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# 2132Final_Practice_Answers - Polytechnic Institute of NYU...

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Polytechnic Institute of NYU MA 2132 Final Practice Answers Fall 2008B Studying from past or sample exams is NOT recommended. If you do, it should be only AFTER you know how to do all of the homework and worksheet problems. These are additional practice problems designed to cover the material, but not necessarily specific to the exam. The final is cumulative, covering all material in the course. In general, 75-80% of the exam is new material. (1) Consider the initial value problem P 0 = te t 2 - P , y (0) = 1 . (a) Use the Euler Method with stepsize h = 0 . 25 to find the approximate value for the solution of the above problem at x = 0 . 5. We will need (0 . 5 - 0)( . 25) = 2 iterations to get the answer. n t n = t 0 + n * h y n f ( t n , P n ) = te t 2 - P P n +1 = P n + h * f ( t n , P n ) 0 0 0 0 * e 0 2 - 0 = 0 P 1 = 0 + (0 . 25)(0) = 0 1 0 . 25 0 0 . 25 * e (0 . 25) 2 - 0 = 0 . 2661 P 2 = 0 + 0 . 25(0 . 2661) = 0 . 0665 2 0 . 5 0 . 0665 n/a n/a (b) Find the exact solution and determine its value at x = 0 . 5. Separable: e P dP = te t 2 dt so P = ln(0 . 5 * e t 2 + C ) Using the IVP (0 , 0), 0 = ln(0 . 5 * 1 + C ) 1 = 0 . 5 + C C = 0 . 5 . So, P = ln(0 . 5 * e t 2 + 0 . 5) and P (0 . 5) = 0 . 1328. The approximation is poor due to the relatively large stepsize. (2) Consider the initial value problem x 2 y 0 = 2 xy - x 3 , y (1) = 3 . (a) Use the Euler Method with stepsize h = 0 . 1 to find the approximate value for the solution of the above problem at x = 1 . 3. n x n y n f ( x n , y n ) = 2 * ( y/x ) - x y n +1 = y n + h * f ( x n , y n ) 0 1 3 2 * 3 / 1 - 1 = 5 y 1 = 3 + (0 . 1)(5) = 3 . 5 1 1 + . 1 = 1 . 1 3 . 5 5 . 2636 y 2 = 3 . 5 + 0 . 1(5 . 2636) = 4 . 02636 2 1 . 2 4 . 02636 5 . 5106 y 3 = 4 . 02636 + 0 . 1(5 . 5106) = 4 . 577 3 1 . 3 4 . 577 n/a n/a 1

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2 (b) Find the exact solution and determine its value at x = 1 . 3. First order linear: You must reformulate the problem as y 0 = 2 y/x - x as in the f ( x n , y n ) column above. The integrating factor is e - 2 /x dx = e - 2 ln | x | = x - 2 . Then x - 2 y 0 - 2 x - 3 y = - x - 1 . ( x - 2 y ) = - ln | x | + C y = - x 2 ln | x | + Cx 2 . Using the IVP (1 , 3), C = 3, so y = - x 2 ln | x | + 3 x 2 . Then y (1 . 3) = 5 . 5134 . (3) Consider the initial value problem xy 0 = y (ln( y/x ) + 1) , y (1) = e. (a) Use the Euler Method with stepsize h = 0 . 1 to find the approximate value for the solution of the above problem at x = 1 . 2. n x n y n f ( x n , y n ) = ( y/x ) * (ln( y/x ) + 1) y n +1 = y n + h * f ( x n , y n ) 0 1 e e * (ln( e ) + 1) = 2 e y 1 = e + 0 . 1 * (2 e ) = 1 . 2 * e = 3 . 2619 1 1 . 1 3 . 2619 6 . 1888 y 2 = 3 . 8808 2 1 . 2 3 . 8808 n/a n/a (b) Find the exact solution and determine its value at x = 1 . 2. Homogeneous: Use y = vx substitution to get v 0 x + v = v (ln( v ) + 1). Note that with the given IVP, x 1 in this problem. Then v 0 = v ln( v ) * x - 1 dv v ln( v ) = dx x ln(ln( v )) = ln( x ) + C ln( v ) = xe C = kx for k = e C . So, v = e kx y = xe kx .
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