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Euler equations having double roots as a solution

Dhamnekar Winod

Active member
Nov 17, 2018
If the Euler equations have double roots as it's solution, second solution will be $y_2(x)=x^r\ln{x}$. what is its proof? or how it can be derived?
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Indicium Physicus
Staff member
Jan 26, 2012
I believe variation of parameters is the usual proof method.

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
The change of variable, $u= \ln(x),$ converts an "Euler type equation" (also known as an "equipotential equation") to a differential equation with constant coefficients. If [tex]ax^2\frac{d^2y}{dx^2}+ bx\frac{dy}{dx}+ cy= 0[/tex] then, with [tex]u= \ln(x)[/tex], [tex]\frac{dy}{dx}= \frac{dy}{du}\frac{du}{dx}= \frac{1}{x}\frac{dy}{du}[/tex] and [tex]\frac{d^2y}{dx^2}= \frac{d}{dx}\left(\frac{1}{x}\frac{dy}{du}\right)= -\frac{1}{x^2}\frac{dy}{du}+ \frac{1}{x}\frac{d}{dx}\frac{dy}{du}= -\frac{1}{x^2}\frac{dy}{du}+ \frac{1}{x^2}\frac{d^2y}{du^2}[/tex].

So [tex]ax^2\frac{d^2y}{dx^2}+ bx\frac{dy}{dx}+ cy= a\frac{d^2y}{du^2}- a\frac{dy}{du}+ b\frac{dy}{du}+ cy= a\frac{d^2y}{du^2}+ (b- a)\frac{dy}{du}+ cy= 0[/tex].

The characteristic equation for that constant-coefficients equation is the same as for the Euler-type equation so both have the same characteristic values. In particular, if the characteristic equation has a double root, r, then the constant-coefficients equation has the general solution [tex]y(u)= Ae^{ru}+ Bue^{ru}[/tex]. Since $u= \ln(x)$ the general solution in terms of $x$ becomes [tex]y(x)= Ae^{r \ln(x)}+ B \ln(x) e^{r \ln(x)}= A e^{\ln(x^r)}+ B \ln(x) e^{\ln(x^r)}= Ax^r+ B\ln(x) x^r.[/tex]
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