Homogconstantcoef

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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 6

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online