Solving second order IVP I want to solve this IVP but the general solution should be the linear combination of $~y_1,~y_2~$ and $~y_3~$, but I only have $~2~$ initial data,how can I solve for $~3~$ unknown ?

And $~y_1, ~y_2,~ y_3~$ are linearly independent on $~(-\infty,1)~$,is it impossible to find $~2~$ vectors to span $~y_1, ~y_2,~ y_3~$ ?

thanks!!!
Consider the homogeneous equation
$$y''(t) + p_1(t)y' + p_2(t)y = 0$$
Then by subtraction of the (linear) inhomogeneous equations, we see that $y_2(t) - y_1(t) = 2e^t$ and $y_3(t) - y_2(t) = 2e^t - t$ solve this homogeneous equation.
Subtracting again we see that $t$ solves the equation, so that $e^t$ and $t$ are independent solutions to the homogeneous equation. Hence the general solution to the homogeneous equation is given by $Ae^t + Bt$ where $A$ and $B$ are arbitrary constants.

The solution to your original inhomogeneous equations is therefore $y_p(t) + Ae^t + Bt$ where $y_p(t)$ is a particular solution. Looking at any of the three given solutions we see $y_p(t) = e^{t^2}$, so the general solution to this equation is given by
$$e^{t^2} + Ae^t + Bt$$
Here $A$ and $B$ are arbitrary constants. If $y(0) = 1$ and $y'(0) = 2$, we plug in and see
$$1 + A = 1$$
$$A + B = 2$$
Hence $A = 0$ and $B = 2$ and we have the solution
$$e^{t^2} + 2t$$
Each of the given solutions has an $e^t$ and an $e^{t^2}$. Thus, the general solution $y(t)$ has "$c_1 e^t + c_2 e^{t^2}$" in it where $c_1, c_2 \in \mathbb{R}$.

The third solution has $-t$ in it. Thus, the general solution $y(t)$ has a polynomial in it call it $At + B$ where $A, B \in \mathbb{R}$ (Q: Why did I choose degree 1?).

There is a first derivative and second derivative. I expect terms of the form $te^{t^2}$ and $t^2 e^{t^2}$. Thus, the general solution $y(t)$ has "$c_3 te^{t^2} + c_4 t^2 e^{t^2}$" in it where $c_3, c_4 \in \mathbb{R}$.

Thus, the general solution is $y(t) = c_1 e^t + c_2 e^{t^2} + c_3 te^{t^2} + c_4 t^2 e^{t^2} + At + B$.

Back these up with propositions in your textbook like linear combinations of solutions are solutions, etc.

Other stuff: If one of the solutions had a $\sin(t)$ in it, then the general solution would have $c_1 \sin(t) + c_2 \cos(t)$ in it.

Note that these coefficients may be zero. So, the general solution could lack a $e^t$ in the sense that its coefficient is zero.