Undergrad Question about the general solution to Hooke's law

[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 ?

 

[SammyS]

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.

 

[Vanadium 50]

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

 

[berkeman]

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...