1. Find a general solution to the following ordinary differential equation: !! + 4 = 32 2. find the eigenvalues and eigenvectors of the following matrix A: 1 2 3 A= 0 5 6 0 0 7 Ans : 1 y ' ' + 4 y=3 sin 2 x auxiliary equation m2+4=0= m=2i ,2i y=C 1 cos 2 x +C2 sin 2 x particular solution y= 1 3 sin 2 x D +4 2 3x 1 sin 2 x 2D 3x cos 2 x . 4 Hence genral solution y=C 1 cos 2 x +C2 sin 2 x 3x cos 2 x 4 2. [ ] 1 0 0 A= 0 1 0 0 0 1 The chracterstice equationis det ( AI )=0 det [ 1 2 3 0 5 6 =0 0 0 7 > ( 1 ) ( 5 )(7 )=0 > =1 ,5,7 Corresponding eigenvector =1 [ [ ] 0 2 3 x 0 4 6 y =0 0 0 8 z z=02 x=3 y choose x=3then y=2 HENCE Corresponding eigenvector =1 3 X= 2 0 Simillarly Corresponding eigenvector =5 [ [ ] 4 2 3 x 0 0 6 y =0 0 0 12 z z=04 x=2 y choose x=1then y =2 HENCE Corresponding eigenvector =5 1 X= 2 . 0 Corresponding eigenvector =7 [ [ ] 8 2 3 x 0 12 6 y =0 0 0 0 z 12 y =6 z 8 x +2 y +3 z=0 choose z =1then y= 1 2 1 x= = 2 8 4 HENCE Corresponding eigenvector =7 1 4 X= 1 . 2 1 . . : 1 +4 = 3 + 4 = 0 => = = 2 , 2 2 + 2 1 3 +4 = = 3 = = 2 1 2 3 4 2 2 2 . 2 + 2 3 4 2 2. 1 = 0 0 2 3 5 6 0 7 det( 1 0 0 )=0 2 5 0 3 6 7 =0 => (1 )(5 )(7 ) = 0 => =1, 5, 7 , =1 0 2 3 0 4 6 0 0 8 =0 =0 2 = 3 =3 = 2 =1 3 = 2 0 =5 4 2 0 0 0 0 3 6 12 =0 =0 4 =2 =1 = 2 =5 1 = 2 . 0 = 7 8 2 0 12 0 0 12 = 6 =1 3 6 0 =0 8 +2 +3 = 0 = 1 2 = 2 1 = 8 4 = 7 1 4 = 1. 2 1 . . Ans : 1 y ' ' + 4 y=3 sin 2 x auxiliary equation m2+4=0= m=2i ,2i y=C 1 cos 2 x +C2 sin 2 x particular solution y= 1 3 sin 2 x D +4 2 3x 1 sin 2 x 2D 3x cos 2 x . 4 Hence genral solution y=C 1 cos 2 x +C2 sin 2 x 3x cos 2 x 4 2. [ ] 1 0 0 A= 0 1 0 0 0 1 The chracterstice equationis det ( AI )=0 det [ 1 2 3 0 5 6 =0 0 0 7 > ( 1 ) ( 5 )(7 )=0 > =1 ,5,7 Corresponding eigenvector =1 [ [ ] 0 2 3 x 0 4 6 y =0 0 0 8 z z=02 x=3 y choose x=3then y=2 HENCE Corresponding eigenvector =1 3 X= 2 0 Simillarly Corresponding eigenvector =5 [ [ ] 4 2 3 x 0 0 6 y =0 0 0 12 z z=04 x=2 y choose x=1then y =2 HENCE Corresponding eigenvector =5 1 X= 2 . 0 Corresponding eigenvector =7 [ [ ] 8 2 3 x 0 12 6 y =0 0 0 0 z 12 y =6 z 8 x +2 y +3 z=0 choose z =1then y= 1 2 1 x= = 2 8 4 HENCE Corresponding eigenvector =7 1 4 X= 1 . 2 1 . . : 1 +4 = 3 + 4 = 0 => = = 2 , 2 2 + 2 1 3 +4 = = 3 = = 2 1 2 3 4 2 2 2 . 2 + 2 3 4 2 2. 1 = 0 0 2 3 5 6 0 7 det( 1 0 0 )=0 2 5 0 3 6 7 =0 => (1 )(5 )(7 ) = 0 => =1, 5, 7 , =1 0 2 3 0 4 6 0 0 8 =0 =0 2 = 3 =3 = 2 =1 3 = 2 0 =5 4 2 0 0 0 0 3 6 12 =0 =0 4 =2 =1 = 2 =5 1 = 2 . 0 = 7 8 2 0 12 0 0 12 = 6 =1 3 6 0 =0 8 +2 +3 = 0 = 1 2 = 2 1 = 8 4 = 7 1 4 = 1. 2 1 .