calculus help with general equations?

i have two problems i'm struggling with.

1) general equation for 6y''+6y=0

2) general solution for differential equation y'= (11x+7y)/x

any help would be greatly appreciated

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  • 9 years ago
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    1) general equation for 6y''+6y=0

    6y''+6y=0

    6d^2y/dx^2 + 6y = 0

    6m^2 + 6 =0

    6m^2 = -6

    m^2 = -1

    m = ± √(-1)

    m= ± i

    y = e^(0x)[c₁cos(x) + c₂sin(x)]

    y= e(0)[c₁cos(x) + c₂sin(x)]

    y= c₁cos(x) + c₂sin(x) answer//

    2) general solution for differential equation y'= (11x+7y)/x

    y'= (11x+7y)/x

    ............11x......7y

    dy/dx = ------ + ------ = 11 + 7(y/x)

    .............x..........x

    v=y/x

    y= vx

    dy/dx = v + xdv/dx

    then

    v + xdv/dx = 11 + 7v

    v - 7v + xdv/dx = 11

    -6v + xdv/dx = 11

    xdv/dx = 11 + 6v

    xdv =(11 + 6v)dx

    ....dv...............dx

    ∫ ------------- =∫ ---------

    ...11 + 6v..........x

    1/6ln(6v + 11) = ln(x) +C

    ln(6v + 11) = 6[ln(x) + C]

    6v + 11 = 6ln(x) + 6C

    6v + 11 = ln(x)^6 + 6C

    6v + 11 = e^ln(x^6) e^(6C)

    6v + 11 = 6Cx^6

    6(y/x) + 11 = 6Cx^6

    [6y/x + 11 = 6Cx^6] x

    6y + 11x = 6Cx^7

    6y = 6Cx^7 - 11x

    y= Cx^7 - 11/6x answer//

  • 9 years ago

    1) dividing by 6 gives y'' + y =0

    This is a simple harmonic differential equation.

    the solution is y= A cos x + Bsin x where A and B are constants

    2) xy'=11x+ 7y Let y=vx where v is a function of x then y' = v + x v'

    y' = (11x + 7vx)/x =11+7v so v + xv' = 11 + 7v ie xv' = (11 + 6v ) ie x dv = (11 + 6v ) dx

    dv/(11 + 6v) = 1/x dx

    by integration, ln(11+6v)=ln x + ln C where C is constant of integration.

    so 11 + 6v =xC so v= (xC-11)/6 so y= vx= x(xC-11)/6

  • 9 years ago

    Here you are talking about differential equations.

    1) y[x] = C[1] Cos[x] + C[2] Sin[x]

    2)y[x] = 11 (-(1/49) - x/7) + E^(7 x) C[1]

    where C[1],C[2] are constants

    Source(s): wolfram mathematica
  • nashim
    Lv 4
    4 years ago

    a) y'' - 4y' + 3y = 0 The function eqn is r^2 - 4r + 3 = 0 => (r - 3)(r - a million) = 0 r = a million and 3 y = Ae^x + B e^(3x) y (0) = 0 ==> A + B = 0 y ' = Ae^x + 3Be^(3x) y ' (0) = 2 ==> A + 3B = 2 2B = 2 ==> B = a million A = -a million y = e^(3x) - e^(x) b) 4y'' + 8y' + 5y = 0 4r^2 + 8r + 5 = 0 r = [ -8 ± ?(sixty 4 - 80) ] /8 = [ -8 ± ?-sixteen ] /8 r = -a million ± (a million/2)i y = e^(-x) [ A sin(x/2) + B cos(x/2) ] y(0) = 3 ==> a million[ 0 + B] = 3 ==> B = 3 y ' = e^(-x) [ A/2 cos(x/2) - B/2 sin(x/2) ] - e^(-x) [A sin(x/2) + B cos(x/2) ] = e^(-x) [ sin(x/2)(- A - B/2) + cos(x/2)( A/2 + B) ] y ' (0) = 0 ==> a million[ 0 + A/2 + B ] = 0 ==> A/2 + B = 0 ==> A/2 = -3 ==> A = -6 y = e^(-x) (3 cos(x/2) - 6 sin(x/2))

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  • 9 years ago

    1.

    Solve this differential equation using the auxiliary equation:

    6m² + 6 = 0

    m² + 1 = 0

    m² = -1

    m = ±i

    y = Asinx + Bcosx

    2.

    Solve this differential equation using an integrating factor:

    y' = (11x + 7y) / x

    dy / dx = (11x + 7y) / x

    dy / dx = 11 + 7y / x

    dy / dx - 7y / x = 11

    dy / dx + P(x)y = f(x)

    P(x) = -7 / x

    f(x) = 11

    I(x) = ℮^[∫ P(x) dx]

    I(x) = ℮^(∫ -7 / x dx)

    I(x) = ℮^(-7ln|x|)

    I(x) = ℮^[ln(1 / x⁷)]

    I(x) = 1 / x⁷

    I(x)y = ∫ I(x)f(x) dx

    y / x⁷ = ∫ 11 / x⁷ dx

    y / x⁷ = -11 / 6x⁶ + C

    y / x⁷ = C - 11 / 6x⁶

    y = Cx⁷ - 11x / 6

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