Show linear dependence directly

[2h2o]

Homework Statement

Show directly that the given functions are linearly dependent on the real line==>find a non-trivial linear combination of the given functions that vanishes identically.

Homework Equations

$$f(x) = 2x, g(x) = 3x^{2}, h(x) = 5x-8x^{2}$$

The Attempt at a Solution

I know that the structure of the solution will be $$C_{1}y_{1}+C_{2}y_{2}+C_{3}y_{3}=0$$, so I find three equations for the three unknown constants.

$$C_{1}2x + C_{2}3x^{2} + C_{3}(5x-8x^{2}) = 0$$...[1]

first derivative

$$2C_{1} + 6C_{2}x + C_{3}(5-16x) = 0$$...[[2]

second derivative

$$0 + 6C_{2} -16C_{3} = 0$$...[3]

Upon which I begin finding the constants:

Equation [3] immediately shows that $$6C_{2}=16C_{3}$$

Then [2] becomes

$$2C_{1}+5C_{3}=0$$

hence $$2C_{1} = -5C_{3}$$

Now $$C_{1}=\frac{-5C_{3}}{2}$$

So $$C_{1}=\frac{-5}{2}*\frac{3C_{2}}{8} = \frac{-15}{16}C_{2}$$

Now everything I try gives me the wrong solutions..matrix, substitution (infinite substitutions since every constant is related to another constant.)

So how do I prove that these functions are L.D.?

Magically (or so it seems) the solution is supposed to be $$15(2x) - 16(3x^{2}) - 6(5x-8x^{2})=0$$

 

[Dick]

Ok, let's recap. You got C1=(-5/2)*C3 and C2=(16/6)*C3. That means you can express C1 and C2 in terms of C3. Ok, let's try it. They gave you a solution where C3=(-6). From your solution that means C1=15 and C2=-16. That checks, right? You could also have put C3 equal to anything else nonzero and gotten a different but valid answer.

 

[2h2o]

Yep, that checks out just fine. So the choice of C3=(-6) was totally arbitrary and that is what was throwing me for a loop. So it turns out then that I had A correct answer all along, just not THE correct answer that they chose to yield nice, neat integer coefficients.

I used a matrix to solve (w/o using a Wronskian) for the coefficients of C1=5/2, C2=(-8/3), then C3=-1. Which works, and I believe to be valid because the "trivial solution" has all of the coefficients as zero--which implies to me that one or "some" coefficients can be zero in the non-trivial solution.

Thanks, Dick. If there is a flaw in my above reasoning, please let me know.

Cheers!

 

[Dick]

No flaws. They just picked one answer to show you. You found another. There's an infinite number of possible choices that work.