# differencial equation

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• Jan 10th 2009, 03:35 AM
KayPee
differencial equation
Solve x^2y"-4xy'+6y=0 to obtain y=

kindly assist me with the full answer and some explanation.

Thanks in advance
• Jan 10th 2009, 04:02 AM
HallsofIvy
That is an "equi-potential" or "Euler type" equation: each derivative is multiplied by a power of x, the degree the same as the order of the derivative.

If you make the substitution x= ln t will convert it to a linear differential equation with constant coefficients for y as a function of t.

For simple equation, you can try a "trial" solution of the form $y= x^r$ for some constant r.

If $y= x^r$, then $y'= rx^{r-1}$ and $y''= r(r-1)x^{r-2}$. Putting those into this equation,

$x^2y''-4xy'+6y= (x^2)r(r-1)x^{r-2}- 4x(r)x^{r-1}+ 6x^r$= $x^r(r^2- r- 4r+ 6)= x^2(r^2- 5r+ 6)= 0$.

In order for that to be 0 for all x, we must have the "characteristic equation", $r^2- 5t+ 6= (r- 3)(r- 2)= 0$ so r= 3 and r= 2.
Both $x^3$ and $x^2$ are solutions to this equation. Since it is a linear homogeneous equation, the general solution is $y(x)= Cx^2+ Dx^3$ where C and D can be any constants.
• Jan 10th 2009, 07:32 AM
KayPee
Thanks
Thanks so much.

I'll try and understand it.