276 Exercises

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Differential Eq

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  • 276

    2

    pi

    pi:

    1. 22 = (4 + 2)

    2. = (1 + 2) 2

    3. + = 0

    4.

    (1 + 2)= 2

    5. 0 =3

    , (1) =3

    6. pi2 2 = 0

    7.

    + = 0,

    3

    2

    =

    8. 2

    +1

    = 0, (1) = 0

    pi .

    1

  • 276

    3

    pi:

    1. 0 3 = 6

    2. 0 2 =

    3. 0 + = , () = 1

    4.

    =

    5. 0 + = 2

    6. 0 3

    2=1

    2

    7.pi

    pi

    = 2 + 3 2

    8. 0 +2

    = , (1) = 0

    9. 0 + 2 = 22

    10. + ( + 1) = 0, (1) = 1

    pi .

    1

  • 276

    4

    pi:

    1.

    3= 244

    2. 0 +

    2=

    3, (1) = 2

    3. 0 + 3 = 22

    4. 0 + = ()3=2, (1) = 4

    5. 1 2

    + = 0

    6. 0 +

    =

    pi3=2

    7. 0 + = 2

    8. 0 3

    = 4 3

    pi

    9.

    =2

    10. + 3 + 83

    = 0

    11.

    +(2+ 1)

    2=2+ 1

    pi .

    1

  • 276

    5

    pi:

    1. 2 =2 + 2

    2. 0 = +pi2 2

    3. 220 = 2 + 2

    4. (4 3) + (2 3) = 0

    5. ( ) + ( + ) = 0

    6.22 2

    + = 0

    7. 2 (2 + 2) = 0,

    : =

    2

    1

    8.2 + 32

    2 = 0, (2) = 6

    :

    2

    2+ 1 = 5

    9. 23+4 + 4

    = 0,

    : 342 + 6 =

    10. 0 =pi2 2+, [ : = ( ), : = = ]

    pi .

    1

  • 276

    6

    :

    1. (+ 2)+ ( + 4) = 0

    2. (+ + 1)+ (2+ 2 1) = 0

    3. ( 1)+ (+ 4 1) = 0

    4. (5+ 2 + 1)+ (2+ + 1) = 0

    5. (+ 2 + 3)+ (2+ 4 1) = 0

    6. (3 + 1) (6 2 3) = 0

    7.

    =+ 3 5

    1

    8. (+ 2) + (1 )0 = 0

    9. (+ )+ ( 2) = 0

    10. (8+ 4 + 1) + (4+ 2 + 1)

    = 0

    pi .

    1

  • 276

    7

    . 2.

    , .

    1. (2 + 2)+ 2 = 0

    2. (2 + )+ (2 + ) = 0

    3. (32 + 2 ) = 0

    4. ( ) = 0

    5. (433 +1

    )+ (342

    1

    ) = 0

    6. (22 + 2) + (22 + 2)0 = 0

    7. (32 23

    3)+ (3 2 + 43 +

    32

    2) = 0

    8.

    (2)

    +

    +

    2

    2

    = 0

    9. (32 2 )+ (2 + 32) = 0

    . pi:

    1.2

    3+(2 32)

    4= 0; (1) = 1

    2. (2 + 2 )+ (2 2 ) = 0; (0) = 3

    3. ( + 2 + 2)+ ( + 2) = 0; (0) = 6

    . ,

    .

    1. (2 + 3)+ (2 + 4) = 0

    2. (1

    2+1

    2)+

    + 1

    3 = 0

    pi .

    1

  • 276

    8

    1.

    0 + pi() = ()

    () = pi()

    2. pi

    ) (2 + 2 + )+ = 0

    ) 22+ (2+ 3) = 0

    ) (2 + 2) = 0

    ) (2 + 2)+ 3 = 0

    ) (1 2)+ 2( ) = 0; = ()

    ) (2 + ) = 0; = ()

    ) (22 33)+ (7 32) = 0; = ()

    ) + ( + + ) = 0; = ()

    ) 22+ (2+ 3) = 0; () =1=pi

    3.

    ) (1 + )+ (1 ) = 0

    ) (4+ 32)+ 2 = 0

    4. (+ 1)

    (2 + 2 + 1) ( + ) = 0

    .

    pi .

    1

  • 276

    9

    pipipi.

