Explore BrainMass

Explore BrainMass

    Solutions to Various First Order ODEs

    Not what you're looking for? Search our solutions OR ask your own Custom question.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    Solve using integrating factor and Bernoullis.

    © BrainMass Inc. brainmass.com December 24, 2021, 9:59 pm ad1c9bdddf
    https://brainmass.com/math/integrals/solutions-various-first-order-odes-430879

    Attachments

    SOLUTION This solution is FREE courtesy of BrainMass!

    1. We have

    x^2 y' + xy = 1.

    Dividing both sides by x^2 we obtain

    y' + y/x = 1/x^2.

    This is a linear ODE with integrating factor given by

    mu(x) = exp(integral(dx/x)) = e^(ln x) = x.

    Multiplying both sides by the integrating factor we obtain

    xy' + y = 1/x
    (xy)' = 1/x.

    Integrating both sides we obtain

    xy = ln x + C.

    Finally, dividing both sides by x we obtain the solution

    y = (ln x + C)/x.

    2. We have

    cos x dy/dx + y sin x = 1.

    Dividing both sides by cos x, we obtain

    dy/dx + y tan x = sec x.

    Once again, this is a linear ODE. The integrating factor is given by

    mu(x) = exp(integral(tan x dx))
    = exp(ln(sec x))
    = sec x.

    Multiplying both sides by mu(x) we obtain

    sec x dy/dx + y tan x sec x = sec^2 x
    (y sec x)' = sec^2 x.

    Integrating both sides we obtain

    y sec x = tan x + C.

    Finally, multiplying both sides by cos x, we obtain the solution

    y = sin x + C cos x.

    3. We have

    y dx - 4(x + y^6) dy = 0.

    We look for an integrating factor mu(y) which makes the above equation exact, i.e.

    (1) y mu(y) dx - 4(x + y^6) mu(y) = 0

    should be exact. We let

    M(x,y) = y mu(y)
    N(x,y) = -4(x + y^6) mu(y)

    We have

    M_y = y mu'(y) + mu(y)
    N_x = -4 mu(y)

    Now in order for (1) to be exact, we need M_y = N_x, i.e.

    y mu'(y) + mu(y) = -4 mu(y)
    y mu'(y) = -5 mu(y)

    Dividing both sides by y mu(y), we obtain

    mu'(y)/mu(y) = -5/y
    (ln mu(y))' = -5/y

    Integrating both sides we obtain

    ln mu(y) = -5 ln y

    Finally, exponentiating both sides we obtain

    mu(y) = 1/y^5.

    Plugging this integrating factor into (1), we obtain

    (2) 1/y^4 dx - 4(x/y^5 + y) dy = 0.

    Thus we have

    M(x,y) = 1/y^4
    N(x,y) = -4x/y^5 - 4y

    We know (2) is exact, so its solution is given by

    C = f(x,y) = integral(M(x,y) dx)
    = x/y^4 + g(y)

    for some function g(y). We have

    f_y = N(x,y)
    -4x/y^5 + g'(y) = -4x/y^5 - 4y

    whence

    g'(y) = -4y
    g(y) = -2y^2

    Therefore the solution to the differential equation is

    x/y^4 - 2y^2 = C.

    4. We have

    L di/dt + Ri = E

    with initial condition i(0) = i_0.

    Dividing both sides by R, we obtain

    di/dt + R/L i = E/L.

    This is a linear ODE with integrating factor given by

    mu(t) = exp(integral(R/L dt))
    = e^(Rt/L).

    Multiplying both sides by mu(t), we obtain

    e^(Rt/L) di/dt + R/L i(t) e^(Rt/L) = E/L e^(Rt/L)
    d/dt(i(t) e^(Rt/L)) = E/L e^(Rt/L).

    Integrating both sides, we obtain

    i(t) e^(Rt/L) = E/R e^(Rt/L) + C.

    Finally, dividing both sides by e^(Rt/L), we obtain the general solution

    i(t) = E/R + C e^(-Rt/L).

    Plugging in the initial condition we obtain

    i(0) = E/R + C = i_0

    whence

    C = i_0 - E/R.

    Therefore the solution to the differential equation with the initial condition is

    i(t) = E/R + (i_0 - E/R) e^(-Rt/L).

    5. We have

    (x + 1) dy/dx + y = ln x

    with initial condition

    y(1) = 10.

    Dividing both sides by x + 1, we obtain

    dy/dx + y/(x+1) = (ln x)/(x+1).

