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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
- hfshawLv 71 decade agoFavorite 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)