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**Example text**

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. I. I. = - - e ax f (D) = eax x2 - - if f" (a) "# 0 f" (a)' Example 1. Solve (D2 - 2D + 5) Y = e- X 42 Linear Differential Equations with Constant Coefficients and Applications Solution.

S. 1995) Solution. The given equation can be written as 1 dy 1 1 1 - - + - - = - log X y2 dx x Y x . 1 putting - - = v y or (1) ~ dy = dv in (1), we get y2 dx dx dv 1 1 - - - v= -logx dx x x (2) which is in the standand form of the linear equation and integrating factor dx =e- 1og x 1 e x =- -p x Multiplying both sides of (2) by the integrating factor and integrating, we get v. or ! Y 1 = (1 + log x) Y - C xy Example 15. S. 1996) Solution. Dividing both sides of the given equation by y, we get x dy + (log y) = x eX y dx - - or 1 dy 1 X - - + - (log y) = e y dx x putting v = log Y or dv 1 dy .

1 equa1 to v putting --n:T y Dividing (1) by yn, we get 1 Put - yn-l = v or y-n + 1 = ~ dy + p. ~l yn dx yn = V Differentiating both sides with respect to x, get (-n+1) y or -n dy dx = dv dx ~ dy = _1_ dv yn dx 1 - n dx Making these substitutions in (2), we have 1 dv - +Pv=Q (1 - n) dx or dv - + P (1 - n) v = Q (1 - n) dx 19 Q (2) A Textbook of Engineering Mathematics Volume - II which is linear in v and can be solve by method of linear differential equation. Example 11. 1994) Solution. Dividing both sides of the given equation by cos2y we get sec2y dy + x (2 tan y) = x 3 dx .