Homogconstantcoef

# Homogconstantcoef - The H.L.D.E with Constant Coefficient...

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The H.L.D.E. with Constant Coefficient Let’s consider the H.L.D.E. with constant coefficients a 0 y n ( ) + a 1 y n ! 1 ( ) + ..... + a n ! 1 " y + a n y = 0 (1) where a 0 , a 1 , …. , a n are constants. Notice that the k-th derivative of e rx is d k dx k e rx ! " # \$ = r k e rx so any derivative of e rx is a multiple of e rx . Assume that the function y = e rx is a solution of equation (1). Substituting y = e rx and its derivatives in equation (1), we get a o r n e rx + a 1 r n ! 1 e rx + a 2 r n ! 2 e rx + + a n ! 1 re rx + a n e rx = 0 factoring out e rx , and dividing both sides by e rx 0, we get a o r n + a 1 r n ! 1 + a 2 r n ! 2 + + a n ! 1 r + a n = 0 (2) Then, the function y = e rx is a solution of (1), if r is a zero of the polynomial (2). Equation (2) is called the characteristic equation of the given H.L.D.E. (1). Therefore, y = e rx is a solution of (1), if r is a solution of the corresponding characteristic equation. Remark : We obtain the corresponding characteristic equation of (1) by replacing the k-th derivative of y by r k ; y itself is consider the derivative order 0, so r 0 = 1 According to the Fundamental Theorem of Algebra, every n-th degree polynomial (such as the characteristic equation) has n zeros, though not necessary distinct and not necessary real numbers. So, four different case arise, according to the nature of the zeros of the characteristic equation: Case 1: the zeros are real numbers and distinct Case 2: the zeros are real numbers and repeated Case 3: the zeros are complex numbers and distinct Case 4: the zeros are complex numbers and repeated Case 1 : Distinct real zeros Suppose that the zeros of the characteristic polynomial are n distinct real numbers: r 1 , r 2 , r 3 , … , r n Then the n functions y 1 = e r 1 x ,y 2 = e r 2 x ,y 3 = e r 3 x , ,y n = e r n x are n distinct solutions of equation (1). They are linearly independent and the general solution of equation (1) can be expressed as: y(x) = c 1 e r 1 x + c 2 e r 2 x + c 3 e r 3 x + + c n e r n x Example : Solve the equation

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1) y’’ + 2y’ – 8y = 0 Characteristic equation: r 2 + 2r – 8 = (r - 2)(r + 4) = 0 The zeros are r 1 = 2, and r 2 = -4, Then y 1 (x) = e 2x and y 2 (x) = e -4x are solutions or the set { e 2x , e -4x } is the fundamental set of solutions. The general solution is: y(x) = c
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Homogconstantcoef - The H.L.D.E with Constant Coefficient...

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