# Solving for C and Y(x) in differential equations

1. Oct 15, 2008

### demersal

1. The problem statement, all variables and given/known data
Solve the separable differential equation Subject to the initial condition y(0) = -10:

7x-3y$$\sqrt{x^{2}+1}$$$$\frac{dy}{dx}$$ = 0

2. Relevant equations
Differential Equations

3. The attempt at a solution
I got up to the point where:
-$$\frac{3}{2}$$y$$^{2}$$=-7$$\sqrt{x^{2}+1}$$+C

I solved for c=-143

but when I plug it back in, I get a mess of an equation in solving for y(x) and each time I plug it in to check my answer, it comes up as wrong. Can anyone show me the mistake I am making? (My answer ends up being (-2/3(-7(x^2+1)^(1/2)-143))^(1/2) or some variation of that).

Thanks!!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 15, 2008

### Staff: Mentor

In your solution equation, solve for y so that you have y = {whatever} + C. That will make it easier to differentiate and to substitute in your DE.

3. Oct 15, 2008

### demersal

Ok, that sounds like a good idea. But I'm having trouble wrapping my head conceptually around the two square roots, since y^2 equals a term with a square root already in it. And does c go under the root? This may seem kind of trivial but I think this is why I've gotten this problem wrong 10 times already haha

4. Oct 16, 2008

### Staff: Mentor

How about multiplying both sides of the equation you ended up with by -2/3? Then you would have y^2 = some stuff + C', where C' = (-2/3) * (-143).

If you want to check your solution, you could take the derivative with respect to x implicitly (i.e., d/dx(y^2) = 2y dy/dx = d/dx(the stuff on the right side)). Then you could solve for dy/dx algebraically and substitute it back into the original equation. Might be a little simpler than a square root of a square root plus a constant. And yes, the constant would have to go under the outer radical. And you would have to consider positive and negative roots.