    This is a linear ODE with integrating factor given by

    mu(x) = exp(integral(dx/(x+1)))
    = exp(ln(x+1))
    = x + 1.

    Multiplying both sides by the integrating factor, we obtain

    (x + 1) dy/dx + y = ln x
    ((x + 1) y)' = ln x.

    Integrating both sides we obtain

    (x + 1) y = x ln x - x + C

    whence

    y(x) = (x ln x - x + C)/(x + 1).

    Plugging in the initial condition we obtain

    10 = y(1) = (C - 1)/2

    whence

    20 = C - 1
    21 = C.

    Therefore the solution satisfying the initial condition is

    y(x) = (x ln x - x + 21)/(x + 1).

    6. We have

    dy/dx = y(xy^3 - 1)

    which we may write as

    dy/dx + y = xy^4.

    This is a Bernoulli ODE with n = 4. To solve it, we first make the change of variables

    w = 1/y^3

    whence

    y = w^(-1/3)
    y' = -1/3 w^(-4/3) w'

    so our differential equation becomes

    -1/3 w^(-4/3) w' + w^(-1/3) = xw^(-4/3).

    Multiplying both sides by -3w^(4/3), we obtain the linear ODE

    w' - 3w = -3x.

    The integrating factor for this ODE is given by

    mu(x) = exp(integral(-3 dx)) = e^(-3x).

    Multiplying both sies by mu(x), we obtain

    (w' - 3w) e^(-3x) = -3x e^(-3x)
    (w e^(-3x))' = -3x e^(-3x).

    Integrating both sides, we obtain

    w e^(-3x) = (x + 1/3)e^(-3x) + C.

    whence

    w = 1/y^3 = x + 1/3 + C e^(3x).

    so

    y = (x + 1/3 + C e^(3x))^(-1/3).

    7. We have

    x^2 dy/dx - 2xy = 3y^4

    with initial condition y(1) = 1/2.

    Dividing both sides by x^2 we obtain

    dy/dx - 2y/x = 3y^2.

    This is a Bernoulli ODE with n = 2, so we make the substitution

    w = 1/y

    whence

    y = 1/w
    y' = -w'/w^2

    and our ODE becomes

    -w'/w^2 - 2/(wx) = 3/w^2.

    Multiplying both sides by -w^2 we obtain

    w' + 2w/x = -3.

    This is a linear ODE with integrating factor

    mu(x) = exp(integral(2 dx/x)) = e^(2 ln x) = x^2.

    Thus we have

    (wx^2)' = -3 x^2
    wx^2 = -x^3 + C
    w = 1/y = -x + C/x^2
    y = 1/(C/x^2 - x)
    = x^2/(C - x^3).

    Plugging in the initial condition we have

    1/2 = y(1) = 1/(C - 1)

    whence C = 3. Thus the solution to the ODE satisfying the initial condition is

    y(x) = x^2/(3 - x^3).

    8. We have

    xy(1 + xy^2) dy/dx = 1

    with initial condition y(1) = 0.

    We make the substitution

    u = xy^2

    whence

    du/dx = y^2 + 2xy dy/dx

    so

    xy dy/dx = 1/2(du/dx - y^2)
    = 1/2(du/dx - u/x).

    Thus we have

    1/2(1 + u)(du/dx - u/x) = 1,

    whence

    du/dx - u/x = 2(1 + u)
    du/dx - u(2 + 1/x) = 2.

    This is a linear ODE with integrating factor

    mu(x) = exp(integral((-2 - 1/x)dx))
    = exp(-2x - ln x)
    = e^(-2x)/x.

    Thus we have

    (ue^(-2x)/x)' = 2e^(-2x)/x
    ue^(-2x)/x = 2 integral(e^(-2x)/x dx) = 2 Ei(2x) + C,

    where Ei is the exponential integral function. Thus we have

    u(x) = xy^2(x) = (2 Ei(2x) + C)x e^(2x)

    whence

    y(x) = e^x sqrt(2 Ei(2x) + C).

    Plugging in the initial condition, we have

    0 = y(1) = e sqrt(2 Ei(1) + C),

    whence C = -2 Ei(1) and the solution to the ODE satisfying the initial condition is

    y(x) = e^x sqrt(2(Ei(2x) - Ei(1)).

    References:

    http://en.wikipedia.org/wiki/Bernoulli_differential_equation
    http://en.wikipedia.org/wiki/Exponential_integral

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 9:59 pm ad1c9bdddf>
    https://brainmass.com/math/integrals/solutions-various-first-order-odes-430879

    ADVERTISEMENT