And it immediately follows that equal to zero we have

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Unformatted text preview: age 215 —————————————————————————— CHAPTER 5. —— Hence the solution of the initial value problem is The solution is bounded, as , if . 2 results 380. Substitution of are given by in the quadratic equation Formally, the roots The roots will be complex, if as , we need The roots will be equal, if zero as long as . The roots will be real and distinct, if . . For solutions to approach zero, In this case, all solutions approach . It follows that . For solutions to approach zero, we need . Hence all solutions approach zero, as 23 . Given that , That is, , as long as . By the chain rule, . Similarly, . Direct substitution results in ________________________________________________________________________ page 216 —————————————————————————— CHAPTER 5. —— , that is, . The associated characteristic equation is it follows that . If the roots , are real and distinct, then . If the roots . Since are real and equal, then . If the roots are complex conjugates, then 24. Based on Prob. , the change of variable , and transforms the ODE into . The associated characteristic equation is Hence , and 26. The change of variable , with roots ,. transforms the ODE into . The associated characteristic equation is , with roots , . Hence Since the right hand side is not a solution of the homogeneous equation, we can use the method of undetermined coefficients to show that a particular solution is . Therefore the general solution is given by , that is, 27. The change of variable transforms the given ODE into ________________________________________________________________________ page 217 —————————————————————————— CHAPTER 5. —— . The associated characteristic equation is , with roots ,. Hence Using the method of undetermined coefficients, let . It follows that the general solution is given by , that is, 28. The change of variable transforms the given ODE into . The solution of the homogeneous equation is The right hand side is not a solution of the homogeneous equation. We can use the method of undetermined coefficients to show that a particular solution is . Hence the general solution is given by , that is, 29. After dividing the equation by , the change of variable into transforms the ODE . The associated characteristic equation is . Hence the general solution is , with complex roots , and therefore . 30. Let and Since . Setting , successive differentiation gives . It follows that , we find that ________________________________________________________________________ page 218 —————————————————————————— CHAPTER 5. —— Given that and are roots of Therefore and differential equation, , for , we have . are linearly independent solutions of the , as long as . ________________________________________________________________________ page 219 —————————————————————————— CHAPTER 5. —— Section 5.6 1. when . Since the three coeff...
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This note was uploaded on 03/11/2014 for the course MA 303 taught by Professor Staff during the Spring '08 term at Purdue.

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