# Solution to a second order differential equation

• I
I have currently been reading a book called 'Mathematical Methods In Physical Sciences'. Whilest I was looking at the differential section I came across a differential which I have never thought about before, which is of the form:
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Directly from book

$$m\left(\frac{d^2y}{dt^2}\right)\pm l\left(\frac{dy}{dt}\right)^2+ky=0$$

where the plus or minus sign must be chosen correctly at each stage of the motion
so that the retarding force opposes the motion. Let us solve the following special
case of this problem. Discuss the motion of a particle which is released from rest at
the point y = 1 when t = 0, and obeys the equation of motion

$$4\left(\frac{d^2y}{dt^2}\right)\pm 2\left(\frac{dy}{dt}\right)^2+y=0$$
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And whilst I was reading how to solve it but it sort of breaks off, it starts explain the Bernoulli method then talks about the transcendental function, as stops. The actual equation in referring damped harmonic motion on a spring except that drag force is now of the form ##F=-bv^2## I assume. So I have tried to do a bit of research and not come up with much, I did think this could be done numerically by series solutions but I am interested to how this equation is solved through transcendental methods, and was wondering if some could explain the the approach on how to solve or even refer me to a bit of literature that could help me understand. This is my first time really tackling this type of differential so it a new concept and one I would like to understand.

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Is ##y## a function of ##x##? Or is it an independent variable?

Is ##y## a function of ##x##? Or is it an independent variable?
In there example y is the height to which the a particle from rest is released at t=0, also sorry I write the equation in the book form to make it clearer.

My approach would be the following: you can try making the substitution ##p = \frac{dy}{dt}##. Using chain rule, you can find that ##\frac{dp}{dt} = \frac{dp}{dy}\frac{dy}{dt} = p \frac{dp}{dy}##. This gives you

$$4p \frac{dp}{dy} \pm 2 p^2 +y = 0$$

This is now a first-order equation. Try and solve it from here. Hope this helps, and hopefully it's what you were looking for.

Divide through by ##4p##.

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