# Differential equation formation

1. Feb 4, 2014

### lionely

1. The problem statement, all variables and given/known data
If α is an arbitrary constant and a a fixed constant show that
xcos α + ysin α = a

is the complete primitive of the equation

(y - xdy/dx)^2 = a^2( 1 + (dy/dx)^2)

2. Relevant equations

3. The attempt at a solution

FIrst I found the first derivative by differentiating implicitly,

cosα + (dy/dx)sinα = 0

But now I'm looking for something to sub to try and get the differential equation and eliminate the arbitrary constants, but I keep getting into a loop and cancelling everything basically...

2. Feb 5, 2014

### Pranav-Arora

Hi lionely!

You have got two equations. Solve for cos(α) and sin(α) and plug them in the expression $\cos^2\alpha + \sin^2\alpha=1$.

3. Feb 5, 2014

### lionely

But if I use those equations I have all I get is cos(α) in terms of x,y and sin(α) or dy/dx. SO when I put it in the identity I still can't get rid of the arbitrary constants.

4. Feb 5, 2014

### pasmith

You goal is to confirm that the left hand side of the ODE is equal to the right hand side. If you work out $y$ and $dy/dx$ and just substitute them into the ODE then everything cancelling is exactly what you want!

5. Feb 5, 2014

### lionely

But I want to move from the primitive to the differential.

6. Feb 5, 2014

### pasmith

Starting from the given solution and trying to manipulate it to obtain the given ODE is not going to work. Your solution is linear; the only way to eliminate the constants of integration is to differentiate twice to obtain $y'' = 0$. That doesn't help you.

There are two approaches you can take:
• Solve the ODE and show that its general solution takes the given form (this can be done; the easiest way is to differentiate both sides with respect to $x$; unfortunately the resulting 2nd order equation admits solutions which don't satisfy the original first order ODE, and these solutions must be identified and eliminated).
• Substitute the given form into the ODE and show that everything cancels (this is straightforward).

Actually, I suppose that if you're asked to show that a primitive is a complete primitive then you should use the first approach. Otherwise you've just shown that it is a primitive.

7. Feb 5, 2014

### lionely

If I can't move from the primitive to the differential equation, how the heck did the author of the book get it? This thing is just puzzling me.

8. Feb 6, 2014

### Pranav-Arora

Let $\cos\alpha=c$ and $\sin\alpha=e$, then you have the following system of linear equations:

$$cx+ey=a$$
$$c+(dy/dx)e=0$$

Solve the above equations for $c$ and $d$.

9. Feb 6, 2014

### lionely

C = (a-ey)/x

and e= -c/(dy/dx) ? I guess.. Then put these in sin^2 a + cos^2a = 1?

10. Feb 6, 2014

### Pranav-Arora

I meant that find $c$ and $e$ in terms of x,y and dy/dx and then plug them in the identity.