Solve (t^2-1)y'' +4ty'+2y=6t, given two particular solutions

[CGandC]

Homework Statement

Solve the DE $\left(t^2-1\right) \ddot{y}+4 t \dot{y}+2 y=6 t$ if its two particular solutions are $y_1=t$ and $y_2=\frac{t^2+t+1}{t+1}$.

Relevant Equations

method of variation of parameters ( possibly utilizing here Wronskian of linear non-homogeneous differential equation of second order order ).

I find a solution in math.exchange site: https://math.stackexchange.com/ques...neral-solution-given-two-particular-solutions

The way I thought about solving the problem is to somehow use the two particular solutions to generate a homogeneous solution, I couldn't figure out how to do so; but, the solutions proposed in the link above leave me with an uneasiness of mind because:

1. The first answer- i.e. that the linear combination of the particular solutions is one fundamental homogenous solution ( in which case, I can use the wronskian to find the other fundamental homogenous solution and then we're finished ) is a little bit tricky because how was I supposed to know that the sum of particular solutions may generate a homogenous solution? is there any theorem backing this?

2. The second answer - i.e. using the substitution $u = (t+1)x$ will enable me to find the homogenous solution regardless of the given information about two particular solutions ( but i'll need information about one particular solution to generate a general solution ), so how does the knowledge of two particular solutions help me here? ( I only need one )

Thanks for the help!

 

[pasmith]

1. The first answer- i.e. that the linear combination of the particular solutions is one fundamental homogenous solution ( in which case, I can use the wronskian to find the other fundamental homogenous solution and then we're finished ) is a little bit tricky because how was I supposed to know that the sum of particular solutions may generate a homogenous solution? is there any theorem backing this?

This is a result from the general theory of linear maps: If $\alpha(x) = \alpha(y) = b$ for linear $\alpha$ then $$\alpha (x - y) = \alpha(x) - \alpha(y) = b - b = 0.$$

 

[CGandC]

This is a result from the general theory of linear maps: If $\alpha(x) = \alpha(y) = b$ for linear $\alpha$ then $$\alpha (x - y) = \alpha(x) - \alpha(y) = b - b = 0.$$

Ah, I see that now, in our case, the linear map is the differentiation operator over vector space of smooth functions over reals? ( because we wish to find such a smooth general solution )