# How to solve this problem? If , `L(y(x))=ax^2 y'' + bxy' + cy` , x>0, where a,b,c are constants, compute `L(ln(x))`

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This equation is actually a common form of differential equation, called a Cauchy-Euler equation.

Here, though, the differential equation is treated as an operator on a function. In order to find the result, we need to figure some things out.

We are given:

` y (x) = ln (x)`

We now need to calculate the first and second derivative to fit the operator's action on the function. We then know these derivatives are as follows:

`y' (x) = 1 / x`

`y'' (x) = -1 / x^2`

Now, to compute `L(ln(x))` we just substitute the values we have above for y, its first derivative, and its second derivative:

`L(ln(x)) = ax^2*-1/x^2 + bx*1/x + clnx`

`L(ln(x)) = -ax^2/x^2 + bx/x + clnx`

Because x is strictly greater than zero, we can cancel all of the x terms in the fractions. We can now simplify:

`L(ln(x)) = -a + b +clnx`

**And there is your answer.** You unfortunately can't simplify much more.

If you wanted to have fun and get everything into a single logarithm you could do the following:

`-a+b+clnx = lne^-a + lne^b +lnx^c = ln(e^(b-a)x^c)`

Again, just for fun! I think the first answer is likely the best.

I hope all that helped you out!