
#1
Oct1613, 10:42 AM

P: 154

Here are two ways:
$$(x{ x }_{ 1 })({ x }{ x }_{ 2 })=0\\ x{ x }_{ 1 }=0\quad \quad \quad \quad x{ x }_{ 2 }=0\\ \\$$$$ \\ { e }^{ x }({ c }_{ 1 }3{ c }_{ 2 })+{ e }^{ 32x }({ c }_{ 5 }{ c }_{ 4 })=0\\ { c }_{ 1 }3{ c }_{ 2 }=0\quad \quad \quad \quad \quad { c }_{ 5 }{ c }_{ 4 }=0$$ Any other ways? 



#2
Oct1613, 10:58 AM

P: 388

"Splitting an equation in two" is not in general a meaningful mathematical operation. Your second example is incorrect: can you see why? (clue: a = b = 0 is not the only solution to a + b = 0).




#3
Oct1613, 11:24 AM

P: 154

What if it's an identity? 



#4
Oct1613, 05:23 PM

P: 337

What other ways can you split an equation into two?The second situation is a statement about the linear independence of ##e^x## and ##e^{32x}##; if ##ae^x+be^{32x}=0## for all ##x## (i.e. ##ae^x+be^{32x}## is the zero function), then ##a=b=0##. It's similar to the statement that if ##a_0+a_1 x+...+a_n x^n=0## for all ##x##, then ##a_0=a_1=...=a_n=0##. 



#5
Oct1613, 06:07 PM

P: 388




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