By Rajesh Pandey
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Additional info for A text book of engineering mathematics Volume 2
31 A Textbook of Engineering Mathematics Volume - II Example 26. S. 2000) Solution. The given equation can be rewritten as y (y + 2X2) dx + X (2X2 - y) dy = 0 which is of the form as given in method VI . Let xhyk be an integrating factor Multiplying the given equation by xhyk, we get (xh yk+2 + 2Xh + 2 yk + 1) dx + (2Xh +3 yk _Xh+1 yk+1) dy = 0 Here M = Xh yk+2 + 2xh +2 yk+ 1 and N = 2xh+3 yk _xh+ 1 yk+1 :. aM = ay aN and ax (k + 2) xhyk +1 + 2 (k + 1) xh +2yk (1) + 3) Xh +2 yk _ (h + 1) Xh yk +1 (2) = 2 (h .
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. 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.
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 .
A text book of engineering mathematics Volume 2 by Rajesh Pandey