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Trig equations question. (1 Viewer)

David_O

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Somebody help please:

Solve for g(t):
g(t)[g(t) + 2sin(t+alpha)] + g'(t)[g'(t)+2cos(t+alpha)] = 0
and prove that there are no other solutions.

I've got g(t) = 0, -2sin(t+alpha), -sin(t+alpha) + 1, -sin(t+alpha) - 1, but I can't prove there aren't any other solutions. Heh, there probably are a few more anyhow.

Thanks.
 

Affinity

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if you differentiate the DE, you get [g + g''][g' + 2sin(t + a)] = 0
then solve for g from there, plug it into the original equation to determine the free coefficients, and those will be the only solutions, others will either fail the original equation of the new one). Ofcourse, this relies on the assumption that you can actually differentiate g', which is not guaranteed.
 

lucifel

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and why are Differential Equations in the 4U section? Kick this to the Appreciating the Elegance section, otherwise you'll scare little kiddies away from maths. forever.
 

acmilan

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dN/dt = kN

DEs are in the 2 unit course :uhhuh:
 
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lucifel

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yeh, you've got a point, but this one isn't. And in the HSC you almost never have to actaully solve the DE's, they always give you an answer, which you simply verify, this question does not comply with that.
 
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Affinity

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lucifel said:
yeh, you've got a point, but this one isn't. And in the HSC you almost never have to actaully solve the DE's, they always give you an answer, which you simply verify, this question does not comply with that.
You solve y'' + w^2 y = 0 amongst many others... they just don't call it DEs
 

mojako

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Affinity said:
You solve y'' + w^2 y = 0 amongst many others... they just don't call it DEs
u also solve dy/dx = f(x).. like dy/dx = sin(x) and so on
 

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