True Cartesian curvature equation, trying to solve it

[c0der]

Homework Statement

Solve the following equation:

v is the dependent variable, x is the independent variable

Homework Equations

\frac{d^2v/dx^2}{(1+\frac{dv}{dx}^2)^{3/2}}=1

The Attempt at a Solution

Hi,

I am trying to solve the following equation:

\frac{d^2v/dx^2}{(1+\frac{dv}{dx}^2)^{3/2}}=1

I used separation of variables as follows:

Let u = \frac{dv}{dx}

\frac{du/dx}{(1+u^2)^{3/2}}=1

Separate the variables and integrate:

\frac{du}{(1+u^2)^{3/2}}=dx

\frac{u}{(1+u^2)^{1/2}}=x + C

u= \sqrt{\frac{(x+C)^2}{1-(x+C)^2}} or -\sqrt{\frac{(x+C)^2}{1-(x+C)^2}}

Why is this not a valid solution when substituting back into the above equation for u and du/dx?

 

[STEMucator]

You have made a mistake somewhere in the final step. Here is what I get:

 

[Mark44]

Homework Statement

Solve the following equation:

v is the dependent variable, x is the independent variable

Homework Equations

\frac{d^2v/dx^2}{(1+\frac{dv}{dx}^2)^{3/2}}=1

The Attempt at a Solution

Hi,

I am trying to solve the following equation:

\frac{d^2v/dx^2}{(1+\frac{dv}{dx}^2)^{3/2}}=1

I used separation of variables as follows:

Let u = \frac{dv}{dx}

\frac{du/dx}{(1+u^2)^{3/2}}=1

Separate the variables and integrate:

\frac{du}{(1+u^2)^{3/2}}=dx

\frac{u}{(1+u^2)^{1/2}}=x + C

u= \sqrt{\frac{(x+C)^2}{1-(x+C)^2}} or -\sqrt{\frac{(x+C)^2}{1-(x+C)^2}}

Why is this not a valid solution when substituting back into the above equation for u and du/dx?

If a fraction is equal to 1, then the numerator must be equal to the denominator.

So this equation: \frac{d^2v/dx^2}{(1+\frac{dv}{dx}^2)^{3/2}}=1
could be written as \frac{d^2v}{dx^2} = {(1+\frac{dv}{dx}^2)^{3/2}}
or, since the Leibniz notation is a bit clumsy here, as this:
v'' = (1 + (v')²)^3/2
The first equation and either of the next two equations say exactly the same thing.

 

[c0der]

Thanks for your replies.

Zondrina: I believe that's the same answer, just in a different form as 1 = [1 - (x + C)^2] / [ 1 - (x+C)^2 ] ?

Mark44: I evaluated the integral in mathematica and matlab, and the same answer is given. I am integrating 1/(1+u^2)^(3/2) over u not x?

 

[Mark44]

Thanks for your replies.

Zondrina: I believe that's the same answer, just in a different form as 1 = [1 - (x + C)^2] / [ 1 - (x+C)^2 ] ?

Mark44: I evaluated the integral in mathematica and matlab, and the same answer is given. I am integrating 1/(1+u^2)^(3/2) over u not x?

Edit: I checked your work and it's fine. Disregard what I said. I have edited my previous post.

 

[c0der]

Thanks for checking. It strikes me as odd that Matlab cannot solve this ODE and that the above procedure I used does not turn out to be a solution when back substituting for u and du/dx:

>> syms u(x)
>> dsolve(diff(u,x)/(1+u^2)^(3/2)==1)
Warning: Explicit solution could not be found.
> In dsolve at 194
ans =
[ empty sym ]

 

[STEMucator]

Solving the equation in post #2 yields the solution:

$$v = \pm \left[- \frac{(x+C)}{\sqrt{\frac{1}{1 - (x+C)^2} - 1}} + K\right] = \mp \left[\frac{(x+C)}{\sqrt{\frac{1}{1 - (x+C)^2} - 1}} - K\right] = \mp \left[\sqrt{1 - (x+C)^2} - K \right]$$

If there were initial conditions, you could even solve for the constants $C$ and $K$.

 

[c0der]

Matlab's simplifier was the problem, I verified the solution by hand, thanks for the help.