Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.

Anonymous asked in Science & MathematicsMathematics · 1 decade ago

Anyone whos good at Calculus look here please!!?

FInd one set of values of the numbers a and b, such that the function y=e^ax + b ln(x) is a solution of the differential equations given below.

a. y'' - ay' = 0

b. y' + y - b/x = 0

1 Answer

  • hfshaw
    Lv 7
    1 decade ago
    Favorite Answer

    An odd question, but one that can be answered.

    If y = exp(ax) + b*ln(x)

    y' = a*exp(ax) + b/x

    y'' = (a^2)*exp(ax) - b/x^2

    Plugging these into the first differential equation gives:

    (a^2)*exp(ax) - b/x^2 - a*(a*exp(ax) + b/x) = 0

    -b/x^2 - a*b/x = 0

    For this equation to be true, the coefficients of each power of x must be identically equal to zero, so we know that b = 0. Because b = 0, though, we can't determine a value for a.

    We now know that:

    y = exp(ax) and y' = a*exp(ax). Plugging these into the second differential equation yields:

    a*exp(ax) + exp(ax) - 0/x = 0

    (a+1)*exp(ax) = 0

    The factor exp(ax) can never equal zero for any finite a or x, so a = -1

    The pair of numbers (a,b) that makes the function a solution to both differential equations is (-1,0), and the function ends up being:

    y(x) = exp(-x)

Still have questions? Get your answers by asking now.