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Unformatted text preview: MATH 2930 Practice Problems  Week 1, Summer 2009 S. Bhat June 9, 2009 Â§ 1.1 #4: We have to verify that the solutions y 1 = e 3 x and y 2 = e 3 x are solutions to the differential equation y 00 = 9 y . If we plug y 1 into the left hand side of the equation, we get: y 00 1 = 9( e 3 x ) = 9 y 1 so y 1 satisfies the differential equation. Similarly, for y 2 , we get: y 00 2 = 9( e 3 x ) = 9 y 2 so y 2 also satisfies the differential equation. #8: Similar to problem 4, we plug in the trial solutions y 1 = cos x cos 2 x and y 2 = sin x cos 2 x into the differential equation y 00 + y = 3 cos 2 x to see if the equation holds. Taking derivatives of y 1 , we get that y 1 = sin x + 2 sin 2 x , and so y 00 1 = cos x + 4 cos 2 x . Plugging this in, we get: y 00 1 + y 1 = cos x + 4 cos 2 x + cos x cos 2 x = 3 cos 2 x so y 1 satisfies the differential equation. Doing the same for y 2 , we have the second derivative y 00 2 = sin x + 4 cos 2 x . Plugging this in: y 00 2 + y 2 = sin x + 4 cos 2 x + sin x cos 2 x = 3 cos 2 x 1 so y 2 also satisfies the differential equation. #19: The problem asks us to do a few things. Letâ€™s take them one by one. First, we have to verify that y ( x ) = Ce x 1 is a solution to the differential equation y = y + 1. Like before, we simply plug our trial solution into both sides of the equation: y = Ce x and y + 1 = Ce x 1 + 1 = Ce x . Thus, the differential equation is satisfied. Now, we have to use the initial condition to figure out what C is. To do so, we simply plug the initial condition y (0) = 5, or, put another way, when x = 0, then y = 5. We thus solve the equation: y (0) = 5 = Ce 1 = C 1 . So, C = 6. Finally, we are asked to graph some typical solutions and highlight the one that is represented by our solution above: y ( x ) = 6 e x 1. Use whichever program you are comfortable with, but itâ€™s a good idea, especially if youâ€™re an engineer, to get used to Matlab. Here is how you can plot the graphs: 2 x = 0.5:0.01:2 y1 = 0*exp(x)  1 y2 = 2*exp(x)  1 y3 = 4*exp(x)1 y4 = 6*exp(x)  1 y5 = 8*exp(x)  1 y6 = 10*exp(x)  1 h = plot(x,y1,x,y2,x,y3,x,y4,x,y5,x,y6) xlabel(â€™xâ€™) ylabel(â€™yâ€™) set(h(4),â€™LineWidthâ€™, 4) The resultant graph is shown here, with the solution satisfying the initial condi tion denoted by the thicker line. ! 0.5 0.5 1 1.5 2 ! 10 10 20 30 40 50 60 70 80 x y 3 #34: â€śThe acceleration dv/dt of a Lamborghini is proportional to the difference between 250 km/h and the velocity of the car.â€ť We know that we have to relate dv/dt on the left side of the equation and the differ ence between 250 km/h and the velocity of the car. The latter term is just 250 v ....
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This note was uploaded on 01/21/2011 for the course MATH 2930 taught by Professor Terrell,r during the Spring '07 term at Cornell.
 Spring '07
 TERRELL,R
 Math

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