To answer this question, we just need to confirm the first two functions of `x`. The third will follow naturally from the linear-operator nature of the derivative.
Let's confirm it works first for `y(x) = x^2`. Let's first calculate the second derivative:
`y'(x) = 2x`
`y''(x) = 2`
Now, we substitute the terms `x^2` for `y` and `2` for `y''`.
`x^2(2) - 2(x^2) = 0`
`2x^2-2x^2 = 0`
That looks right. We just confirmed that `x^2` is a solution to the D.E.
Now, let's confirm for `y(x) = x^-1=1/x`. Our derivatives will be:
`y'(x) = -1/x^2`
`y''(x) = 2/x^3`
Substituting the above functions into the D.E.:
`x^2(2/x^3) - 2(1/x) = 0`
`2/x - 2/x = 0`
This looks good, too! Now, based on the linear nature of the derivative (see link below), we might just go ahead and say that `c_1x^2 + c_2x^-1` is a general homogeneous solution for the D.E. However, let's just confirm it for fun. Let's have `y_1 = x^2` ad `y_2 = x^-1`.
Let's let `y = c_1x^2 + c_2x^-1 = c_1(y_1)+c_2(y_2)`
Now, let's put this equation into the differential equation and take advantage of the lineararity of the derivative:
`x^2y'' - 2y = 0`
`x^2(d^2/(dx)^2)(c_1y_1+c_2y_2) - 2 (c_1y_1+c_2y_2)=0`
Now, here's where the linearity comes into play:
`x^2(c_1y_1'' + c_2y_2'') - 2c_1y_1 - 2c_2y_2 =0`
`x^2c_1y_1'' + x^2c_2y_2'' - 2c_1y_1-2c_2y_2 = 0`
`c_1(x^2y_1'' - 2y_1) + c_2(x^2y_2'' - 2y_2) = 0`
based on what we showed before for `y_1` and `y_2`, we know that the full terms inside the parentheses are 0:
`c_1*0 + c_2*0 = 0`
Therefore, the linear combination of the two functions is also a solution to the D.E.!
Hope that helps!