Solving Non-Linear Differential Equations for Free Fall Due to Gravity

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The discussion revolves around solving a non-linear second-order differential equation related to a body falling under gravity, where gravitational acceleration is not constant. The initial equation for speed, dv/dt = -g - kv, is straightforward, but the challenge arises when extending it to z'' + kz' + GMz^-2 = 0. The poster attempts to solve the equation by separating terms and using complementary and particular solutions, but encounters difficulties due to the non-exponential form of the right-hand side. Suggestions include transforming the equation to show the relationship between velocity and height and finding an integrating factor to make the equation exact. The conversation emphasizes the need for a different approach to tackle the non-linear aspects of the problem.
Pete69
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Homework Statement



I was given a problem to solve for the speed of a body falling under gravity [equation (1)] where g is acceleration due to gravity, which was easy enough.. but then i thought i would extend it to the case where g is non-constant, and so arrived at equation (2), (where where z is the height above Earth [z'=dz/dt and z=dv/dt and z^-2 means z to power -2], and M is the mass of the Earth and G is the gravitational constant)

Homework Equations



(1) : dv/dt = - g - kv

(2) : z'' + kz' + GMz^-2 = 0

The Attempt at a Solution



I believe this is a non-linear second order DE?? i attempted to solve by setting

z'' + kz' = 0

and solving the complimentary equation, which was OK, but when i came to solve for the particular integral

z'' + kz' = -GMz^-2

i ran into problems, as after substituting in the D and Q operators (http://silmaril.math.sci.qut.edu.au/~gustafso/mab112/topic12/ ), i could not use the First Shift Theorem, as the RHS is not in an exponential form...

Any ideas anyone?
 
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Pete69 said:
I believe this is a non-linear second order DE?? i attempted to solve by setting

Yes, but rather than get into that let's look at your first order equation for v.

Do the following.

1.) Show that \frac{dv}{dt}=v\frac{dv}{dz}.
2.) Insert g=\frac{GM}{z^2} into the equation.
3.) Find an integrating factor that makes this equation exact (it can be done).
4.) Solve.
 
Pete69 said:

Homework Statement



I was given a problem to solve for the speed of a body falling under gravity [equation (1)] where g is acceleration due to gravity, which was easy enough.. but then i thought i would extend it to the case where g is non-constant, and so arrived at equation (2), (where where z is the height above Earth [z'=dz/dt and z=dv/dt and z^-2 means z to power -2], and M is the mass of the Earth and G is the gravitational constant)

Homework Equations



(1) : dv/dt = - g - kv

(2) : z'' + kz' + GMz^-2 = 0

The Attempt at a Solution



I believe this is a non-linear second order DE?? i attempted to solve by setting

z'' + kz' = 0

and solving the complimentary equation, which was OK, but when i came to solve for the particular integral
There is no "complementary equation" nor is there a "particular integral". Those are both concepts in linear differential equations where the non-homogeneous part is a function of the independent variable only.

z'' + kz' = -GMz^-2

i ran into problems, as after substituting in the D and Q operators (http://silmaril.math.sci.qut.edu.au/~gustafso/mab112/topic12/ ), i could not use the First Shift Theorem, as the RHS is not in an exponential form...

Any ideas anyone?
 
Last edited by a moderator:
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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