# Thread: General Solution of this ODE

1. ## General Solution of this ODE

Hi,

I just need to know where to start in solving this ODE. I know it has something to do with the Method of Undetermined Coefficients, but I'm really stuck. The question is as follows:

Find the general solution of:

$x^2y'' + 2xy' + 2y = x^2 + Bx + C$

where A and B are constants.

Thanks a lot!

2. This a Cauchy-Euler Equation

which is

$ar^2 + (b-a)r + c =0$

$r^2 + r +2 = 0$

so

$r = -\frac{1}{2} \pm \frac{\sqrt{7}}{2}i$

$y = Ax^{-\frac{1}{2}}\cos{\frac{\sqrt{7}}{2}} + Bx^{-\frac{1}{2}}\sin{\frac{\sqrt{7}}{2}}$

3. From my understanding a Cauchy-Euler equation of the 2nd order has the form: Ay''+By+C=0 which is certainly not the case here. How do you arrive at that solution?

4. You "understanding" is completely wrong. Ay"+ By+ C= 0 is a special case of the "constant coefficients" equation with Ay"+ 0y'+ By= -C. The general Cauchy-Euler equation of the 2nd order is $Ax^2y"+ Bxy'+ Cy= f(x)$".

You can solve the homogeneous equation as Super Member (and he certainly is!) indicates and then try a "specific solution" of the form y= ux^2+ vx+ w for u, v, and w constants to be determined.

More generally, the substitution t= ln(x) converts a "Cauchy-Euler" equation to an equation with constant coefficients.