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Unformatted text preview: MAP 2302, Fall 2010 — Midterm 1 Review Solutions 1 The exam will cover sections 1.2, 1.3, 2.2, 2.3, 2.4, and 2.6. All topics from this review sheet or from the suggested exercises are fair game. 1 Give explicit solutions to the initial value problem dy dx = xy 3 with y (0) = 1, y (0) = 1 / 2 , and y (0) = − 2. Then determine the domains of each of these solutions. Solution: First, we separate the differential equation and solve it: integraldisplay y 3 dy dx dx = integraldisplay xdx, integraldisplay y 3 dy = x 2 2 + C, y 2 − 2 = x 2 2 + C. By our standard abuse of constants, we then get y 2 = − 1 x 2 + C , so y = ± radicalbigg − 1 x 2 + C . Now we want to match the various initial conditions. For each, we need to decide whether to take the + or the − of the ± , and then we need to determine C : initial condition ± C solution y (0) = 1 + C = − 1 y ( x ) = radicalBig 1 x 2 1 y (0) = 1 / 2 + C = − 4 y ( x ) = radicalBig 1 x 2 4 y (0) = − 2 − C = − 1 / 2 y ( x ) = − radicalBig 1 x 2 1 / 4 We now need to determine the domains of these solutions. For all of them, we must be careful not to get 0 in the denominator or to take the square root of a negative number. For example, for the first solution, we have a problem when x = 1 (division by 0) or when  x  > 1 (square root of a negative number), so the domain is  x  < 1. solution domain y ( x ) = radicalBig 1 x 2 1  x  < 1 y ( x ) = radicalBig 1 x 2 4  x  < 2 y ( x ) = − radicalBig 1 x 2 1 / 4  x  < 1 / 2 MAP 2302, Fall 2010 — Midterm 1 Review Solutions 2 2 Show that every separable firstorder differential equation can easily be converted into an exact equation. Solution: A firstorder differential equation is separable if it can be put in the form dy dx = g ( x ) p ( y ) . To test for exactness, we put this equation into the form M ( x,y ) + N ( x,y ) dy dx = 0. In this case we see that M ( x,y ) = − g ( x ) and N ( x,y ) = 1 p ( y ) . Now we only need that ∂M ∂y = ∂N ∂x , which is true because both partials are 0. 3 For each of the following differential equations, indicate whether they are separable, linear, or can easily be converted into an exact equation. Note that some equations may be more than one type, while others may not be any of these types. Then, solve the equations which are separable, linear, or exact. a. dy dx = 2 xy x 2 + y 2 . Solution: This equation is not separable, because there is no way 1 to write it in the form dy dx = g ( x ) p ( y ). The equation is also not linear, because the y 2 term prevents us from putting it in the form dy dx + P ( x ) y = Q ( x ). To test for exactness we put the equation in the form M ( x,y ) + N ( x,y ) dy dx = 0: 2 xy + ( x 2 + y 2 ) dy dx = 0 ....
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This note was uploaded on 12/10/2011 for the course MAP 2302 taught by Professor Tuncer during the Fall '08 term at University of Florida.
 Fall '08
 TUNCER

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