AEM_3e_Chapter02

# 40 to see that the equations are not equivalent

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40. To see that the equations are not equivalent consider dx = ( x/y ) dy . An integrating factor is µ ( x, y ) = y resulting in y dx + x dy = 0. A solution of the latter equation is y = 0, but this is not a solution of the original equation. 41. The explicit solution is y = (3 + cos 2 x ) / (1 x 2 ). Since 3 + cos 2 x > 0 for all x we must have 1 x 2 > 0 or 1 < x < 1. Thus, the interval of definition is ( 1 , 1). 42. (a) Since f y = N ( x, y ) = xe xy +2 xy +1 /x we obtain f = e xy + xy 2 + y x + h ( x ) so that f x = ye xy + y 2 y x 2 + h ( x ). Let M ( x, y ) = ye xy + y 2 y x 2 . (b) Since f x = M ( x, y ) = y 1 / 2 x 1 / 2 + x ( x 2 + y ) 1 we obtain f = 2 y 1 / 2 x 1 / 2 + 1 2 ln x 2 + y + g ( y ) so that f y = y 1 / 2 x 1 / 2 + 1 2 ( x 2 + y ) 1 + g ( x ). Let N ( x, y ) = y 1 / 2 x 1 / 2 + 1 2 ( x 2 + y ) 1 . 43. First note that d x 2 + y 2 = x x 2 + y 2 dx + y x 2 + y 2 dy. Then x dx + y dy = x 2 + y 2 dx becomes x x 2 + y 2 dx + y x 2 + y 2 dy = d x 2 + y 2 = dx. 51

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2.4 Exact Equations The left side is the total differential of x 2 + y 2 and the right side is the total differential of x + c . Thus x 2 + y 2 = x + c is a solution of the differential equation. 44. To see that the statement is true, write the separable equation as g ( x ) dx + dy/h ( y ) = 0. Identifying M = g ( x ) and N = 1 /h ( y ), we see that M y = 0 = N x , so the differential equation is exact. 45. (a) In differential form we have ( v 2 32 x ) dx + xv dv = 0. This is not an exact form, but µ ( x ) = x is an integrating factor. Multiplying by x we get ( xv 2 32 x 2 ) dx + x 2 v dv = 0. This form is the total differential of u = 1 2 x 2 v 2 32 3 x 3 , so an implicit solution is 1 2 x 2 v 2 32 3 x 3 = c . Letting x = 3 and v = 0 we find c = 288. Solving for v we get v = 8 x 3 9 x 2 . (b) The chain leaves the platform when x = 8, so the velocity at this time is v (8) = 8 8 3 9 64 12 . 7 ft/s . 46. (a) Letting M ( x, y ) = 2 xy ( x 2 + y 2 ) 2 and N ( x, y ) = 1 + y 2 x 2 ( x 2 + y 2 ) 2 we compute M y = 2 x 3 8 xy 2 ( x 2 + y 2 ) 3 = N x , so the differential equation is exact. Then we have ∂f ∂x = M ( x, y ) = 2 xy ( x 2 + y 2 ) 2 = 2 xy ( x 2 + y 2 ) 2 f ( x, y ) = y ( x 2 + y 2 ) 1 + g ( y ) = y x 2 + y 2 + g ( y ) ∂f ∂y = y 2 x 2 ( x 2 + y 2 ) 2 + g ( y ) = N ( x, y ) = 1 + y 2 x 2 ( x 2 + y 2 ) 2 . Thus, g ( y ) = 1 and g ( y ) = y . The solution is y y x 2 + y 2 = c . When c = 0 the solution is x 2 + y 2 = 1. (b) The first graph below is obtained in Mathematica using f ( x, y ) = y y/ ( x 2 + y 2 ) and ContourPlot[f[x, y], { x, -3, 3 } , { y, -3, 3 } , Axes > True, AxesOrigin > { 0, 0 } , AxesLabel > { x, y } , Frame > False, PlotPoints > 100, ContourShading > False, Contours > { 0, -0.2, 0.2, -0.4, 0.4, -0.6, 0.6, -0.8, 0.8 } ] The second graph uses x = y 3 cy 2 y c y and x = y 3 cy 2 y c y . In this case the x -axis is vertical and the y -axis is horizontal. To obtain the third graph, we solve y y/ ( x 2 + y 2 ) = c for y in a CAS. This appears to give one real and two complex solutions. When graphed in Mathematica however, all three solutions contribute to the graph. This is because the solutions involve the square root of expressions containing c . For some values of c the expression is negative, causing an apparent complex solution to actually be real.

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