57321-0136147054_08

57321-0136147054_08 - Section 8.1 8.1.1 Separate the...

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Section 8.1 8.1.1: Separate the variables: dy dx = 2 y ; 1 y dy = 2 dx ; ln y = C + 2 x ; y ( x ) = Ae 2 x . 3 = y (1) = Ae 2 : A = 3 e 2 . Therefore y ( x ) = 3 exp(2 x 2) = 3 e 2 x 2 . 8.1.2: Given: dy dx = 3 y , y (5) = 10: 1 y dy = 3 dx ; ln y = C 3 x ; y ( x ) = Ae 3 x . 10 = y (5) = Ae 15 ; A = 10 e 15 . y ( x ) = 10 exp(15 3 x ) . 8.1.3: Given: dy dx = 2 y 2 , y (7) = 3: 1 y 2 dy = 2 dx ; 1 y = C 2 x ; y ( x ) = 1 C 2 x . 3 = y (7) = 1 C 14 : C = 43 3 . y ( x ) = 1 43 3 2 x = 3 43 6 x . 8.1.4: Given: dy dx = 7 y , y (0) = 6: 2 y dy = 14 dx ; y 2 = C + 14 x. 36 = [ y (0)] 2 = C : C = 36 . y 2 = 36 + 14 x : y ( x ) = 36 + 14 x . We chose the nonnegative square root in the last step because y (0) > 0. 8.1.5: Given: dy dx = 2 y 1 / 2 , y (0) = 9: y 1 / 2 dy = 2 dx ; 2 y 1 / 2 = C + 2 x ; 820

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y 1 / 2 = A + x ; y ( x ) = ( A + x ) 2 . y (0) = 9 : A = 3 . Therefore y ( x ) = ( x + 3) 2 . 8.1.6: Given: dy dx = 6 y 2 / 3 , y (1) = 8: y 2 / 3 dy = 6 dx ; 3 y 1 / 3 = 6 x + C. y (1) = 8 : C = 0 . y 1 / 3 = 2 x ; y ( x ) = 8 x 3 . 8.1.7: Given: dy dx = 1 + y , y (0) = 5: 1 1 + y dy = 1 dx ; ln(1 + y ) = x + C ; 1 + y = Ae x ; y ( x ) = Ae x 1 . 5 = y (0) = A 1 : A = 6 . Therefore y ( x ) = 6 e x 1. 8.1.8: Given: dy dx = (2 + y ) 2 , y (5) = 3: 1 ( y + 2) 2 dy = 1 dx ; 1 y + 2 = C x ; y ( x ) = 2 + 1 C x . 3 = y (5) = 2 + 1 C 5 ; 1 C 5 = 5; C = 26 5 . Therefore y ( x ) = 2 + 1 26 5 x = 2 + 5 26 5 x = 52 + 10 x + 5 26 5 x = 10 x 47 26 5 x . 8.1.9: Given: dy dx = e y , y (0) = 2: e y dy = 1 dx ; e y = x + C ; y ( x ) = ln( x + C ) . 2 = y (0) = ln C : 821
C = e 2 . y ( x ) = ln( x + e 2 ) . 8.1.10: Given: dy dx = 2 sec y , y (0) = 0: cos y dy = 2 dx ; sin y = 2 x + C. y (0) = 0 : 0 = 0 + C ; C = 0; sin y = 2 x. Therefore y ( x ) = sin 1 (2 x ). 8.1.11: If the slope of y = g ( x ) at the point ( x, y ) is the sum of x and y , then we expect y = g ( x ) to be a solution of the differential equation dy dx = x + y. (1) Some solutions of this differential equation with initial conditions y (0) = 1, 0 . 5, 0, 0 . 5, and 1 are shown next. The figure is constructed with the same scale on the x - and y -axes, so you can confirm with a ruler that the solution curves agree with the differential equation in (1). -1 1 x -1 1 y 8.1.12: If the line tangent to the graph of y = g ( x ) at the point ( x, y ) meets the x -axis at the point ( x/ 2 , 0), then y = g ( x ) should be a solution of the differential equation dy dx = y 0 x 1 2 x = 2 y x . (1) The general solution of the equation in (1) is y ( x ) = Cx 2 . Some particular solutions (with c = 3 . 5, 1 . 5, 0 . 5, 0, 0 . 5, 1 . 5, and 3 . 5) are shown next. The figure was constructed with the same scale on the x - and y -axes, so you can confirm with a ruler that the solution curves agree with the differential equation in (1). 822

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-1 1 x -1 1 y 8.1.13: If every straight line normal to the graph of y = g ( x ) passes through the point (0 , 1), then we expect that y = g ( x ) will be a solution of the differential equation dy dx = x 1 y . (1) The general solution of this equation is implicitly defined by x 2 + ( y 1) 2 = C where C > 0. Some solution curves for C = 0 . 16, C = 0 . 49, and C = 1 are shown next. Note that two functions are solutions for each value of C . Because the figure was constructed with the same scale on the x - and y -axes, you can confirm with a ruler that the solution curves agree with the differential equation in (1).
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