# Euler's Equations (DiffEq)

1. Nov 2, 2014

### _N3WTON_

1. The problem statement, all variables and given/known data
Part 1: Show that Euler's equation has solutions of the form $x^r$. This can be found by obtaining the characteristic equation $ar(r-1) + br + c = 0$
Part 2: Solve the following Euler equation: $x^{2}y'' + xy' = 0$

2. Relevant equations
Euler's Equation:
$ax^{2}y'' + bxy' + cy = 0$

3. The attempt at a solution
For part 1, I didn't have much of a problem, I let $y(x) = x^{r}$. Therefore:
$y'(x) = rx^{r-1}$
$y''(x) = r(r-1)x^{r-2}$
Substituting these values into the Euler equation:
$ax^{2}(r)(r-1)x^{r-2} + bx(r)x^{r-1} + cx^{r} = 0$
$ax^{2}[r(r-1)x^{r-2}] + bx[rx^{r-1}] + c[x^{r}] = 0$
$a(r^{2}-r)x^{r} + brx^{r} + cx^{r} = 0$
$a(r^{2}-r) + br + c = 0$
$ar(r-1) + br + c$
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:
$x^{2}y'' + xy' = 0$
$r(r-1) + r = 0$
$r^{2} - r + r = 0$
$r^{2} = 0$
So the answer has repeated roots with $r = 0$. So the solution given in the back of the book is:
$Y(x) = c_1 + c_{2}ln(x)$
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.

2. Nov 2, 2014

### RUber

I would look at the assumption you used for the derivatives.
$y' = x^{r-1}$
If r = 0, you either have the trivial solution, or something else that satisfies that assumptions.

3. Nov 2, 2014

### _N3WTON_

Thanks for the advice...I was wondering, could I also try to derive that equation using reduction of order, similar to how the solution $Y(x) = c_{1}e^{\frac{-bx}{2a}} + c_{2}xe^{\frac{-bx}{2a}}$ is derived?

4. Nov 2, 2014

### LCKurtz

Here's an argument that shows you where the $\ln x$ comes from. For simplicity I will call your differential equation $ax^2y''+bxy'+cy = 0$ as $L(y)=0$ where $L$ is the differential operator. When you substitute $y = x^r$ into that equation you got $L(x^r) = x^rp(r)$ where $p(r)$ is the characteristic polynomial. Now consider the case where $r$ is a double root, as you have in your problem. $r$ being a double root means that $p(r)=0$ and $p'(r)=0$. Now let's differentiate $L(x^r) = x^rp(r)$ with respect to $r$. If we differentiate the left side, since the order of taking derivatives doesn't matter we get$$\frac \partial {\partial r}L(x^r) = L(\frac \partial {\partial r}x^r) = L(x^r\ln x)$$If we differentiate the right side with respect to $r$ we get$$(x^r \ln x) p(r) + x^r p'(r) = 0$$since $r$ is a double root. So putting this together we have$$L(x^r\ln x) = 0$$which tells you $x^r\ln x$ is a solution of the DE in addition to $x^r$ when $r$ is a double root.

5. Nov 2, 2014

### _N3WTON_

Fantastic...thank you so much for showing me that. Surprisingly, that argument is hard to find online and I was having some trouble performing it by myself..