Solution of Constant Coefficients ODE

Transcription

1 Solution of Constant Coefficients ODE Department of Mathematics IIT Guwahati

2 Thus, k r k (erx ) r=r1 = x k e r 1x will be a solution to L(y) = 0 for k = 0, 1,..., m 1. So, m distinct solutions are e r 1x, xe r 1x,, x m 1 e r 1x. Theorem: If P (r) = 0 has the real root r 1 occurring m times and the remaining roots r m+1, r m+2,..., r n are distinct, then the general solution of L(y) = 0 is y(x) = (C 1 + C 2 x + C 3 x C m x m 1 )e r 1x +C m+1 e r m+1x + + C n e rnx, where C 1, C 2,..., C n are arbitrary constants.

3 Example: Consider y (4) 8y + 16y = 0. In this case, r 1 = r 2 = 2 and r 3 = r 4 = 2. The general solution is y = (C 1 + C 2 x)e 2x + (C 3 + C 4 x)e 2x. Case III (Complex roots): If α + iβ is a non-repeated complex root of P (r) = 0 so is its complex conjugate. Then, both e (α+iβ)x and e (α iβ)x are solution to L(y) = 0. Then, the corresponding part of the general solution is of the form e αx (C 1 cos(βx) + C 2 sin(βx)).

4 Theorem: If P (r) = 0 has non-repeated complex roots α + iβ and α iβ, the corresponding part of the general solution is e αx (C 1 cos(βx) + C 2 sin(βx)). If α + iβ and α iβ are each repeated roots of multiplicity m, then the corresponding part of the general solution is e αx [ (C 1 + C 2 x + C 3 x C m x m 1 ) cos(βx) +(C m+1 + C m+2 x + + C 2m x m 1 ) sin(βx) ], where C 1, C 2,..., C 2m are arbitrary constants. Example: Consider y (4) 2y + 2y 2y + y = 0. Here, r 1 = r 2 = 1, r 3 = i and r 4 = i. The general solution is y = (C 1 + C 2 x)e x + (C 3 cos x + C 4 sin x).

5 Particular solution of constant coefficients ODE Method of undetermined coefficients: A simple procedure for finding a particular solution(y p ) to a non-homogeneous equation L(y) = g, when L is a linear differential operator with constant coefficients and when g(x) is of special type: That is, when g(x) is either a polynomial in x, an exponential function e αx, trigonometric functions sin(βx), cos(βx) or finite sums and products of these functions.

6 Case I. For finding y p to the equation L(y) = p n (x), where p n (x) is a polynomial of degree n. Try a solution of the form y p (x) = A n x n + + A 1 x + A 0 and match the coefficients of L(y p ) with those of p n (x): L(y p ) = p n (x). Remark: This procedure yields n + 1 linear equations in n + 1 unknowns A 0,..., A n.

7 Example: Find y p to L(y)(x) := y + 3y + 2y = 3x + 1. Try the form y p (x) = Ax + B and attempt to match up L(y p ) with 3x + 1. Since equating L(y p ) = 2Ax + (3A + 2B), 2Ax + (3A + 2B) = 3x + 1 = A = 3/2 and B = 7/4. Thus, y p (x) = 3 2 x 7 4.

8 Case II: The method of undetermined coefficients will also work for equations of the form L(y) = ae αx, where a and α are given constants. Try y p of the form y p (x) = Ae αx and solve L(y p )(x) = ae αx for the unknown coefficients A. Example: Find y p to L(y)(x) := y + 3y + 2y = e 3x. Seek y p (x) = Ae 3x. Then L(y p ) = 9Ae 3x + 3(3Ae 3x ) + 2(Ae 3x ) = 20Ae 3x. Now, L(y p ) = e 3x = 20Ae 3x = e 3x = A = 1/20. Thus, y p (x) = (1/20)e 3x.

9 Case III: For an equation of the form try y p of the form L(y) = a cos βx + b sin βx, y p (x) = A cos βx + B sin βx and solve L(y p ) = a cos βx + b sin βx for the unknowns A and B. Example: Find y p to L(y) := y y y = sin x. Seek y p (x) of the form y p (x) = A cos x + B sin x. Then L(y p ) = sin x = A = 1/5, B = 2/5. Thus, y p (x) = 1 cos x 2 sin x. 5 5

10 Example: Find y p to L(y) := y y 12y = e 4x. Note that y h (x) = c 1 e 4x + c 2 e 3x. Try finding y p with the guess y p (x) = Ae 4x as before. Since e 4x is a solution to the corresponding homogeneous equation L(y) = 0, we replace this choice of y p by y p (x) = Axe 4x. Since L(xe 4x ) 0, there exists a particular solution of the form y p (x) = Axe 4x. Remark: If L(y p ) = 0 then replace y p (x) by xy p (x). If L(xy p ) = 0 the replace xy p by x 2 y p and so on. Thus, employing x s y p, where s is the smallest nonnegative integer such that L(x s y p ) 0.

11 Form of y p : g(x) = p n (x) = a n x n + + a 1 x + a 0, y p (x) = x s P n (x) = x s {A n x n A 1 x + A 0 } g(x) = ae αx, y p (x) = x s Ae αx g(x) = a cos βx + b sin βx, y p (x) = x s {A cos βx + B sin βx} g(x) = p n (x)e αx, y p (x) = x s P n (x)e αx g(x) = p n (x) cos βx + q m (x) sin βx, where q m (x) = b m x m + + b 1 x + b 0 and p n (x) as above. y p (x) = x s {P N (x) cos βx + Q N (x) sin βx}, where Q N (x) = B N x N + + B 1 x + B 0, P N (x) = A N x N + + A 1 x + A 0 and N = max(n, m).

12 g(x) = ae αx cos βx + be αx sin βx, y p (x) = x s {Ae αx cos βx + Be αx sin βx} g(x) = p n (x)e αx cos +q m (x)e αx sin βx, y p (x) = x s e αx {P N (x) cos βx + Q N (x) sin βx}, where N = max(n, m). Note: 1. The nonnegative integer s is chosen to be the smallest integer so that no term in y p is a solution to L(y) = P n (x) or P N (x) must include all its terms even if p n (x) has some terms that are zero. Similarly for Q N (x). *** End ***