Give general solution to differential equation Watch

Jack93o
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#21
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(Original post by atsruser)
I'm not sure what you mean by "seems unlikely". Unlikely that the OP would have to perform the second integral? If so, possibly, but that's the correct result regardless.

The next step is:

\int dt = \pm \frac{1}{2\sqrt{2}} \int \frac{dx}{\sqrt{x^2+k^2}} \Rightarrow t = \pm \frac{1}{2\sqrt{2}} \sinh^{-1}(x/k) + d

which I think is about all we can say without knowing more about the initial conditions. This type of integral only comes up in the FP syllabus, though.

It's just occurred to me though that the question may make more sense if it was meant to be \frac{d^2x}{dt^2}=-8x which has sin/cos solutions. (It's the equation for SHM).
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I don't quite understand what you've done there. Anyway it seems to me like this question could be done by 'inspection'.

let x = something involving t, which if differentiated, would remain intact. This could only mean one thing: use exponential

ok, so the differential equation is: d^2 x / dt^2 = 8x

lets make x = Ae^(mt)

m is + or - square root of 8, we'll get to this again at the end

differentiating x, gives Ame^(mt)

differentiating again, gives A(m^2)e^(mt), where m^2 = 8

therefore the general solution is:

x(t) = Ce^(+ve square root of 8)t + De(-ve square root of 8)t

C and D are arbitrary constants

I'm pretty sure thats correct, can someone double check this?

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james22
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(Original post by Jack93o)
I don't quite understand what you've done there. Anyway it seems to me like this question could be done by 'inspection'.

let x = something involving t, which if differentiated, would remain intact. This could only mean one thing: use exponential

ok, so the differential equation is: d^2 x / dt^2 = 8x

lets make x = Ae^(mt)

m is + or - square root of 8, we'll get to this again at the end

differentiating x, gives Ame^(mt)

differentiating again, gives A(m^2)e^(mt), where m^2 = 8

therefore the general solution is:

x(t) = Ce^(+ve square root of 8)t + De(-ve square root of 8)t

C and D are arbitrary constants

I'm pretty sure thats correct, can someone double check this?
That works, because you have 2 linearly independent solutions it actually proves that it is the general solution as well.
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fuzzybear
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which method should I use, atsruser or jack93o's?
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Mr M
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(Original post by fuzzybear)
which method should I use, atsruser or jack93o's?
Neither if you are not a Further Mathematician.
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