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    Solutions to Various First Order ODEs

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    Solve using integrating factor and Bernoullis.

    © BrainMass Inc. brainmass.com June 1, 2020, 10:38 pm ad1c9bdddf
    https://brainmass.com/math/integrals/solutions-various-first-order-odes-430879

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    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, ...

    Solution Summary

    We solve various types of first order ordinary differential equations.

    $2.19

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