Determining the properties of solutions of a second order ODE The set of all solutions defined for $t>0$ of one of the following ODE is a two dimensional vector space. Which one
$$y''-7y'+4y=1$$
$$t^2y''-7y'+4y=\sin(t)$$
$$t^2y'-7y=0$$
$$t^2y''-7y'=0$$
$$t^2y''-7y^2=0$$
What i tried
A two dimensional vector space means that the two solutions are linearly independent of each other. So does it mean that i have to solve for all the above equations to find their solutions and then to test for linear independence for each of the solutions that i got. I'm thinking whether is there a better and more efficient method of tackling the above problem, without even needing to solve for the solutions.I believe one method was to determine the Wronskian directly form the above equations and if the Wronskian is not equal to $0$, then we have proved that the solutions are linearly independent.
 A: No, trying to solve all of these ODEs is definitely overthinking the problem. The question is asking you to classify the ODEs and think about what conditions are necessary for the solutions to form a two-dimensional linear space.
Some food for thought:


*

*Does the ODE have to be homogeneous?

*Does it have to be linear?

*What order does it have to be? Or rather, what order can't it be?


and then eliminate the answers that cannot possible work.
EDIT: Let's look for instance at the first ODE. Suppose $y(t)$ is a solution. If the ODE had a two-dimensional linear space of solutions, then $\alpha y(t)$ would have to be a solution for any scalar $\alpha$ (there would also have to exist a second, linearly-independent solution, but that doesn't really matter here). Let's pick $\alpha=2$. Plugging in gives
$$2y''-14y'+8y = 2(y''-7y'+4y) = 2(1) = 2 \neq 1$$
so $2y$ is not a solution. This means that the space of solutions to the first ODE is not linear, and we can rule it out.
See if you can think of the general condition that is needed in order for the space of solutions to be linear (unlike this example).
