- #1
_N3WTON_
- 351
- 3
Homework Statement
Part 1: Show that Euler's equation has solutions of the form [itex]x^r[/itex]. This can be found by obtaining the characteristic equation [itex] ar(r-1) + br + c = 0 [/itex]
Part 2: Solve the following Euler equation: [itex] x^{2}y'' + xy' = 0[/itex]
Homework Equations
Euler's Equation:
[itex] ax^{2}y'' + bxy' + cy = 0 [/itex]
The Attempt at a Solution
For part 1, I didn't have much of a problem, I let [itex] y(x) = x^{r} [/itex]. Therefore:
[itex] y'(x) = rx^{r-1} [/itex]
[itex] y''(x) = r(r-1)x^{r-2} [/itex]
Substituting these values into the Euler equation:
[itex] ax^{2}(r)(r-1)x^{r-2} + bx(r)x^{r-1} + cx^{r} = 0 [/itex]
[itex] ax^{2}[r(r-1)x^{r-2}] + bx[rx^{r-1}] + c[x^{r}] = 0 [/itex]
[itex] a(r^{2}-r)x^{r} + brx^{r} + cx^{r} = 0 [/itex]
[itex] a(r^{2}-r) + br + c = 0 [/itex]
[itex] ar(r-1) + br + c [/itex]
This is the characteristic equation, so that portion of the problem is solved. For part 2, I am not really having a problem formatting the problem, but I am having a problem understanding the given solution. Here is what I did:
[itex] x^{2}y'' + xy' = 0 [/itex]
[itex] r(r-1) + r = 0 [/itex]
[itex] r^{2} - r + r = 0 [/itex]
[itex] r^{2} = 0 [/itex]
So the answer has repeated roots with [itex] r = 0 [/itex]. So the solution given in the back of the book is:
[itex] Y(x) = c_1 + c_{2}ln(x) [/itex]
I looked around online and verified that this is indeed the answer. However, I am having some trouble understanding where the natural log came from, there is no explanation given in my book and I also can't seem to find a derivation online. Probably I am missing something obvious, but I was hoping someone could help me out and give an explanation, thanks.