How Can I Solve a Differential Equation in Physics Homework?

[Daniel5423]

Homework Statement

In my physics homework, I ran into a differential equation. I am attempting to solve this differential equation for y(x).

Homework Equations

y''[x] = -C/(y[x]^3) - y[x]

C is a constant

The Attempt at a Solution

dy^2/dx^2 = -C y[x]^-3 - y[x]
(1)/(-Cy[x]^-3 - y[x]) dy^2= dx^2
(-1/4)(ln(y^4 + C) dy = x dx

Now, I have to solve the integral for dy.

I was unable to find the integral for dy when it had a logarithm, so I manipulated the equation
ln(y^4 + C) dy = -4x dx
Now, use the e exponent.
(y^4 + C) e^dy = e^(-4x dx)

Now, do the integral.

(y^4 + C) e^y = e^-2x^2

Now, I can not solve for y in terms of x. If I try to use a natural logarithm to get rid of the e, it puts the other y term within the logarithm. I believe I did the differential equation incorrectly. Could someone please help me and give me a hint of what I did incorrectly? Thanks in advance.

 

[Ray Vickson]

Homework Statement

In my physics homework, I ran into a differential equation. I am attempting to solve this differential equation for y(x).

Homework Equations

y''[x] = -C/(y[x]^3) - y[x]

C is a constant

The Attempt at a Solution

dy^2/dx^2 = -C y[x]^-3 - y[x]
(1)/(-Cy[x]^-3 - y[x]) dy^2= dx^2
(-1/4)(ln(y^4 + C) dy = x dx

Now, I have to solve the integral for dy.

I was unable to find the integral for dy when it had a logarithm, so I manipulated the equation
ln(y^4 + C) dy = -4x dx
Now, use the e exponent.
(y^4 + C) e^dy = e^(-4x dx)

Now, do the integral.

(y^4 + C) e^y = e^-2x^2

Now, I can not solve for y in terms of x. If I try to use a natural logarithm to get rid of the e, it puts the other y term within the logarithm. I believe I did the differential equation incorrectly. Could someone please help me and give me a hint of what I did incorrectly? Thanks in advance.

You cannot do the type of manipulation you performed when you went from
$$\frac{d^y}{dx^2} = -\frac{c}{y^3} - y$$
to
$$\frac{1}{-c y^{-3} - y} dy^2 = dx^2 \; \Leftarrow \text{false!}$$

For one thing, you should have $d^2y$, not $dy^2$, and for another thing, $dx^2$ does not really mean anything. In fact, neither $d^2y$ nor $dx^2$ have any real meaning here---they are just notation used for writing a second derivative.

One way you could try to solve the problem would be to use time $t$ (instead of $x$) as your independent variable, then view the problem as one of motion $y = y(t)$ in a force field that depends on $y$. If the force is derivable from a potential $V(y)$, the total of kinetic + potential energy is conserved (that is, will remain constant). That should allow you to find a first-order differential equation for $dy/dt$ whose right-hand-side is a function of $y$, so you have a first-order separable DE that can be solved in principle. Unfortunately, the solution will be "implicit", in the form $F(y) = t$ for some complicated function $y$, so it will probably not be possible to solve for $y$ explicitly in terms of $t$.

 

[Orodruin]

Hint: Multiply both sides by $y'(x)$. This is what will allow you to solve the problem by integrating twice.

 

[Ray Vickson]

Hint: Multiply both sides by $y'(x)$. This is what will allow you to solve the problem by integrating twice.

That produces exactly the same DE as does the "energy-conservation" method suggested in #2. I guess the advantage of your approach is that it can be applied by somebody who has never taken Physics and so does not know about kinetic energy and potential functions, etc. Of course, that method is exactly how one derives the energy conservation result!

 

[Orodruin]

That produces exactly the same DE as does the "energy-conservation" method suggested in #2. I guess the advantage of your approach is that it can be applied by somebody who has never taken Physics and so does not know about kinetic energy and potential functions, etc. Of course, that method is exactly how one derives the energy conservation result!

Well yes. The basic idea is to obtain something that is a total derivative and therefore directly integrable - meaning that what you have the total derivative of is a constant of motion. In many physics problem that constant of motion is total energy, but the method itself I would argue is independent of this interpretation. Of course, it may help someone who has taken the appropriate physics courses to think in those terms.