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Unformatted text preview: Chapter 13 More Differential Equations 13.1 Consider the differential equation dy dt = a by where a , b are constants. (a) Show that the function y ( t ) = a b Ce bt satisfies the above differential equation for any constant C . (b) Show that by setting C = a b y we also satisfy the initial condition y (0) = y . Remark: You have now shown that the function y ( t ) = parenleftBig y a b parenrightBig e bt + a b is a solution to the initial value problem (i.e differential equation plus initial condition) dy dt = a by, y (0) = y . 13.2 For each of the following, show the given function y is a solution to the given differential equation. (a) t · dy dt = 3 y , y = 2 t 3 . (b) d 2 y dt 2 + y = 0, y = 2 sin t + 3 cos t . (c) d 2 y dt 2 2 dy dt + y = 6 e t , y = 3 t 2 e t . v.2005.1  September 4, 2009 1 Math 102 Problems Chapter 13 13.3 Show the function determined by the equation 2 x 2 + xy y 2 = C , where C is a constant and 2 y negationslash = x , is a solution to the differential equation ( x 2 y ) dy dx = 4 x y . 13.4 Find the constant C that satisfies the given initial conditions. (a) 2 x 2 3 y 2 = C , y  x =0 = 2. (b) y = C 1 e 5 t + C 2 te 5 t , y  t =0 = 1 and dy dt  t =0 = 0. (c) y = C 1 cos( t C 2 ), y  t = π 2 = 0 and dy dt  t = π 2 = 1. 13.5 Friction and terminal velocity The velocity of a falling object changes due to the acceleration of gravity, but friction has an effect of slowing down this acceleration. The differential equation satisfied by the velocity v ( t ) of the falling object is dv dt = g kv where g is acceleration due to gravity and k is a constant that represents the effect of friction. An object is dropped from rest from a plane. (a) Find the function v ( t ) that represents its velocity over time. (b) What happens to the velocity after the object has been falling for a long time (but before it has hit the ground)? 13.6 Alcohol level Alcohol enters the blood stream at a constant rate k gm per unit time during a drinking session. The liver gradually converts the alcohol to other, nontoxic byproducts. The rate of conversion per unit time is proportional to the current blood alcohol level, so that the differential equation satisfied by the blood alcohol level is dc dt = k sc where k , s are positive constants. Suppose initially there is no alcohol in the blood. Find the blood alcohol level c ( t ) as a function of time from t = 0, when the drinking started....
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This note was uploaded on 05/19/2011 for the course MATH 102 Math 102 taught by Professor Allard during the Winter '09 term at UBC.
 Winter '09
 Allard

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