    1. (+ 2 1)+ 3(+ 2) = 0(: + 3 + = (+ + 2))

    2. (1 + 3 ) 2 = 0(: = ; : 4 = 4 1)

    3. = 2 2 20 + 2 = 2 2: () =

    + 2

    2

    4. =

    0 = (2 + )

    3 2 ; > 0

    : () =

    2

    1 +

    5. = (0)2

    000 = (0)4 (0) = 1; 0(0) = 1

    (: () = 1 )

    6. 30 + 32 = (3)2 + 1: () =

    (+ )

    3

    7. 0 = (+ )

    8. 0 = ( + 5)2

    9. 0 = + ( + 1)2 + 1

    10. 220 + 2 = (: = 22)

    . 1114 .

    11. 00 = 0 + (0)2

    12. 00 = 2(0)2

    13. 00 = 1 + (0)2

    14. 00 = 0 + (0)2

    15. 00 + 2(0)3 = 0

    16. 00 = 0

    ( 1516:

    0 = pi; 00 =pi

    =pi

    =pi

    pi

    5)

    pi .

    1

  • Math 276 Differential Equations

    Exercises 10

    Higher Order Equations - Part I

    1. Show that ex and ex are solutions of y y = 0. Hence show that sinhx and coshxare also solutions.

    (Hint: sinhx =ex ex

    2, coshx =

    ex + ex

    2)

    2. Verify that 1 andx are solutions of yy+ (y)2 = 0 over (0,) but that c1+ c2

    x

    is not a solution. Why?

    3. Verify that f1 and f2 are solutions and determine in which interval they form a fun-

    demantal set of solutions by computing W (f1, f2).

    a) y 2y + y = 0; f1 = ex and f2 = xex

    b) y y y = 0; f1 = ex and f2 = e2x

    c) x2y x(x+ 2)y + (x+ 2)y = 0; f1 = x and f2 = xex

    4. Find the homogeneous linear differential equation of order 3 (with continuous coeffi-

    cients) which has the given set as a fundemantal set

    a) x, x2, ex

    b) x, sinx, cosx

    5. Show that

    a) e3x, ex and e4x are solutions to the differential equation

    y 6y + 5y + 12y = 0b){e3x, ex, e4x

    }is the fundamental set of solutions of the above equation

    Write the general solution of the equation.

    6. Examine whether the given functions linearly independent in their domain of definition

    a) 4, x b)1, 2, x, x2 c) x, 2x, x2

    d) ex, xex, x2ex e) sinx, cosx, cos 2x f) 1, sinx, cos 2x

    g) 1, arcsinx, arccosx h) pi, arcsinx, arccosx i) 1/x, e1/x

    j) x, lnx k) e3x, ex, e4x

    7. Determine the intervals for which a unique solution of each of the following linear

    differential equations satisfying the initial conditions y(x0) = y0, y(x0) = y1 where

    x0 is any point in the interval, is certain to exist.

    a) xy + 3y = x b) exy + xy + y = tanx

    c) (1 + x2)y + 4y = ex d) x(x 1)y + 3xy + 4y = 2e) y + 6y + 7y = 2 sinx f) y + (cosx)y + (3 ln |x|) y = 0

    Prepared by U.Yksel

  • 276

    1) ) (4+ 2)+ ( + 2) = 0

    ) pi (1) = 2

    2) 0 = 3) (32 + )+ ( 43) = 0 .4) pi

    . .

    ) ( + 4) = 0) (4 2)+ = 0) (22 )+ (2 2) = 0) 2+ (2 3 3) = 0; (2) = 0) (2 + 34 ) (43 2 + ) = 0

    5) 0 2 = 3 46) (23 + 3) 32 = 0

    7) 0 =5+ 4

    2 8) 0 + = 3

    9) = 0 +

    (0)

    2+ 1

    10) 00 + 20 = 4

    11) 1 + 00 + (0)2= 0

    12) 00 + 20 + 2 = 0; (0) = 2; 0(0) = 513) 200 50 + 2 = 014)

    23 2 5 2 = 0

    15) 2 + 6 + 25

    = 0

    16) pi

    ) 2; 3+ 2; 1) ; ; 2

    17) = 00 + 0 = 0 .

    18) 200 20 + 2 = 3:) = 2 pi .

    ) .

