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Unformatted text preview: HOMEWORK 2: MATH 265, L Keshet ( Final version ) Due in class on September 29, 2010 NOTE: Most problems on this assignment are straightforward. Problem 4 may take a bit more time and effort. Problem 1: In each case, solve the following second order ODEs for y ( t ): (a) y ′′ + 2 y ′ 3 y = 0 , and y (0) = 1 , y ′ (0) = 2 (b) y ′′ 9 y ′ + 20 y = 0 and y (0) = 1 , y ′ (0) = 0 (c) y ′′ 2 y ′ + 5 y = 0 and y (0) = 1 , y ′ (0) = 1. (d) y ′′ 2 y = 0 and y (0) = 0 , y ′ (0) = 2. Solution to Problem 1: (a) The characteristic equation is r 2 + 2 r 3 = 0. This factors into ( r + 3)( r 1) = 0 so has solutions r = 1 , 3. The general solution is thus y ( t ) = C 1 e t + C 2 e − 3 t . We find C 1 , C 2 from the initial conditions. We need to find y ′ ( t ) by differentiating y ( t ): we get y ′ ( t ) = C 1 e t 3 C 2 e − 3 t . Using the initial conditions, we have y (0) = 1 , ⇒ 1 = C 1 + C 2 and y ′ (0) = 2 ⇒ 2 = C 1 3 C 2 . Solving these equations leads to C 2 = 1 / 4 and C 1 = 5 / 4 so the solution is y ( t ) = 5 4 e t + 1 4 e − 3 t (b) The characteristic equation is r 2 9 r + 20 = ( r 4)( r 5) = 0, so the roots are r = 4 , 5 and the general solution is y ( t ) = C 1 e 4 t + C 2 e 5 t . Then the initial conditions mean that y (0) = 1 = C 1 + C 2 and y ′ (0) = 0 = 4 C 1 + 5 C 2 . Solving these for C 1 , C 2 leads to y ( t ) = 5 e 4 t 4 e 5 t . (c) The characteristic equation is r 2 2 y + 5 = 0. This has the complex roots r = 1 ± 2 i so the general solution is y ( t ) = C 1 e t sin(2 t ) + C 2 e t cos(2 t ). We also need the derivative y ′ ( t ) = e t ( C 1 sin(2 t ) + 2 C 1 cos(2 t ) + C 2 cos(2 t ) 2 C 2 sin(2 t )). Then we use the initial conditions: 1 = y (0) = C 2 and 1 = (2 C 1 + C 2 ) (we used the facts that e = 1 , sin(0) = 0 , cos(0) = 1). This tells us that C 2 = 1 , C 1 = 0 so the solution is y ( t ) = e t cos(2 t ). (d) Characteristic equation: r 2 2 = 0 so r = ± √ 2 and y ( t ) = C 1 e √ 2 t + C 2 e − √ 2 t . The derivative: y ′ ( t ) = √ 2 C 1 e √ 2 t √ 2 C 2 e − √ 2 t . Using the I.C’s: y (0) = 0 = C 1 + C 2 and y ′ (0) = 2 = √ 2 C 1 √ 2 C 2 . Solving for constants leads to y ( t ) = √ 2 2 e √ 2 t √ 2 2 e − √ 2 t . Problem 2: Consider the differential equation ay ′′ + by ′ + cy = 0. Suppose that the two functions y = f 1 ( t ) and y = 1 2 [ f 2 ( t ) + f 1 ( t )] are both solutions to this equation. Show that the function f 2 ( t ) is also a solution....
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 Winter '10
 LEAHKESHET
 Math, Differential Equations, Equations, Quadratic equation, Elementary algebra

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