# Solution -> Second Order ODE?

1. Mar 28, 2008

### Tom1

Hi, I am trying to decide whether y(t) = c1t^2 + c2 t^−1, where c1 and c2 are arbitrary constants, is the general solution of the diﬀerential equation (t^2)y'' − 2y = 0 for t > 0 and justify the answer, but I don't really know how to approach it from this "side" of the problem.

Any suggestions would be greatly appreciated.

Last edited: Mar 28, 2008
2. Mar 28, 2008

### Hootenanny

Staff Emeritus
To check that the proposed solution is a solution of the ODE, one only needs to substitute the proposed solution into the ODE and verify that the equation holds.

3. Mar 28, 2008

### sutupidmath

Like Hootenanny said, i am just going to give more details:$$y(t)=c_1t^{2}+c_2t^{-1}$$

$$y'(t)=2c_1t-c_2t^{-2},\ \ \ y''(t)=2c_1+2c_2t^{-3}$$

$$t^{2}y''-2y=0$$ Now all you need to do is plug in for y'', and y. That is

$$t^{2}(2c_1+2c_2t^{-3}-2(c_1t^{2}+c_2t^{-1})=.....$$

NOw if this equals 0, then we conclude that :$$y(t)=c_1t^{2}+c_2t^{-1}$$
is a solution of

$$t^{2}y''-2y=0$$

4. Mar 28, 2008

### HallsofIvy

Staff Emeritus
sutupidmath showed that that is a solution to the differenital equation. The fact that it is the general solution follows from the fact that this is a second order linear differential equation- so its solution set is a two dimensional vector space- and the two given functions are independent- and so form a basis for the solution space.