Calculus with Analytic Geometry by edwards & Penney soln ch9

Calculus with Analytic Geometry

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Section 9.1 C09S01.001: Separate the variables: dy dx =2 y ; 1 y dy dx ; ln y = C +2 x ; y ( x )= Ae 2 x . 3= y (1) = Ae 2 : A =3 e 2 . Therefore y ( x ) = 3exp(2 x 2 )=3 e 2 x 2 . C09S01.002: Given: dy dx = 3 y , y (5) = 10: 1 y dy = 3 dx ;l n y = C 3 x ; y ( x Ae 3 x . 10 = y (5) = Ae 15 ; A = 10 e 15 .y ( x 10exp(15 3 x ) . C09S01.003: Given: dy dx y 2 , y (7 : 1 y 2 dy = 2 dx ; 1 y = C 2 x ; y ( x 1 C 2 x . y (7) = 1 C 14 : C = 43 3 ( x 1 43 3 2 x = 3 43 6 x . C09S01.004: Given: dy dx = 7 y , y (0) = 6: 2 ydy =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. C09S01.005: Given: dy dx y 1 / 2 , y (0 )=9 : y 1 / 2 dy dx ;2 y 1 / 2 = C x ; y 1 / 2 = A + x ; y ( x )=( A + x ) 2 . y )=9: A . 1

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Therefore y ( x )=( x +3) 2 . C09S01.006: Given: dy dx =6 y 2 / 3 , y (1) = 8: y 2 / 3 dy dx ;3 y 1 / 3 x + C. y (1 )=8: C =0 . y 1 / 3 =2 x ; y ( x )=8 x 3 . C09S01.007: Given: dy dx =1+ y , y (0 )=5 : 1 1+ y dy =1 dx ; ln(1 + y )= x + C ; y = Ae x ; y ( x Ae x 1 . 5= y (0) = A 1: A . Therefore y ( x )=6 e x 1. C09S01.008: 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) = 1 C 5 ; 1 C 5 =5 ; C = 26 5 . Therefore y ( x 1 26 5 x = 5 26 5 x = 52+10 x +5 26 5 x = 10 x 47 26 5 x . C09S01.009: Given: dy dx = e y , y )=2 : e y dy dx ; e y = x + C ; y ( x )=ln ( x + C ) . 2= y C : C = e 2 .y ( x ( x + e 2 ) . C09S01.010: Given: dy dx = 2sec y , y (0) = 0: cos ydy dx ; sin y x + C. y (0) = 0 : 0 = 0 + C ; 2
-1.5 -1 -0.5 0.5 1 1.5 -1.5 -1 -0.5 0.5 1 1.5 C = 0; sin y =2 x. Therefore y ( x ) = sin 1 (2 x ). C09S01.011: 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 diFerential equation dy dx = x + y. (1) Some solutions of this diFerential equation with initial conditions y (0) = 1 . 5, 1, 0 . 5, 0, 0 . 5, and 1 are shown next. The ±gure is constructed with the same scale on the x - and y -axes, so you can con±rm with a ruler that the solution curves agree with the diFerential equation in (1). C09S01.012: 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 diFerential 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 . 5, 1 . 5, and 3 . 5) are shown next. The ±gure was constructed with the same scale on the x - and y -axes, 3

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-1.5 -1 -0.5 0.5 1 1.5 -1.5 -1 -0.5 0.5 1 1.5 -1 -0.5 0.5 1 0.5 1 1.5 2 so you can confrm with a ruler that the solution curves agree with the diFerential equation in (1). C09S01.013: I± every straight line normal to the graph o± y = g ( x ) passes through the point (0 , 1), then we expect that y = g ( x ) will be a solution o± the diFerential equation dy dx = x 1 y . (1) The general solution o± this equation is implicitly defned by x 2 +( y 1) 2 = C where C> 0. Some solution curves ±or C =0 . 16, C . 5, and C = 1 are shown next. Note that two ±unctions are solutions ±or each value o± C . Because the fgure was constructed with the same scale on the x - and y -axes, you can confrm with a ruler that the solution curves agree with the diFerential equation in (1).
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Calculus with Analytic Geometry by edwards &amp; Penney...

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