Question

Find a general solution to the differential x^2y''-2y=5

Answer 1

Characteristic equation for general solution of it is \(r^2 -2=0\), hence \(r=\pm\sqrt2\)
Therefore, the general solution is \(y= C_1e^{\sqrt2}+C_2e^{-\sqrt2}\)

Answer 2

@OOOPS the coefficients aren't constant, we can't extract a characteristic equation.

Answer 3

What we CAN do, though, since this is an equation of Euler-Cauchy form, is set \(y=x^r\). Then \(y''=r(r-1)x^{r-2}\), so we get the associated homogeneous equation
\[r(r-1)x^r-2x^r=0~~\implies~~r(r-1)-2=0~~\implies~~r=-1,2\]
so we find two general solutions \(y=\dfrac{C_1}{x}+C_2x^2\).

You can determine the particular solution to the nonhomogenous equation via variation of parameters since you know two of the fundamental solutions.

Answer 4

Thanks for making it clear @SithsAndGiggles