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Questions with likely obvious answers, but I don't have the required intuition to go with the flow.

Consider $a+be^x + ce^{-x} = 0$. To solve it for the constants, we can try out different values of $x$ to get a system of $3$ equations and simply use a calculator (given technique). Why are we allowed to stick various values of $x$ into the equation? Is that because the equation is valid for all $x$? I mean the fact(?) that this equation holds for all $x$ allows us to stick any value of $x$ into the equation whose byproduct is the isolation of constants(just a bonus). Does that make sense?

Consider $k_1(1) + k_2(t^2 - 2t) + 5k_3(t - 1)^2 = (k_2 + 5k_3)t^2 +(-k_2 - 10k_3)t + (k_1 + 5k_3) = 0.$ To solve the LHS of the equation for the constants, we can let $(k_2 + 5k_3) = (-k_2 - 10k_3) = (k_1 + 5k_3) = 0$ and solve the equality for $k_i.$ Call this maneuver $X.$ Why are we allowed to do $X$? Is it because $X$ is one of the solutions to $(k_2 + 5k_3)t^2 +(-k_2 - 10k_3)t + (k_1 + 5k_3) = 0$ which allows us to consider $X$? Hopefully, this makes sense.

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    On a side note, since this equation should valid for al $x$, you can use differentiation with respect to $x$. If you differentiate it twice, you get that $be^x+ce^{-x}=0,$ hence immediately $a=0$. If you differentiate it only once, then $be^x-ce^{-x}=0$, which, combined with the previsou eqality gives you $b=c=0$. – 2017-01-25

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Yes, it is because the equation is valid for all values of $x$. You can think of that equation as representing infinitely many true facts, each of the same form. We would write this as $$(\forall x\in\mathbb{R})(a+be^x+e^{-c}=0)$$ which is read as "for all $x$ in the Real numbers, $a+be^x+ce^{-x}=0$." Since this is true for all $x$, it is true for each specific value of $x$, such as $7$ and $\frac{4}{11}\pi$. It's worth noting that the only values of $a,b,c$ that satisfy this is $a=b=c=0$

This is known as Universal Instantiation.

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Maybe I did not understand your problem. If you want to find $a,b,c$ so that $a+be^x+ce^{-x}=0$ for all $x\in \mathbb{R}.$ Then the only choice is $a=b=c=0$. Indeed the equation is equivalent to $ae^x+be^{2x}+c=0$ letting $x$ goes to $-\infty$ we get $c=0$ repeating the same we get $a=0.$

If $a,b,c$ are given then set $X=e^x$ and solve the equation $aX+bX^2+c=0.$

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If

$$a+be^x+ce^{-x}=0$$ is known to hold for some values of $x$, this forms a system of equations which you can try and solve for $a,b,c$.

When the number of equations equals the number of unknowns and the equations are independent, you can expect a finite number of solutions. Under some conditions, this number is one.

When there are less equations, there can be an infinity of solutions (i.e. the system is undeterminate).

When there are more equations (I repeat, independent ones), there is no solution, the system is overdeterminate.

In your case, if the equation holds for "all" $x$, the system has no solution except the trivial one $a=b=c=0$. (Actually, you cannot form more than three independent equations.)

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In both examples, you have chosen three functions that are linearly independent. (In the first example, you have $f_1(x)=1$, $f_2(x)=e^x$ and $f_3(x)=e^{-x}$. In the second example, you have $f_1(t)=t^2$, $f_2(t)=t$ and $f_3(t)=1$.) This linear independency ensures that there exist three values $x_1$, $x_2$ and $x_3$ for which the system of linear equations $$ af_1(x_1)+bf_2(x_1)+cf_3(x_1) = 0 \\ af_1(x_2)+bf_2(x_2)+cf_3(x_2) = 0 \\ af_1(x_3)+bf_2(x_3)+cf_3(x_3) = 0 $$ has a unique solution, which is obviously $a=b=c=0$. It can happen that your functions are not linearly independent, in which case you will not find a unique solution. So if you do not find a unique solution when you solve your homogeneous system of equations, the functions might be linearly dependent, or you had bad luck with your choice of $x_1$, $x_2$ and $x_3$. But if you find a solution and it is unique, then you only need to cross check if it solves the initial equation for all values $x$. There cannot be any other candidates, because each solution of the initial equation must also hold for your choice of $x_1$, $x_2$ and $x_3$.