2 where 4 4 7 4 first nd general solution of y ay

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Unformatted text preview: ion of y00 1 Aycos(x) A = 2 1 3 cos(x). = 2 where 4 4 7. 4. First find general solution of y = Ay where A = 4 7 3 4 Clearly this is satisfied for 1 6= = 0. Characteristic equation is: det all x 0 if we choose ) =2 +26 2 + 90. So y1 , y2 are linearly =0 1 2 14 4 7 dependent on x . Characteristic equation0is: det = 0 ) + 6 + 9 = 0. 4 7 ThusOn x < 0 equation = ) becomes (b) eigenvalues are: = 3, 3. Thus eigenvalues are: ( 4 3, 4 . 3 v1 0 1 Eigenvector for = 3: = 0 ) v = v . Choose v = 1 . 44 4 v x2 x2 v 1 = 0 )3v2 = v1 . Choose v = 1 . Eigenvector for = 3: 4 +2 1 4 v2 0 of v. ) 1 = 0. 1 All other eigenvectors of = 34arecos(x) multiples 3 cos(xGeneralized eigenvector for = 3: scalar All other eigenvectors of1 = 3 are scalar multiples of v. Generalized eigenvector for = 3: 1 4 4 u1 1 4 . satisfied for all x 1 + u if Choose u 4 Clearly this is = 1 ) u1 = 4< 0 2 . we choose = 1 =0 3 2 6= 0. So y1 , y2 are linearly 4 u1 1 4 4 u2 =< 0. ) u1 = 1 + u2 . Choose u = 4 . 4 4 dependent 2on x u 1 0 General4solution: 1 General solution: (c) Comparing parts (a) and (b),x we1see that tox s...
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