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# 0027 - It fol-lows that p =(x2 1'3”exp(3x2/2 and the-nee...

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Unformatted text preview: It fol-lows that p =(x2+1)'3”exp(3x2/2) and the-nee that Dx-(J’ ' (x2 + IT“ exp(3x2f2)) : 51-“? +4)—5r2’ y - (x2 +1)--‘r2 exp(3x2 f2) = —2(x2 +4)M + C, y(x) = —2exP(3x2/2)+C(x2+1)-‘”cxp(—3x1/2). Finally, y(0) = 1 implies that C = 3 so the desired particular solution is -y(x) = —2exp(3x2'x2)+3 (1:2 +1)“ EXPO-312 I2). 26. With x’=dx/dy, the differential equation is y"x’+4y2x =1. Then with y as the independent variable we calculate .00») =‘ BXPUWNdyJ = 8““ = y“; D_v(x'y‘) I y 4 1 2 1 C x- = — +C; = —+~— y 2 y x02) 2y2 yl. 27. With A": dx/dy, the differential equation is x'—x= ye". Then with y as the independent variable We calculate 9(3)) = BXPU(-1)dy) a e'J'; Dy(x-e”') = y x-e” :13y2+C; 25(3)) : Gyg-l—C)?‘ 28. With x’=dxldy, the differential equation is (1+y2)x’—2yx= 1. Then with y as the independent variable we calculate 90’) = 6XpU(—2yf(l+y2)d3i) = 9““‘3'2’4’ = (1+y2y1 DJ, (x-(l+y2)_1) : (”ﬂ—z An integral table (or trigonometric substitution) now yields x2 : ——dy—2=—1— y2+tan_1y+C x(y) = %[y+(l+y2)(tan‘1y+C)] Section 1.5 27 ...
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