1) x = 1 is an ordinary point of the D.E :(x-1)y - 2xy + 6y...

 

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1) x = 1 is an ordinary point of the D.E :(x²-1)y" - 2xy + 6y = 0 ( a₁ 2) If y = = 0 anx" is the series solution of the D.E: y" + y'=0 then a₂ = ( ) 2n=0 100 3) If y = 0 anx" is the series solution of the D.E: y" - 2y = 0, then the recurrence equation we get is an+2 = an , n = 0,1,2,... ( ) (n+2)(n+1) 4) The eigen values of the matrix, (3) 5) The general solution of the system X'(t) = (3) X is X(t) = C₁₂ (²₁) e²³² + c₂ (¹₁) e-t. e3t are 1 F 3i. ( ) 6) The general solution of the linear system X' (t) = (13) X is X(t) = ₁₁ (₁¹) e²t + c₂[(¹) te²t +(²¹]. C1 solution is X(t) = 4(-2 sin2t) e-t. cos ( 7) Given that, the general solution of the system; X' (t) = AX is cos 2t X(t) = C₁ (²052) e-t + c₂ (-2sin2t) e-t .If X (0) = (2) sin 2t cos 2 8) The matrix; ( has repeated eigen values. (-34) ) ( ) then the particular 9) The eigen vector could be the zero vector. 10) If L(h(t)) = H(s) then L(h""(t)) = s³H(s) - s²H(0) - s H'(0) - H"(0) . ( ) 1) x = 1 is an ordinary point of the D.E :(x²-1)y" - 2xy + 6y = 0 ( a₁ 2) If y = = 0 anx" is the series solution of the D.E: y" + y'=0 then a₂ = ( ) 2n=0 100 3) If y = 0 anx" is the series solution of the D.E: y" - 2y = 0, then the recurrence equation we get is an+2 = an , n = 0,1,2,... ( ) (n+2)(n+1) 4) The eigen values of the matrix, (3) 5) The general solution of the system X'(t) = (3) X is X(t) = C₁₂ (²₁) e²³² + c₂ (¹₁) e-t. e3t are 1 F 3i. ( ) 6) The general solution of the linear system X' (t) = (13) X is X(t) = ₁₁ (₁¹) e²t + c₂[(¹) te²t +(²¹]. C1 solution is X(t) = 4(-2 sin2t) e-t. cos ( 7) Given that, the general solution of the system; X' (t) = AX is cos 2t X(t) = C₁ (²052) e-t + c₂ (-2sin2t) e-t .If X (0) = (2) sin 2t cos 2 8) The matrix; ( has repeated eigen values. (-34) ) ( ) then the particular 9) The eigen vector could be the zero vector. 10) If L(h(t)) = H(s) then L(h""(t)) = s³H(s) - s²H(0) - s H'(0) - H"(0) . ( )