Lecture 9 - Reduction of Order Repeated Roots of the...

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Reduction of Order Repeated Roots of the Characteristic Equation
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So far… We’ve Learned How To Solve Second Order Linear Homogeneous ODEs with Constant Coefficients Characteristic Equation We’ve seen distinct real roots, complex conjugate roots
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Characteristic Equation A y 00 + B y 0 + C y = 0 Gives the Characteristic Equation A r 2 + B r + C = 0 If roots are real and distinct, r 1 r 2 and General solution: y = C 1 e r 1 t + C 2 e r 2 t If roots are complex, a + b i a - b i and General solution: Case 1: Case 2: y = B 1 e a t cos ( b t ) + B 2 e a t sin ( b t )
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Characteristic Equation A y 00 + B y 0 + C y = 0 Gives the Characteristic Equation A r 2 + B r + C = 0 What if there’s only one root, r = - B ± p B 2 - 4 AC 2 A and B 2 - 4 AC = 0 ? r 1 = - B 2 A
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Characteristic Equation A y 00 + B y 0 + C y = 0 Gives the Characteristic Equation A r 2 + B r + C = 0 What if there’s only one root, Then we know one solution: y = e r 1 t How do we find another? r 1 = - B 2 A
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Reduction of Order If we have a linear homogeneous second order equation y 00 + p ( t ) y 0 + q ( t ) y = 0 and we know one solution y 1 How do we find another?
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Reduction of Order y 00 + p ( t ) y 0 + q ( t ) y = 0 One solution y 1 In general, finding solutions is difficult
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Reduction of Order y 00 + p ( t ) y 0 + q ( t ) y = 0 One solution y 1 Guess second solution has the form: v ( t ) y 1 ( v ( t ) y 1 ) 0 ( v ( t ) y 1 ) 00 v 0 ( t ) v 00 ( t ) y 1 0 y 1 00 = y 1 + v ( t ) y 1 = 2 + v 0 ( t ) y 1 0 + v ( t ) Insert guess into ODE y 00 + p ( t ) q ( t ) y 0 + = 0 y
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Reduction of Order y 00 + p ( t ) y 0 + q ( t ) y = 0 One solution y 1 Guess second solution has the form: v ( t ) y 1 ( v ( t ) y 1 ) 0 ( v ( t ) y 1 ) 00 v 0 ( t ) v 00 ( t ) y 1 0 y 1 00 = y 1 + v ( t ) y 1 = 2 + v 0 ( t ) y 1 0 + v ( t ) Insert guess into ODE + p ( t ) q ( t ) + = 0 v ( t ) y 1 ( v ( t ) y 1 ) 00 v 0 ( t ) y 1 0 y 1 + v ( t ) ( )
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Reduction of Order y 00 + p ( t ) y 0 + q ( t ) y = 0 One solution
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