2010 asm - Solutions for Assignment #1 1.2 (7) Note that...

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Unformatted text preview: Solutions for Assignment #1 1.2 (7) Note that the general solution of the differential equation is p ( t ) = 900 + Ce t 2 . (a) We need to find t such that p ( t ) = 0. First we solve for C . Since p (0) = 850 we get 850 = 900 + C , so C =- 50. We therefore need to find t such that 900- 50 e t 2 = 0. This means solving e t 2 = 18 . We take the logarithm of both sides and eventually get t 5 . 8 months. (b) If p (0) = p we solve for C to get C = p- 900. We then need to find t such that 900+( p- 900) e t 2 = 0, which means solving e t 2 = 900 900- p . We take the logarithm of both sides and eventually get t = 2ln 900 900- p months. (b) Remember that 1 years is 12 months so we need to find p that such p (12) = 0, i.e., 900 + ( p- 900) e 12 2 = 0 . This gives p = 900( e 6- 1). 1.3 (6) The equation is a linear equation of order 3. 1.3 (11) We simply need to substitute y 1 and y 2 into the differential equation and verify the right hand side equals the left hand side. We begin with y 1 ( t ) = t 1 / 2 . We first calculate the first two derivatives: y 1 = 1 2 t- 1 / 2 , y 00 1 =- 1 4 t- 3 / 2 . We then substitute these into the differential equation: l.h.s. = 2 t 2 y 00 1 + 3 ty 1- y 1 =- 1 2 t 1 / 2 + 3 2 t 1 / 2- t 1 / 2 = 0 = r.h.s., so y 1 is a solution. Similarly for y 2 ( t ) = t- 1 , we first calculate its first two derivatives: y 2 =- t- 2 , y 00 2 = 2 t- 3 , And then substitute into the differential equation: l.h.s. = 2 t 2 y 00 2 + 3 ty 2- y 2 = 4 t- 1- 3 t- 1- t- 1 = 0 = r.h.s., so y 2 is also a solution. 1 2.1 (17) We first find the general solution of the differential equation y- 2 y = e 2 t . This is already of the form y + p ( t ) y = g ( t ) with p ( t ) =- 2. This means we take our integrating factor ( t ) to be ( t ) = e R (- 2) dt = e- 2 t . If we multiply the equation by we know we can solve it as follows. d dt ( e- 2 t y ) = e- 2 t e 2 t d dt ( e- 2 t y ) = 1 Z d dt ( e- 2 t y ) dt = Z dt e- 2 t y = t + C y = Ce 2 t + te 2 t . We have shown that y = Ce 2 t + te 2 t , where C is an arbitrary constant, is the general solution of the differential equation y- 2 y = e 2 t . We now use the initial condition y (0) = 2 to determine the constant C . At t = 0 we have C + 0 = 2 C = 2 . The solution of the initial value problem is therefore y ( t ) = 2 e 2 t + te 2 t . 2.1 (31) We first find the general solution of the differential equation y- 3 2 y = 3 t + 2 e t . We take the integrating factor ( t ) to be ( t ) = e R (- 3 2 ) dt = e- 3 2 t . We then solve the equation: d dt ( e- 3 2 t ) = e- 3 2 t (3 t + 2 e t ) d dt ( e- 3 2 t y ) = 3 te- 3 2 t + 2 e- 1 2 t Z d dt ( e- 3 2 t y ) dt = Z 3 te- 3 2 t + 2 e- 1 2 t dt e- 3 2 t y =- 2 te- 3 2 t- 4 3 e- 3 2 t- 4 e- 1 2 t + C y = Ce 3 2 t- 2 t- 4 3- 4 e t ....
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This note was uploaded on 05/24/2011 for the course MAT 244 taught by Professor Christinasaleh during the Spring '10 term at University of Toronto- Toronto.

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2010 asm - Solutions for Assignment #1 1.2 (7) Note that...

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