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Math Help - Differential Equation

  1. #1
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    Differential Equation

    Working on second order equations

    The equation is y'' - 6y' + 5y = 0

    I have to find the general solution and the particular solution with the condition y(0) = 0, y'(0) = 4


    Need some help with this problem


    Thanks
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by mathlg View Post
    Working on second order equations

    The equation is y'' - 6y' + 5y = 0

    I have to find the general solution and the particular solution with the condition y(0) = 0, y'(0) = 4


    Need some help with this problem


    Thanks
    y'' - 6y' + 5y = 0

    try y = e^mx

    we get the characteristic equation:

    m^2 - 6m + 5 = 0
    => (m - 5)(m - 1) = 0
    => m = 5 or m = 1

    so our general solution is of the form:

    y = Ae^5x + Be^x .............general solution

    now we have to find the particular solution with the conditions
    y(0) = 0, y'(0) = 4

    now if y = Ae^5x + Be^x
    => y' = 5Ae^5x + Be^x

    since y(0) = 0 and y'(0) = 4, we have the simultaneous equations

    Ae^0 + Be^0 = 0
    5Ae^0 + Be^0 = 4

    that is

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

    => 4A = 4 .............(3) = (2) - (1)
    => A = 1

    but A + B = 0
    => 1 + B = 0
    => B = -1

    so the particular solution is:

    y = e^5x - e^x ...............particular solution

    any questions?
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  3. #3
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    ok I see how you got that. Let me try to do the next one and see if I do it correct.

    This problem is Y'' - 10y' + 25y = 0
    and again I have to fiind the general solution and the particular solution with the condition y(0) = 0, y'(0) = 4


    First step:

    m^2 - 10m + 25m = 0

    => (m - 25)(m + 15) = 0
    => m = 25 or m = -15

    y = Ae^25x + Be^-15x


    Is this correct so far before I go on?????
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by mathlg View Post
    ok I see how you got that. Let me try to do the next one and see if I do it correct.

    This problem is Y'' - 10y' + 25y = 0
    and again I have to fiind the general solution and the particular solution with the condition y(0) = 0, y'(0) = 4


    First step:

    m^2 - 10m + 25m = 0

    => (m - 25)(m + 15) = 0
    => m = 25 or m = -15

    y = Ae^25x + Be^-15x


    Is this correct so far before I go on?????
    I'm afraid not. You seem to have trouble with quadratic eqautions.

    we have m^2 - 10m + 25 = 0

    we want two numbers whose product is +25 and whose sum is -10
    -5 and -5 have this property, since (-5)(-5) = 25 and -5 + (-5) = -10

    so we get
    m^2 - 10m + 25 = 0
    => (m - 5)(m - 5) = 0
    => (m - 5)^2 = 0
    => m = 5
    this is a double root, so we need to add an x to one of the e's

    that is, y = Ae^5x + Bxe^5x

    now continue


    (if you still have problems with solving quadratics, tell me)
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  5. #5
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    Yes I have always been kind of bad with that. I always use the formula but did not that time.

    The general solution is y = Ae^5x + Bxe^5x

    Step two:

    0 = Ae^0 + B0e^0
    A = 0

    Y' = B(e^5x + 5xe^5x)
    =Be^5x( 1 + 5x)
    4 = B

    particular solution is Y = 4xe^5x


    Is this correct?????
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by mathlg View Post
    Yes I have always been kind of bad with that. I always use the formula but did not that time.

    The general solution is y = Ae^5x + Bxe^5x

    Step two:

    0 = Ae^0 + B0e^0
    A = 0

    Y' = B(e^5x + 5xe^5x)
    =Be^5x( 1 + 5x)
    4 = B

    particular solution is Y = 4xe^5x


    Is this correct?????
    yes, you are correct. good job!
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  7. #7
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    Hey thanks for helping me on that...

    You asked me to see if I need help on quadratics.

    On this other question I had it says:

    Find the Particular solution of the equation y' - y = x^2 with the condition
    y(0) = -1


    Do I have to use quadratics on this problem??

    Not really sure what to do or if it is like the others I just did.
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by mathlg View Post
    Hey thanks for helping me on that...

    You asked me to see if I need help on quadratics.

    On this other question I had it says:

    Find the Particular solution of the equation y' - y = x^2 with the condition
    y(0) = -1


    Do I have to use quadratics on this problem??

    Not really sure what to do or if it is like the others I just did.
    this one is a bit different, here we have a first order inhomogeneous differential equation. now that may sound bad, but in differential equations that's just another way of saying "one of the easiest kinds of problems you will get." now this particular one is kind of complicated, and you'll see why, but it could be worst.

    first we need to find an integrating factor. experience will tell you in this case e^-x, so we will multiply through by that. If you have problems with finding integrating factors, tell me. however, my recommendation would be to read your text book, it's kind of complicated to explain if you are new to the concept.

    => (e^-x)y' - (e^-x)y = (e^-x)x^2
    but now we realize, that on the right, we have what we would have gotten if we had differentiated (e^-x)y implicitly using the chain rule, so we have

    d/dx[(e^-x)y] = (e^-x)x^2 ........now we integrate both sides
    => (e^-x)y = int{(e^-x)x^2}dx .....we have to use integration by parts for this

    so int{(e^-x)x^2}dx = (-e^-x)x^2 + 2*int{(e^-x)x}dx .....we have to do this guy by parts again
    .............................= (-e^-x)x^2 + 2[(-e^-x)x + int{e^-x}dx]
    .............................= (-e^-x)x^2 + 2[(-e^-x)x - e^-x] + C
    .............................= (-e^-x)x^2 + 2x(-e^-x) - 2e^-x + C
    .............................=(-e^-x)(x^2 + 2x - 2) + C .......ah, we have a quadratic afterall

    so now let's go back to our equation (told you this one was complicated)

    => (e^-x)y = (-e^-x)(x^2 + 2x - 2) + C
    divide both sides by e^-x, we get
    y = -(x^2 + 2x - 2) + Ce^x .......................general solution

    Now let's find the particular solution:

    y(0) = -1
    => -1 = -(0 + 0 - 2) + Ce^0
    => -1 = 2 + C
    => C = -3

    so y = -(x^2 + 2x - 2) - 3e^x
    => y = -(x^2 + 2x - 2 + 3e^x) ...........particular solution
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