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Unformatted text preview: MAT 127: Calculus C, Fall 2010 Solutions to Midterm I Problem 1 (10pts) Consider the four differential equations for y = y ( x ) : (a) y ′ = x (1 + y 2 ) (b) y ′ = y (1 + x 2 ) (c) y ′ = e x + y (d) y ′ = e − x − y . Each of the four diagrams below shows a solution curve for one of these equations: I y x II y x III y x IV y x Match each of the diagrams to the corresponding differential equation (the match is onetoone): diagram I II III IV equation b c a d Answer Only: no explanation is required. Explanation is on the next page . Of the four diagrams, the most distinctive is I ; it shows the graph of the constant function y ( x )=0. Plugging in this function into the four equations, we get (a) 0 ? = x (1 + 0 2 ) (b) 0 ? = 0(1 + x 2 ) (c) 0 ? = e x +0 (d) 0 ? = e − x − . Of these four potential equalities, only (b) is satisfied for all x ; so diagram I must correspond to (b). Of the three remaining graphs, the most distinctive is III ; the slope of the graph at (0 , 0) there is 0. This is the case for the slope of equation (a) at (0,0) (because 0(1+0 2 )=0), but the slopes for the other two remaining equations are always positive (because e x is always positive). So diagram III must correspond to (a). The two remaining graphs look rather similar. One distinguishing feature is that the graph in II becomes steeper as x,y increase, while the graph in IV less steep as x,y increase. Considering the two remaining equations, e x + y becomes larger as x,y increase (thus making the slope of the graph steeper), while e − x − y becomes smaller as x,y increase (thus making the slope of the graph less steep). So, diagram II must correspond to (c), while diagram IV must correspond to (d). Alternatively, diagram II depicts the graph of a function with y ′′ > 0, while diagram IV depicts the graph of a function with y ′′ < 0. If y = y ( x ) is any function satisfying equation (c), then by the Chain Rule y ′′ = ( y ′ ) ′ = ( e x + y ) ′ = e x + y · ( x + y ) ′ = e x + y · (1 + y ′ ) = e x + y · (1 + e x + y ) > , because e x + y > 0. On the other hand, if y = y ( x ) is any function satisfying equation (d), then by the Chain Rule y ′′ = ( y ′ ) ′ = ( e − x − y ) ′ = e − x − y · ( − x − y ) ′ = − e − x − y · (1 + y ′ ) = − e − x − y · (1 + e − x − y ) < , because e − x − y > 0. So, II must correspond to (c), while IV must correspond to (d). Grading: correct − repeats 0 1 2 3 4 points 2 5 9 10 2 Problem 2 (15pts) (a; 7pts) Show that the function y ( x ) = x e − 2 x is a solution to the initialvalue problem y ′′ + 4 y ′ + 4 y = 0 , y = y ( x ) , y (0) = 0 , y ′ (0) = 1 ....
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This note was uploaded on 01/25/2012 for the course MAT 127 taught by Professor Guanyushi during the Fall '07 term at SUNY Stony Brook.
 Fall '07
 GuanYuShi
 Calculus, Equations

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