Solve the second order Differential equation xy” + y’ = 0; where y1 = ln x is a solution.?

given that the solution is Y=ln x * v(x)

Find V(x)

2 Answers

  • xy'' + y' = 0

    y1 = lnx

    Y = v(x) * y1

    Y = v ln(x) ( i wanna make it simpler but keeping in mind that v is a function of x)

    dY/dx = v (1/x) + ln(x) *dV/dx

    Y' = v/x + v'ln(x)

    Y'' = 2v'/x - v/x² + v''ln(x)

    replace in the original diff equa; xy'' + y' =- 0

    xy'' + y' = 0

    x{2v'/x - v/x² + v''ln(x) } + v/x + v'ln(x) = 0

    2v' - v/x + xv''ln(x) + v/x + v'ln(x) = 0

    2v' + xv''ln(x) + v'ln(x) = 0

    let u = v'

    ```````````````` keeping in mind that u = u(x)

    u' = v''

    2v' + xv''ln(x) + v'ln(x) = 0

    2u + xu' ln(x) + uln(x) = 0

    xu'lnx = -ulnx - 2u

    `````````````````````````` separable

    u' (xlnx) = -u(2 + lnx)

    u' / u = -(2 + lnx) / (xlnx)

    (1/u) du/dx = -(2 + lnx) / (xlnx)

    (1/u) du = -(2 + lnx) / (xlnx) dx

    integrate both sides to get

    ∫ (1/u) du = -∫ (2 + lnx) / (xlnx) dx

    u = C / (xln²x)

    but u = v'

    so

    v(x) = A + B/lnx

  • Like whitesox suggested, there comes a factor the position you purely can't get a closed form answer. The (3x^2)^(x^2) is amazingly difficult to combine. yet the position did a majority of these human beings come from. you do no longer have any contacts.

Leave a Reply

Your email address will not be published. Required fields are marked *

Related Posts