ch11 - 11.1 SOLUTIONS 761 CHAPTER ELEVEN Solutions for...

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11.1 SOLUTIONS 761 CHAPTER ELEVEN Solutions for Section 11.1 Exercises 1. (a) (III) An island can only sustain the population up to a certain size. The population will grow until it reaches this limiting value. (b) (V) The ingot will get hot and then cool off, so the temperature will increase and then decrease. (c) (I) The speed of the car is constant, and then decreases linearly when the breaks are applied uniformly. (d) (II) Carbon-14 decays exponentially. (e) (IV) Tree pollen is seasonal, and therefore cyclical. 2. Since y = x 3 , we know that y 0 = 3 x 2 . Substituting y = x 3 and y 0 = 3 x 2 into the differential equation we get 0 = xy 0 - 3 y = x (3 x 2 ) - 3( x 3 ) = 3 x 3 - 3 x 3 = 0 . Since this equation is true for all x , we see that y = x 3 is in fact a solution. 3. In order to prove that y = A + Ce kt is a solution to the differential equation dy dt = k ( y - A ) , we must show that the derivative of y with respect to t is in fact equal to k ( y - A ) : y = A + Ce kt dy dt = 0 + ( Ce kt )( k ) = kCe kt = k ( Ce kt + A - A ) = k ( ( Ce kt + A ) - A ) = k ( y - A ) . 4. If P = P 0 e t , then dP dt = d dt ( P 0 e t ) = P 0 e t = P. 5. We know that at time t = 0 the value of y is 8. Since we are told that dy/dt = 0 . 5 y , we know that at time t = 0 the derivative of y is . 5(8) = 4 . Thus as t goes from 0 to 1, y will increase by 4, so at t = 1 , y = 8 + 4 = 12 . Likewise, at t = 1 , we get dy/dt = 0 . 5(12) = 6 so that at t = 2 , we obtain y = 12 + 6 = 18 . At t = 2 , we have dy/dt = 0 . 5(18) = 9 so that at t = 3 , we obtain y = 18 + 9 = 27 . At t = 3 , we have dy/dt = 0 . 5(27) = 13 . 5 so that at t = 4 , we obtain y = 27 + 13 . 5 = 40 . 5 . Thus we get the values in the following table t 0 1 2 3 4 y 8 12 18 27 40 . 5 6. Since y = x 2 + k , we know that y 0 = 2 x . Substituting y = x 2 + k and y 0 = 2 x into the differential equation, we get 10 = 2 y - xy 0 = 2( x 2 + k ) - x (2 x ) = 2 x 2 + 2 k - 2 x 2 = 2 k. Thus, k = 5 is the only solution.
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762 Chapter Eleven /SOLUTIONS 7. If Q = Ce kt , then dQ dt = Cke kt = k ( Ce kt ) = kQ. We are given that dQ dt = - 0 . 03 Q, so we know that kQ = - 0 . 03 Q. Thus we either have Q = 0 (in which case C = 0 and k is anything) or k = - 0 . 03 . Notice that if k = - 0 . 03 , then C can be any number. 8. If y satisfies the differential equation, then we must have d (5 + 3 e kx ) dx = 10 - 2(5 + 3 e kx ) 3 ke kx = 10 - 10 - 6 e kx 3 ke kx = - 6 e kx k = - 2 . So, if k = - 2 the formula for y solves the differential equation. 9. If y = sin 2 t, then dy dt = 2 cos 2 t, and d 2 y dt 2 = - 4 sin 2 t. Thus d 2 y dt 2 + 4 y = - 4 sin 2 t + 4 sin 2 t = 0 . 10. If y = cos ωt, then dy dt = - ω sin ωt, d 2 y dt 2 = - ω 2 cos ωt. Thus, if d 2 y dt 2 + 9 y = 0 , then - ω 2 cos ωt + 9 cos ωt = 0 (9 - ω 2 ) cos ωt = 0 . Thus 9 - ω 2 = 0 , or ω 2 = 9 , so ω = ± 3 . 11. Differentiating and using the fact that d dt (cosh t ) = sinh t and d dt (sinh t ) = cosh t, we see that dx dt = ωC 1 sinh ωt + ωC 2 cosh ωt d 2 x dt 2 = ω 2 C 1 cosh ωt + ω 2 C 2 sinh ωt = ω 2 ( C 1 cosh ωt + C 2 sinh ωt ) . Therefore, we see that d 2 x dt 2 = ω 2 x. 12. Differentiating x 2 + y 2 = r 2 implicitly, with r a constant, gives 2 x + 2 y dy dx = 0 . Solving for dy/dx , we get dy dx = - 2 x 2 y = - x y . Problems 13. (a) If y = Cx n is a solution to the given differential equation, then we must have x d ( Cx n ) dx - 3( Cx n ) = 0 x ( Cnx n - 1 ) - 3( Cx n ) = 0 Cnx n - 3 Cx n = 0 C ( n - 3) x n = 0 .
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11.2 SOLUTIONS 763 Thus, if C = 0 , we get y = 0 is a solution, for every n . If C 6 = 0 , then n = 3 , and so y = Cx 3 is a solution.
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