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Math 5

Math 5 - i “Wit?— 30 The homogeneous equation y’ =...

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Unformatted text preview: i “Wit?— 30. The homogeneous equation, y’ = —3y has solution yh(t) = e‘3’. We look for a particular solution in t, the form yp(t) = v(t)yh (t), where v is an unknown» function. Since y; = v’yh + vyi = ”th — 3W}: = v'yh — By,” and y; = -3y,, + 4, we have 11’ = 4/yh : 4e3r, Integrating we see that v(t) = 4e3t /3, and 4 4 M!) = v(t)yh(t) = 363t .e—ar = _3_' The general solution is 4 3'0) : Ypa) “l“ £301“): 3‘ + Ce—Bt. 6. dF ~_— (1/x + 2xy3)dx + (1/Y + 3x2y2)dy 10. With P = 1—— ysinx and Q = cosx, we see that so the equation is exact. We solve by setting FOR”: [P(x,y)dx=/(1—ysinx)dx =r+yeosx+¢(y). To ﬁnd 45, we differentiate 3 F Q06, y) = —— = COSX +¢’(y)- 3y Thus 45’ = 0, so we can take 4) = 0. Hence the solution is F(x, y) = x + ycosx = C. 36. x2 2 F(x,y) = —(1/2)1n<" “2) + arctan(y/x)—1nx = C 40. The homogeneous equation x’ = (2/ t2)x has solu- tion xh (t) = 6—2” . We look for a particular solution of the form xp(t) = v(t)xh(t), where v is an un— known function. Since x; = v’x;l + inc;t = v'xh + 212xh/t2 = v'xh +2xp/t2, and x; : 2xP/t2 +1/t2, we have 1/ : l/(tth) = e2/’/t2. Integrating we ﬁnd that v(t) : —62/‘/2, and xp (t) = —1/2. The general solution is x(t) = xp(t) + th(t) = «1/2 + Ce“2/‘. Since 36(1) 2 0, we ﬁnd that C = (22/2, and the solution is 1 _ x<r> = §(-1+em W). 12. With P = x and Q _ y X2 + _ “7* , WC 00m- pute y W a P —2xy a Q 3y_m/7*"a? so the equation is exact. To ﬁnd the solution we integrate F(x7y) = / P(x,y)dx x 4%" =‘m+¢(y). To ﬁnd (I), we differentiate Q(x,y)=3l: y a ”7—— = ¢’()')~ y w/x +y2 Thus (15’ = 0, so we can take (15 = 0. Hence the solution is F(x, y) = t/xz + 2 = C. 40. y(x) :- x ln(C + 21m x) meﬁﬁ‘lu 4 ea waNimywunwinwn “WNW w u mmpwmmmmm— «mm :mm cmwumu it . ., i _ t ., ...
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