show for no continuous coefficient functions, these two can be solutions of ODE for equation $y^"+ p(t) y^{'}+q(t)y=0$, I want to show there are no continuous functions $p(t)$ and $q(t)$ s.t both $y_1(t)=e^t$ and $y_2(t)=t^2$ are solutions of the equation.
I began with the idea that the linear combination of the two solutions above is a solution of the equation. so $y(y)=c_1e^t+c_2t^2$ is a solution. I then took the first and second derivative and substitute in general equation. but it ended up with no particular form to help answer this question. then I believe my initial idea is not correct in the first place. 
 I also thought of taking the wronskin of two solutions, which is $e^t.t.(2-t)$ which can be non zero. so they can be the solutions of the equation!
would you please help me  with this question . what should I try?
 A: We may exploit the hypothesized known form of the solutions to investigate what form the coefficients of the differential equation must take, and on which open intervals they are defined, and in so doing show that, while there exists no globally defined ordinary differential equation on all of $\Bbb R$ having $e^t$ and $t^2$ as solutions, there are regions where such equations do in fact exist.  To wit:                                                                                                                                                                                                           
Given the equation
$y''(t) + p(t) y'(t) + q(t)y(t) = 0 \tag 0$
with the solution
$y = e^t, \tag{0.5}$
we find upon substituting (0.5) into (0)
$e^t + p(t)e^t + q(t)e^t = 0; \tag 1$
cancelling $e^t$ and performing a little algebra yields
$p(t) + q(t) = -1; \tag 2$
similarly, inserting
$y = t^2 \tag{2.5}$
into (0) we obtain
$2 + p(t)(2t) + q(t)(t^2) = 0; \tag 3$
a little more algebra:
$2tp(t) + t^2q(t) = -2; \tag 4$
now from (2),
$q(t) = - 1 - p(t), \tag{4.5}$
whence (4) yields
$2tp(t) + t^2(-1 - p(t)) = -2; \tag 5$
yet more algebra:
$(2t - t^2)p(t) = t^2 - 2, \tag 6$
and we can isolate $p(t)$:
$p(t) = \dfrac{t^2 - 2}{2t - t^2}; \tag 7$
returnig to (4.5):
$q(t) = -1 - p(t) = - 1 - \dfrac{t^2 - 2}{2t - t^2}$
$= \dfrac{ t^2- 2t - t^2 + 2}{2t - t^2} = \dfrac{2 - 2t}{2t - t^2} = \dfrac{2t - 2}{t^2 - 2t}; \tag 8$
the original equation is thus
$y'' + \dfrac{t^2 - 2}{2t - t^2} y'(t) +  \dfrac{2t - 2}{t^2 - 2t} y(t) = 0, \tag 9$
and we may inquire if both $e^t$ and $t^2$ are in fact solutions:  with $y(t) = e^t$ we ask
$e^t + \dfrac{t^2 - 2}{2t - t^2} e^t +  \dfrac{2t - 2}{t^2 - 2t}e^t \overset{?}{=} 0, \tag{10}$
or
$1 + \dfrac{t^2 - 2}{2t - t^2} + \dfrac{2t - 2}{t^2 - 2t}   \overset{?}{=} 0; \tag{11}$
the left-hand side reduces to
$1 + \dfrac{t^2 - 2}{2t - t^2} + \dfrac{2t - 2}{t^2 - 2t} = 1 + \dfrac{t^2 - 2t}{2t - t^2} = 1 +(-1) = 0!!! \tag{12}$
with $y = t^2$:
$2 + \dfrac{t^2 - 2}{2t - t^2} (2t) +  \dfrac{2t - 2}{t^2 - 2t}(t^2) \overset{?}{=} 0, \tag{13}$
whence
$2 + \dfrac{2t^3 - 4t}{2t - t^2}  +  \dfrac{2t^3 - 2t^2}{t^2 - 2t} \overset{?}{=} 0, \tag{14}$
$2 + \dfrac{2t^2 - 4t}{2t - t^2} = 2 + (-2) = 0 !!!\tag{15}$
We thus see that both $e^t$ and $t^2$ are bona fide solutions to (9) on $\Bbb R \setminus \{0, 2 \}$, i.e., away from the singularities of $p(t)$ and $q(t)$.  The fact that $p(t)$ and $q(t)$ cannot be continuous at $0, 2$ implies there is no equation of the general form (0), which has coefficients continuous on all of $\Bbb R$ (or for that matter on any open interval containing $0$ or $2$), which both these functions $e^t$ and $t^2$ solve.
There is a sense in which the initial value problem defined by Lutz Lehmann in his answer does not rally exist, for as we have seen the equation (0) cannot in fact be defined at $0$ if it is to have solutions $e^t$ and $t^2$.  
Robert Israel arrived at the only possible correct $p(t)$ and $q(t)$, and apparently left it to the reader to realize they are not continuous on all of $\Bbb R$.
I mention these things because I found this problem to be subtle and somewhat confusing, and I wish if possible to clarify the situation for my readers and myself.  I hope I have done so here.
Like Professor Henry Jones, Sr., I too seek illumination. Fiat Lux! 
A: Substitute $t^2$ and $\exp(t)$ into the differential equation, and solve for $p(t)$ and $q(t)$.  You should get
$$  p \left( t \right) =-{\frac {{t}^{2}-2}{ \left( t-2 \right) t
}},\ q \left( t \right) ={\frac {2t-2}{ \left( t-2 \right) t}}
$$
A: If $y(t)=t^2$ is a solution, then it solves the initial value problem (IVP) $y(0)=y'(0)=0$. With continuous coefficients the only solution of that IVP however is the zero function, thus your situation is impossible.
