MA240_2009FALL_EXAM1__[0]

# MA240_2009FALL_EXAM1__[0] - MA240 Ordinary Differential...

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Unformatted text preview: MA240 Ordinary Differential Equations The Cooper Union, Fall 2009 Prof. Alex Casti Midterm Exam Solutions Problem 1 (20 points) Consider the following linear initial value problem: x 2 dy dx + xy = 1 y ( 1 ) = 1 (a) Solve for y ( x ) . There should be no free parameters (constants) in your final solution. Notice that the ODE can be rewritten as x dy dx + y = 1 x → d dx ( xy ) = 1 x Z x 1 d dx ( xy ) dx = Z x 1 dx x xy- y ( 1 ) = ln x ( use IC ) xy- 1 = ln x → y = 1 x ( 1 + ln x ) . (b) Describe the largest interval I on the x-axis (containing the initial data point) over which the solution exists. The solution y = x- 1 ( 1 + ln x ) is singular at x = 0, so we must exclude that point. Because we want a solution domain that contains x = 1, where the initial condition is given, we choose I = ( , ∞ ) . Problem 2 (10 points) Suppose the autonomous ODE dy dx = ye y- 9 y e y Find the critical points and classify each of them as asymptotically stable, unstable, or semi-stable. Justify your answers. The critical points correspond to solutions of ye y- 9 y e y = → y ( e y- 9 ) = . Thus, there are two critical points (critical lines) y c = 0 and y c = ln9. To assess the stability of these solutions, we investigate the sign of the derivative dy dx just below and just above the lines y = y c in the xy-plane. Consider first y c = 0. For 0 < y < ln9, we have dy dx < 0, which means that any initial condition y ( ) = y , with 0 < y < ln9, will lie on a solution curve that decreases toward the critical line at y = 0. If y < 0, we see that dy dx > 0, since dy dx = ( ye- y )( e y- 9 ) is the product of two negative terms. Therefore, any initial condition y < 0 will be on a solution curve that increases toward the critical line y = 0. From this we conclude that the critical point y c = 0 is asymptotically stable (an attractor )....
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MA240_2009FALL_EXAM1__[0] - MA240 Ordinary Differential...

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