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Ebook Ordinary differential equations - Student solution manual

Nonhomogeneous ordinary differential equations can be solved if the general solution to the homogenous version is known, in which case the undetermined coefficients method or variation of parameters can be used to find the particular solution. | William A. Adkins, Mark G. Davidson ORDINARY DIFFERENTIAL EQUATIONS Student Solution Manual January 23, 2012 Springer Chapter 1 Solutions Section 1.1 1. The rate of change in the population P (t) is the derivative P ′ (t). The Malthusian Growth Law states that the rate of change in the population is proportional to P (t). Thus P ′ (t) = kP (t), where k is the proportionality constant. Without reference to the t variable, the differential equation becomes P ′ = kP 3. Torricelli’s law states that the change in height, h′ (t) is proportional to the square root of the height, h(t). Thus h′ (t) = λ h(t), where λ is the proportionality constant. 5. The highest order derivative is y ′′ so the order is 2. The standard form is y ′′ = t3 /y ′ . 7. The highest order derivative is y ′′ so the order is 2. The standard form is y ′′ = −(3y + ty ′ )/t2 . 9. The highest order derivative is y (4) so the order is 4. Solving for y (4) gives the standard form: y (4) = 3 (1 − (y ′′′ )4 )/t. 11. The highest order derivative is y ′′′ so the order is 3. Solving for y ′′′ gives the standard form: y ′′′ = 2y ′′ − 3y ′ + y. 13. The following table summarizes the needed calculations: 3 4 1 Solutions ty ′ (t) Function y1 (t) = 0 y2 (t) = 3t y3 (t) = −5t y4 (t) = t 3 ′ ty1 (t) ′ ty2 (t) ′ ty3 (t) ′ ty4 (t) y(t) =0 y1 (t) = 0 = 3t y2 (t) = 3t = −5t = 3t y3 (t) = −5t 3 y4 (t) = t3 To be a solution, the entries in the second and third columns need to be the same. Thus y1 , y2 , and y3 are solutions. 15. The following table summarizes the needed calculations: y ′ (t) Function ′ y1 (t) ′ y2 (t) ′ y3 (t) y1 (t) = 0 y2 (t) = 1 y3 (t) = 2 y4 (t) = 1 1−e2t =0 =0 =0 ′ y4 (t) = 2e2t (1−e2t )2 2y(t)(y(t) − 1) 2y1 (t)(y1 (t) − 1) = 2 · 0 · (−1) = 0 2y2 (t)(y2 (t) − 1) = 2 · 1 · 0 = 0 2y3 (t)(y3 (t) − 1) = 2 · 2 · 1 = 4 1 2y4 (t)(y4 (t) − 1) = 2 1−e2t 1 = 2 1−e2t 2t e 1−e2t = 1 1−e2t 2t 2e (1−e2t )2 −1 Thus y1 , y2 , and y4 are solutions. 17. The following table .