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Extra resources for A Text Book of Engineering Mathematics. Volume II

Example text

When the equation is n n I D + at D - +.......... ) and any particular integral of the equation f(D) y = Q :. P. I. A particular integral of the differential equation f(D) y = Q is given by _1_ Q f(D) Methods of finding Particular integral (A) Case I. , when Q is of the form of eax, where a is any constant and f(a) :1; 0 we know that 41 A Textbook of Engineering Mathematics Volume - II D2 (eax) = a2 eax (e ax) = a 3 e ax Dn (e ax) = an e ax :. I. : f(a) is constant e ax f(D) e ax = _1_ f(a) eax if f(a) "# 0 , Case II.

2 e,ax -~ (~) (i cos ax - sin ax) . sIn ax 1 x . ---::---:-- cos ax = - sm ax .. D2 + a 2 2a Example 4. Solve (D2 + D + 1) Y = sin 2x Solution. Here the auxiliary equation is m 2 + m + 1 = 0 which gives m :. P. 1. = 2 21 sin 2x replacing D2 by - 22 (-2) +D+1 = -1- D-3 . 2x sm 1 (D - 3) (D + 3) 21 D -9 (D + 3) sin 2x (D + 3) sin 2x = ~ 21 -9 (D + 3) sin 2x = -~ (D + 3) sin 2x = -1 [D (sin 2x) + 3 sin 2x] 13 13 = -~ [2 cos 2x + 3 sin 2x] Since D means differentiation with respect to x 13 :. The complete solution is 46 Linear Differential Equations with Constant Coetficients and Applications x2 y = e- / [C 1 cos (% xJ3) + C 2 sin (% xJ3)] -113 (2 cos 2x + 3sin 2x) Example 5.

Replacing X and Y in the solution so obtained by x-h and y-k [from (3)] respectively We can get the required solution is terms of x and y, the original variables. A Special Case When a:b =a':b' In this case we cannot solve the equations given by (4) above and the differential equation is of the form dy dx = ax + by + C kax + kby + C (5) In this case the differential equation is solve by putting v = ax + by (6) Differentiating both sides of (6) with respect to x, we get dv = a + b dy or dy = ! (dV _ a) dx dx dx b dx :.