Lesson 11a -Cauchy Euler

Thenourtwoindependent e i cos isin

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Unformatted text preview: ns and ”. Then our two independent e i Cos ( ) iSin( ) solutions are: (remember that ) y x Cos ( ln( x)) Why? y x Sin( ln( x)) y x ( i ) y x ( i ) We know when we have two distinct roots, the solution was y c1 x r1 c2 x r2 y c1 x r1 c2 x r2 c1 x ( i ) c2 x ( i ) c1 x x i c2 x x i c1 x e ln( x ) i c2 x e ln( x ) i x [c1Cos (ln( x) ) ic1Sin(ln( x) ) c2Cos ( ln( x) ) ic2 Sin( ln( x) )] x [c1Cos (ln( x) ) ic1Sin(ln( x) ) c2Cos (ln( x) ) ic2 Sin(ln( x) )] x [(c1 c2 )Cos (ln( x) ) (c1 ic2 ) Sin(ln( x) )] y x [C1Cos (ln( x) ) C2 Sin(ln( x) )] Strat: Second Order Cauchy‐Euler Step 1: Recognize that it is in the form: ax 2 y ' 'bxy ' cy 0 Determine the solution(s) to indicial equation: ar2+(b‐a)r+c=0 Case 1: If they are both real and distinct use: Case 2: If it is real and unique use: y c1 x c2 x r1 r2 y c1 x r c2 x r ln x i i Case 3: If they are imaginary and work out to and then use: y x [C1Cos (ln( x) ) C2 Sin(ln( x) )] Example 1 Solve the following IVP when y(1)=7, and y(4)=7: 2 x 2 y ' '3 xy ' y 0 Indicial Equation: 2r2+(3‐2)r‐1=0 Quadratic formula gives: r 1 9 1 1, 4 2 This is Case 1 (two real) this makes our general solution: y c1 x 1 c2 x 1 2 For the IVP we sub in y(1)=7 and y(4)=7 to get to get: 1 1 1 21 7c2 7 c1 (4) c2 (4) 2 7 c1 (1) 1 c2 (1) 2 1 3 c2 7 c1 2c2 7 c1 c2 4 7 c2 c1 28 c1 8c2 28 7 c2 8c2 7 3 c1 4 c1 1 y 4 x 3x 1 2 Example 2 Solve the following second order ODE: x 2 y ' '3 xy'4 y 0 Auxiliary Equation: r2‐4r+4=0 Quadratic formula gives: r2 This is Case 2 (one real) this makes our general solution: y c1 x c2 ln( x) x 2 2 Example 3 Solve the following second order ODE x 2 y ' '2 xy '2 y 0 Auxiliary Equation: r2‐6r+13=0 Quadratic formula gives: r 2 4 1 i 2 This is Case 3 (two imaginary) this makes our general solution: y x(c1Cos (ln( x)) c2 Sin(ln( x))...
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