Solve x^2y"-4xy'+6y=0 to obtain y=

kindly assist me with the full answer and some explanation.

Thanks in advance

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- Jan 10th 2009, 03:35 AMKayPeedifferencial 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 AMHallsofIvy
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 $\displaystyle y= x^r$ for some constant r.

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

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

In order for that to be 0 for all x, we must have the "characteristic equation", $\displaystyle r^2- 5t+ 6= (r- 3)(r- 2)= 0$ so r= 3 and r= 2.

Both $\displaystyle x^3$ and $\displaystyle x^2$ are solutions to this equation. Since it is a linear homogeneous equation, the general solution is $\displaystyle y(x)= Cx^2+ Dx^3$ where C and D can be any constants. - Jan 10th 2009, 07:32 AMKayPeeThanks
Thanks so much.

I'll try and understand it.