1. ## Euler equation

Hi. I have this problem, which says: The equation $x^2y''+pxy'+qy=0$ (p and q constants) is called Euler equation. Demonstrate that the change of variable $u=\ln (x)$ transforms the equation to one at constant coefficients.

I haven't done much. I just normalized the equation: $y''+\displaystyle\frac{p}{x}y'+\displaystyle\frac{ q}{x^2}y=0$

Then $P(x)=\displaystyle\frac{p}{x}$ and $Q(x)=\displaystyle\frac{q}{x^2}$

What should I do now? I thought instead of doing $x=e^u$, then $y''+\displaystyle\frac{p}{e^u}y'+ \displaystyle\frac{q}{e^{2u}} y=0$ may be this is the right way, cause it seems more like following the problem suggestion.

2. Originally Posted by Ulysses
Hi. I have this problem, which says: The equation $x^2y''+pxy'+qy=0$ (p and q constants) is called Euler equation. Demonstrate that the change of variable $u=\ln (x)$ transforms the equation to one at constant coefficients.

I haven't done much. I just normalized the equation: $y''+\displaystyle\frac{p}{x}y'+\displaystyle\frac{ q}{x^2}y=0$

Then $P(x)=\displaystyle\frac{p}{x}$ and $Q(x)=\displaystyle\frac{q}{x^2}$

What should I do now? I thought instead of doing $x=e^u$, then $y''+\displaystyle\frac{p}{e^u}y'+ \displaystyle\frac{q}{e^{2u}} y=0$ may be this is the right way, cause it seems more like following the problem suggestion.
$x = e^u$

Thus
$dx = e^u~du$

$\frac{d}{dx} = e^{-u}\frac{d}{du}$

So
$y' = \frac{d}{dx}y = e^{-u} \frac{d}{du}y$

Now to find y'' we apply the differential operator twice:
$y'' = \frac{d}{dx} \left ( \frac{d}{dx} \right ) y = \left ( e^{-u} \frac{d}{du} \left ( e^{-u} \frac{d}{du} \right ) \right ) y$

Can you take it from here? (Remember to use the product rule!)

-Dan

Edit: This is just about the last step. Once we get this we can plug it into the equation, which then simplifies enormously.

3. I'll try, thank you!

4. I'm a bit confused with so many operators

$y'' = \frac{d}{dx} \left ( \frac{d}{dx} \right ) y = \left ( e^{-u} \frac{d}{du} \left ( e^{-u} \frac{d}{du} \right ) \right ) y=\left ( e^{-u} \left ( -e^{-u}\frac{d}{du}+e^{-u}\frac{d^2}{d^2u} \right ) \right ) y$

5. Originally Posted by Ulysses
I'm a bit confused with so many operators

$y'' = \frac{d}{dx} \left ( \frac{d}{dx} \right ) y = \left ( e^{-u} \frac{d}{du} \left ( e^{-u} \frac{d}{du} \right ) \right ) y=\left ( e^{-u} \left ( -e^{-u}\frac{d}{du}+e^{-u}\frac{d}{du} \right ) \right ) y$
Close:
$y'' = \frac{d}{dx} \left ( \frac{d}{dx} \right ) y = \left ( e^{-u} \frac{d}{du} \left ( e^{-u} \frac{d}{du} \right ) \right ) y=\left ( e^{-u} \left ( -e^{-u}\frac{d}{du}+e^{-u}\frac{d^2}{du^2} \right ) \right ) y$

So
$y'' = -e^{-2u}\frac{dy}{du} + e^{-2u}\frac{d^2y}{du^2}$

and
$y' = e^{-u}\frac{dy}{du}$

Just to be clear the expressions on the LHS, y' and y'', are in terms of x ie y'(x) and y''(x) whereas the RHS is describing the derivatives in terms of u.

Now change x into $e^{u}$ in the differential equation and plug in your new expessions for y' and y''. What does the new equation look like?

-Dan

6. $\frac{d^2y}{du^2}+\frac{dy}{du}(p-1)+qy=0$ nice, so it has constant coefficients in u, right?

Thank you very much sir

7. Originally Posted by Ulysses
$\frac{d^2y}{du^2}+\frac{dy}{du}(p-1)+qy=0$ nice, so it has constant coefficients in u, right?

Thank you very much sir
Yup.
-Dan