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MIT18_03S10_2ex

MIT18_03S10_2ex - 2 Higher-order Linear ODEs 2A...

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�� 2. Higher-order Linear ODE’s 2A. Second-order Linear ODE’s: General Properties 2A-1. On the right below is an abbreviated form of the ODE on the left: (*) y + p ( x ) y + q ( x ) y = r ( x ) Ly = r ( x ) ; where L is the differential operator : L = D 2 + p ( x ) D + q ( x ) . a) If u 1 and u 2 are any two twice-differentiable functions, and c is a constant, then L ( u 1 + u 2 ) = L ( u 1 ) + L ( u 2 ) and L ( c u ) = c L ( u ) . Operators which have these two properties are called linear . Verify that L is linear, i.e., that the two equations are satisfied. b) Show that if y p is a solution to (*), then all other solutions to (*) can be written in the form y = y c + y p , where y c is a solution to the associated homogeneous equation Ly = 0. 2A-2. a) By eliminating the constants, find a second-order linear homogeneous ODE whose x + c 2 e 2 x general solution is y = c 1 e . b) Verify for this ODE that the IVP consisting of the ODE together with the initial conditions y ( x 0 ) = y 0 , y ( x 0 ) = y y 0 , y constants 0 0 is always solvable. 2A-3. a) By eliminating the constants, find a second-order linear homogeneous ODE whose y = c 1 x + c 2 x 2 general solution is . b) Show that there is no solution to the ODE you found in part (a) which satisfies the initial conditions y (0) = 1 , y (0) = 1 . c) Why doesn’t part (b) contradict the existence theorem for solutions to second-order linear homogeneous ODE’s? (Book: Theorem 2, p. 110.) 2A-4. Consider the ODE y �� + p ( x ) y + q ( x ) y = 0. a) Show that if p and q are continuous for all x , a solution whose graph is tangent to the x -axis at some point must be identically zero, i.e., zero for all x . b) Find an equation of the above form having x 2 as a solution, by calculating its derivatives and finding a linear equation connecting them. Why isn’t part (a) contradicted, since the function x 2 has a graph tangent to the x axis at 0? 1
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�� 2 18.03 EXERCISES 2A-5. Show that the following pairs of functions are linearly independent, by calculating their Wronskian. m 1 x m 2 x mx a) e , e , m 1 = m 2 b) e , xe mx (can m = 0?) 2A-6. Consider y 1 = x 2 and y 2 = x | x | . (Sketch the graph of y 2 .) a) Show that W ( y 1 , y 2 ) 0 (i.e., is identically zero). b) Show that y 1 and y 2 are not linearly dependent on any interval ( a, b ) containing 0. Why doesn’t this contradict theorem 3b, p. 116 in your book? 2A-7. Let y 1 and y 2 be two solutions of y �� + p ( x ) y + q ( x ) y = 0. dW a) Prove that = p ( x ) W , where W = W ( y 1 , y 2 ), the Wronskian. dx b) Prove that if p ( x ) = 0, then W ( y 1 , y 2 ) is always a constant. c) Verify (b) by direct calculation for y �� + k 2 y = 0 , k = 0, whose general solution is y 1 = c 1 sin kx + c 2 cos kx . 2B. Reduction of Order 2B-1. Find a second solution y 2 to y �� 2 y + y = 0, given that one solution is y 1 = e x , by three methods: a) putting y 2 = ue x and determining u ( x ) by substituting into the ODE; b) determining W ( y 1 , y 2 ) using Exercise 2A-7a, and from this getting y 2 ; c) by using the general formula y 2 = y 1 1 2 e p dx dx .
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