# Question about the general solution to Hooke's law

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• dainceptionman_02
dainceptionman_02
ok, so usually one of the first equations in Diff Eq is F = -kx, which is the second order differential equation mx'' = -kx, where they give you the only general solution in the universe as Acoswt + Bsinwt. I was wondering, why can't you just separate the equation and get m∫x''/x = -∫kt'' which should be m(xlnx - x) = -kt^2/2 + C ?

dainceptionman_02 said:
ok, so usually one of the first equations in Diff Eq is F = -kx, which is the second order differential equation mx'' = -kx, where they give you the only general solution in the universe as Acoswt + Bsinwt. I was wondering, why can't you just separate the equation and get m∫x''/x = -∫kt'' which should be m(xlnx - x) = -kt^2/2 + C ?
That does not solve the differential equation: ##\displaystyle \quad m \ddot{x} = -kx \ \ .##

What do you even mean by m∫x''/x = -∫kt'' ?

Your solution seems to have interpreted this as

##\displaystyle \quad \quad m \int \left( \int \frac 1 x dx \right) dx = -k \int \left( \int dt \right) dt \ \ ,##

which is so very wrong.

Last edited:
If you plug your solution back into the original equation, it doesn't satisfy it.

dainceptionman_02
SammyS said:
That does not solve the differential equation: ##\displaystyle \quad m \ddot{x} = -kx \ \ .##

What do you even mean by m∫x''/x = -∫kt'' ?

Your solution seems to have interpreted this as ##\displaystyle
m \int \left( \int \frac 1 x dx \right) dx = -k \int \left( \int dt \right) dt \ \ ,## which is so very wrong.
I'm getting dizzy...

SammyS

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