Lecture 4 Notes

Lecture 4 Notes - Today Independence of functions and the...

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Today • Independence of functions and the Wronskian • Distinct roots of the characteristic equation • Review of complex numbers • Complex roots of the characteristic equation
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Homog. eq. with constant coeff. (Section 3.1) • For the general case, , by assuming we get the characteristic equation : ay °° + by ° + cy =0 y ( t )= e rt ar 2 + br + c
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Homog. eq. with constant coeff. (Section 3.1) • For the general case, , by assuming we get the characteristic equation : • There are three cases. ay °° + by ° + cy =0 y ( t )= e rt ar 2 + br + c
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Homog. eq. with constant coeff. (Section 3.1) • For the general case, , by assuming we get the characteristic equation : • There are three cases. i. Two distinct real roots: b 2 - 4ac > 0. ( r 1 r 2 ) ay °° + by ° + cy =0 y ( t )= e rt ar 2 + br + c
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Homog. eq. with constant coeff. (Section 3.1) • For the general case, , by assuming we get the characteristic equation : • There are three cases. i. Two distinct real roots: b 2 - 4ac > 0. ( r 1 r 2 ) ii.A repeated real root: b 2 - 4ac = 0. ay °° + by ° + cy =0 y ( t )= e rt ar 2 + br + c
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Homog. eq. with constant coeff. (Section 3.1) • For the general case, , by assuming we get the characteristic equation : • There are three cases. i. Two distinct real roots: b 2 - 4ac > 0. ( r 1 r 2 ) ii.A repeated real root: b 2 - 4ac = 0. iii.Two complex roots: b 2 - 4ac < 0. ay °° + by ° + cy =0 y ( t )= e rt ar 2 + br + c
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Homog. eq. with constant coeff. (Section 3.1) • For the general case, , by assuming we get the characteristic equation : • There are three cases. i. Two distinct real roots: b 2 - 4ac > 0. ( r 1 r 2 ) ii.A repeated real root: b 2 - 4ac = 0. iii.Two complex roots: b 2 - 4ac < 0. • For case i, we get and . ay °° + by ° + cy =0 y ( t )= e rt ar 2 + br + c y 1 ( t e r 1 t y 2 ( t e r 2 t
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Homog. eq. with constant coeff. (Section 3.1) • For the general case, , by assuming we get the characteristic equation : • There are three cases. i. Two distinct real roots: b 2 - 4ac > 0. ( r 1 r 2 ) ii.A repeated real root: b 2 - 4ac = 0. iii.Two complex roots: b 2 - 4ac < 0. • For case i, we get and . • Do our two solutions cover all possible ICs? That is, can we use them to form a general solution ? ay °° + by ° + cy =0 y ( t )= e rt ar 2 + br + c y 1 ( t e r 1 t y 2 ( t e r 2 t
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Independence and the Wronskian (Section 3.2) • Example: Suppose and are two solutions to some equation. Can we solve ANY initial condition . . with these two solutions? • Solve this system for C 1 , C 2 ... • Can’t do it. Why? y 2 ( t )= e 2 t 3 y 1 ( t e 2 t +3 y (0) = y 0 ,y ° (0) = v 0
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Independence and the Wronskian (Section 3.2) • Example: Suppose and are two solutions to some equation. Can we solve ANY initial condition . . with these two solutions? • Solve this system for C 1 , C 2 ... • Can’t do it. Why? y 2 ( t )= e 2 t 3 y 1 ( t e 2 t +3 y (0) = y 0 ,y ° (0) = v 0 y ( t C 1 e 2 t +3 + C 2 e 2 t 3
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Independence and the Wronskian (Section 3.2) • Example: Suppose and are two solutions to some equation. Can we solve ANY initial condition . . with these two solutions?
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Lecture 4 Notes - Today Independence of functions and the...

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