How to derive Euler DE for n=0

[yungman]

$x^2 y'' + x y' + n^2 y = 0 \;$ Is Euler equation and solution is $y=x^m$

I understand the three cases with different solution. But my question is if n=0.

If I use $y=x^m \;\Rightarrow\; m(m-1)+m=0 \;\Rightarrow m^2 = 0 \;\Rightarrow m=0$

That would not work. I know the answer is $y=C_1 ln(x) + C_2$

Can anyone show me or give me hint how to derive this, I can't find it in my book.

Thanks

 

[yungman]

I got it, I tried to delete this post and can't. Moderator, please delete the entire thread for me.

Thanks

 

[Office_Shredder]

Just as a general rule of thumb, any time you get a solution of x⁰ that isn't really a solution, you should check to see if ln(x) is a solution. ln(x) differentiates with respect to the power rule in the way that x⁰ is "supposed to" (i.e. give x^-1)

 

[yungman]

Just as a general rule of thumb, any time you get a solution of x⁰ that isn't really a solution, you should check to see if ln(x) is a solution. ln(x) differentiates with respect to the power rule in the way that x⁰ is "supposed to" (i.e. give x^-1)

Yeh, right after I posted, I remember getting the second solution by reduction of order!

 

[HallsofIvy]

Note, by the way, that the euler type equation, $a_nx^n y^{(n)}+ \cdot\cdot\cdot+ a_1x y'+ a_0y= f(x)$
can be converted to an equation with constant coefficients by substituting t= ln(x) as independent variable.

Since, if m is a double root of the characteristic equation for an equation with constant coefficients, the solution involves both $e^{mt}$ and $te^{mt}$, solutions for the Euler type equation will involve $e^{mln(x)}= e^{x^m}= x^m$ and $ln(x)x^m$.