4 Contents Homogenous second order linear differential equations Homogenous linear differential equations with constant coefficients SAM Mathematics for Engineers Lecture 4 SAM — Seminar in Abstract Mathematics [Version 20130226] is created by Zurab Janelidze at Mathematics Division, Stellenbosch Univeristy SAM Mathematics for Engineers Lecture 4 Contents Homogenous second order linear differential equations Homogenous linear differential equations with constant coefficients 1 Homogenous second order linear differential equations 2 Homogenous linear differential equations with constant coefficients Non-constant coefficients SAM Mathematics for Engineers Lecture 4 Contents Homogenous second order linear differential equations Homogenous linear differential equations with constant coefficients Consider a homogenous second order linear equation y 00 + b(x)y 0 + c(x)y = 0. (1) Given one solution y1 of this equation such that ∀x (y1 (x) 6= 0), we can find another solution y2 in the form y2 = y1 u, such that y2 is linearly independent from y1 . Indeed, the substation y = uy1 (with y 0 = u 0 y1 + uy10 and y 00 = u 00 y1 + 2u 0 y10 + uy100 ) gives u 00 y1 + 2u 0 y10 + uy100 + bu 0 y1 + buy10 + cuy1 = 0 which simplifies to u 00 y1 + (2y10 + by1 )u 0 = 0. We can find a solution of this equation by Rthe substitution w = u 0 and using separation of variables to get w = e − b(x)dx y1−2 . Finally, Z y2 = y1 u = y1 e− R b(x)dx y1−2 dx which is easily seen to be linearly independent from y1 . Constant coefficients SAM Mathematics for Engineers Lecture 4 Suppose now that in the equation y 00 + by 0 + cy = 0 Contents Homogenous second order linear differential equations Homogenous linear differential equations with constant coefficients b and c are constants. Then we can find two linearly independent solutions by substitution y = e mx where m ∈ R. Indeed, this yields m2 + bm + c = 0. Suppose m1 and m2 are the two roots of this quadratic equation. When m1 6= m2 are real roots, the functions y1 = e m1 x and y2 = e m2 x are linearly independent. When m1 = m2 = m then the discriminant of the quadratic equation must be 0, so we will get m = − b2 . After this, direct verification shows that we can take y2 = xe mx as a solution linearly independent from y1 = e mx . When m1 and m2 are complex, say m1 = a + bi (then m2 = a − bi), we can take y1 to be the real part of e mx = e ax cos bx + ie ax sin bx and y2 to be the imaginary part of e mx . Homogenous linear differential equations with constant coefficients SAM Mathematics for Engineers Lecture 4 Consider a homogenous linear differential equation y (n) + an−1 y (n−1) + · · · + a1 y 0 + a0 y = 0 Contents Homogenous second order linear differential equations Homogenous linear differential equations with constant coefficients where an−1 , . . . , a0 are constants. The technique of finding linearly independent solutions of such an equation in the case when n = 2 straightforwardly generalizes to arbitrary n. In detail, if m is a solution of the corresponding equation mn + an−1 mn−1 + · · · + a1 m + a0 = 0 having multiplicity k, then it produces k linearly independent solutions y1 = e mx , y1 = xe mx , . . . , y1 = x k−1 e mx . When m happens to be a complex number m = a + bi, we simply take real and imaginary parts of these solution to get 2k linearly independent solutions — the other k of these should be deemed to be on the account of the complex conjugate of m which will also be a root of the above equation since it has real coefficients.
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