57321-0136147054_08

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

This preview shows pages 1–5. Sign up to view the full content.

Section 8.1 8.1.1: 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 . 8.1.2: 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 10 exp(15 3 x ) . 8.1.3: Given: dy dx y 2 , y (7) = 3: 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 . 8.1.4: 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. 8.1.5: Given: dy dx y 1 / 2 , y (0) = 9: y 1 / 2 dy dx ;2 y 1 / 2 = C x ; 820

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 dx ;3 y 1 / 3 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 ;l n ( 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. 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) = 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 . 8.1.9: Given: dy dx = e y , y (0) = 2: e y dy 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 =2s e c y , y (0) = 0: cos ydy =2 dx ;s i n y x + C. y (0) = 0 : 0 = 0 + C ; C =0 ; s in y x. Therefore y ( x )=s 1 (2 x ). 8.1.11: If the slope of y = g ( x )atthepo int( x, y )isthesumo f x and y , then we expect y = g ( x )tobea solution of the diFerential equation dy dx = x + y. (1) Some solutions of this diFerential equation with initial conditions y (0) = 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). -1 1 x -1 1 y 8.1.12: If the line tangent to the graph of y = g ( x )atthepoint( 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, 0 . 5, 1 . 5, and 3 . 5) are shown next. The ±gure was constructed with the same scale on the x y -axes, so you can con±rm with a ruler that the solution curves agree with the diFerential equation in (1). 822

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
-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 diFerential equation dy dx = x 1 y . (1) The general solution of this equation is implicitly de±ned by x 2 +( y 1) 2 = C where C> 0. Some solution curves for C =0 . 16, C . 49, and C = 1 are shown next. Note that two functions are solutions for each value of C . Because the ±gure was constructed with the same scale on the x -and y -axes, you can con±rm with a ruler that the solution curves agree with the diFerential equation in (1).
This is the end of the preview. Sign up to access the rest of the document.

This document was uploaded on 11/28/2010.

Page1 / 112

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

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online