    : () = 1+ 22 +

    3

    2:

    1

  • 19) 0 = ; (=2) =

    20) = 0 + 2 2 = 1 2 .(: )

    21)

    ) (5 7 + 1) + (+ 1) = 0) (+ + 1)+ (2+ 2 1) = 0

    22) pi ()

    :

    ) 2 + 3 + 2 = 0

    ) ( + 1)( + 2) = 0

    ) (2 + 4)(2 + 9) = 0

    23) pi

    pi :

    ) 1 = 1; 2 = 2

    ) 1 = 3 2; 2 = 3 + 2; 3 =pi2

    ) 1 = 3; 2 = 5; 3 = 7

    ) 1 = 2; 2 = 2; 3 = 2; 4 = 2 + 3; 5 = 2 3; 6 = 3 + 4; 7 = 3 4;8 = 2 + 3; 9 = 2 3; 10 = 3; 11 = 3; 12 = 4

    24) pi :

    ) ; ) 3; 3) 1; ; )

    2; 4 5; 4 5

    pi .

    2

  • Exercise Sheet 18Power Series Solutions about Ordinary Points

    1. Find a general solution for the following differential equations usingpower series about given x0.

    (a) y 4y = 0, x0 = 0.(ans: y = a0

    k=0(2x)2k

    (2k)!+ a1

    k=0(2x)2k+1

    (2k)!)

    (b) y(1x)y = 0,x0 = 1(ans:y = a0[1 +

    k=1(1)k 1.4(3k2)

    (3k)!(x 1)3k

    ]+

    a1[x +

    k=1(1)k 2.5(3k1)

    (3k+1)!(x 1)3k+1

    ])

    (c) y + 2xy 4y = 0, y(0) = 2, y(0) = 0.(ans: y = 2 + 4x2).

    2. Find the general solution valid near the origin and state the region ofvalididty of the solution.(ie. possible interval of convergence)

    (a) (1+4x2)y8y = 0, (ans:y = a0(1+4x2)+a1

    n=0(1)n+1.2n+1.x2n+1

    4n21

    valid for |x| < 1/2)

    (b) (x2+4)y+2xy12y = 0, (ans:y = a0[1 +

    n=13(1)n(n+1)x2n

    22n(2n1)(2n3)

    ]+

    a1(x + 5/12 x3), valid for |x| < 2).

    Good Luck!!!

    1

  • Exercise Sheet 19 Power Series Solutions about Regular Singular

    Points

    1. Locate and classify the Singular points of each D.E.

    (a) x(x 2)2y + xy + y = 0

    (b) (x2 3x)y + (x + 2)y + y = 0

    (c) x3y + (xy) + xy = 0

    2. For each differential equation below find 2 linearly independent solu-tions valid near the origin(x > 0)(Case r1 r2 6=integer)

    (a) 2x(x + 1)y + 3(x + 1)y y = 0 (ans:y1 +

    n=1(1)nxn

    4n21, y2 =

    x1/2 + x1/2)

    (b) 2xy+5(12x)y5y = 0 (ans:y1 = 1+3

    n=15nxn

    n!(2n+1)(2n+3), y2 =

    x3/2 + 10x1/2)

    (c) 3xy + (2 x)y 2y = 0 (ans:y1 =

    n=0(3n+4)xn+1/3

    4.3n.n!, y2 = 1 +

    n=1(n+1)xn

    2.5.8(3n1))

    3. Find a solution for each D.E. below which is valid for x > 0(Caser1 r2 = 0)

    (a) xy+y+xy = 0(Bessels equation of index zero)(ans:y1 =

    n=0(1)nx2n

    22n(n!)2)

    (b) x2y + x(x 1)y + (1 x)y = 0 (ans:y1 = x)

    (c) x2y x(1 + x)y + y = 0 (ans: y1 =

    n=0xn+1

    n!= xex

    4. Find a solution for each D.E. below which is valid for x > 0(Caser1 r2 =integer)

    (a) (non-logarithmic case,i.e. small r gives general solution)

    xy(3+x)y+2y = 0(ans:y = a0(1+2/3x+1/3x2)+a4

    n=424(n3)xn

    n!

    (b) (non-logarithmic case,i.e. small r gives general solution)x(1x)y3y+2y = 0 (Solve this about x = 1(ans:y = a0(x

    3

    3x2 + 9/2x1) + 6a3

    n=3(3)n3xn3

    n!)

    (c) (logarithmic case)

    x2y + x(1 x)y (1 + 3x)y = 0 (ans: y1 =

    n=0(n+3)xn+1

    n!)

    Good Luck!!!

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