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# 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

- 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)