Linearly Dependent Differential Equations

Weatherkid11
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Show directly that the following functions are linearly dependent on the real line. That is, find a nontrivial linear combination of the funtions that vanishes indetically.
f(x)=17, g(x)= 2sin^2 x, h(x)= 3cos^2 x

Do you just take the 1st and 2nd derivatives and do the determinate?? I am so confused on how to do this problem. Thanks
 
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You have to find a nontrivial linear combination of the functions that vanishes identically. In other words, you have to find constants A, B, C, not all zero, so that
A*f(x) + B*g(x) +C*h(x) = 0
Use trig.
 
So basically something like: 3g(x) + 2h(x) =6sin²x + 6cos²x = 6, but then where do I go from there?
 
Then you use f(x) to make it 0.
 
ok, I got that A17 has to equal -6 to make it zero, so then that would mean A=-6/17, so the final answer would be (-6/17)17 + 3(2sin²x)+2(3cos²x)=0, Correct? And thanks for the help
 
Yes, correct.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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